linalg.xml

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<book xml:id="linalg"><title> Manitoba linear algebra </title>
<frontmatter>
   <titlepage>
      <author>
         <personname>Michael Doob (editor)</personname>
         <department> Department of Mathematics </department>
         <institution> The University of Manitoba 
         </institution>
         <email>Michael_Doob@UManitoba.ca</email>
      </author>
      <credit>
      <title>
      </title>
      <author>
         <personname>
         </personname>
      </author>
      </credit>
      <date>
         <today />
      </date>
   </titlepage>
</frontmatter>

<chapter>
  <title>Systems of Linear Equations</title>
  
  <section><title>Review of two equations in two unknowns </title>
    
    <subsection><title>The standard method for finding the solution </title>
      <p>Suppose we want to find <em>all</em> solutions to the equations</p>
      <p>
      <md>
      <mrow>2x+3y \amp = 5</mrow>
      <mrow>3x+2y \amp= 7</mrow>
      </md>
      </p>
      
      <p>The <em>standard</em> technique is to manipulate one or both of the equations until
      either <m>x</m> has the same coefficient in both equations 
      or <m>y</m> has the same coefficient, and then 
      subtract to eliminate one variable. Since there is only one variable left,
      its value can be found; with this information, the value for the other variable
      can be found.</p>
     
      <p>In the case above, we can multiply both sides of the second equation by
      <m>\frac23</m>,
      to get
      <me>
      2x+\tfrac43 y= \tfrac{14}3,
      </me>
      and subtracting from the first equation gives
      <me>
      \frac53y = \frac13,
      </me>
      or
      <me>
      y=\frac15.
      </me>
      </p>

      <p>
      Using the first equation and the (now) known value of <m>y</m>, we find that
      <me>
      x=\frac{11}5.
      </me>
      </p>
      
      <p>
      Hence we see that there is exactly one pair of values for 
      <m>x</m>
      and 
      <m>y</m> that simultaneously satisfy both equations:
      <m>x=\frac{11}5</m> and <m>y=\tfrac15.</m> We can also write this
      as <m>(x,y)=(\frac{11}5,\frac15).</m>
      In this case we say that there is a <em>unique</em> solution (in mathematics, the term
      unique means <em>exactly one</em>).
      </p>
    </subsection>
    
    <subsection><title>The geometric method of finding the solution</title> 
      
      <p>The set of equations solved in the previous section can also be
      viewed geometrically. The points in the <m>x</m>-<m>y</m>
      plane satisfying one of the equations lie on a  straight line, and
      any point satisfying both of the equations must lie on both lines.
      The plot of the two lines looks like this:</p>
      
      <!---see here -->
      <figure>
      <caption>Intersection of two lines in the plane</caption>
      <image xml:id="TwoLinesInPlane" width="40%">
      <asymptote>
      import graph;
      unitsize(30);
      real xmin=-1; real xmax=4;
      xaxis(LeftTicks);yaxis(LeftTicks);

      real f1(real x) {return (5-2*x)/3;}
      real f2(real x) {return (7-3*x)/2;}

      draw((xmin,f1(xmin))--(xmax,f1(xmax)),green);
      draw((xmin,f2(xmin))--(xmax,f2(xmax)),red);
      dot((11/5,1/5));
      label("\((\frac{11}5,\frac15)\)",(11/5,1/5),NE);
      </asymptote>
      </image>
      </figure>

      <p>The point of intersection is <m>(x,y)=(\frac{11}5,\frac15)</m>,
      just as before.</p>
    </subsection>
 </section> 
  
  <section><title>Equations with multiple solutions or no solutions</title>
    <subsection><title>Equations with no solutions</title> 
     <p>
     Suppose we want to find all solutions to the equations
      <md>
      <mrow>2x+4y \amp = 5</mrow>
      <mrow>-x-2y \amp = 1</mrow>
      </md>
     Using the <em>standard</em> approach we multiply the second equation by <m>2</m> and add it
      to the first one to eliminate the variable <m>x</m>.  This leaves us
      with the equation
      0=7
      This is certainly an equality that is not valid. What happened? We can see by
      multiplying the second equation by <m>-2</m>. We then have 
      <md>
      <mrow>2x+4y \amp = 5</mrow>
      <mrow>2x+4y \amp = -2</mrow>
      </md>
     </p>

      <p>
      Whatever <m>x</m> and <m>y</m> are, the value of <m>2x+4y</m> can't be 
      <m>5</m> and <m>-2</m> at the same time. So there are no solutions.
      What happens if we try to graph these two equations? Here is what we get:
     </p>
     
     <figure>
     <caption>Two Parallel lines in the Plane</caption>
     <image xml:id="ParallelLinesInPlane" width="70%">
     <asymptote>
      import graph;
      unitsize(40);
      real xmin=-2; real xmax=4;
      xaxis(LeftTicks);yaxis(LeftTicks);

      real f1(real x) {return (5-2*x)/4;}
      real f2(real x) {return (1+x)/-2;}

      draw((xmin,f1(xmin))--(xmax,f1(xmax)),green);
      draw((xmin,f2(xmin))--(xmax,f2(xmax)),red);
      </asymptote>
      </image>
     </figure>
      
      <p>
      The geometry of the situation is now clear: the two lines are parallel and so there is no 
      point on both lines (indeed, both lines have slope <m>-\tfrac12</m>).
      When we have equations with no common solution, they are called <em>inconsistent</em>.
      </p>
    </subsection>
    
    <subsection><title> Equations with more than one solution </title>
      
      <p>Now let's alter the situation of the previous section slightly. 
      We consider the pair of equations

      <md>
      <mrow>2x+4y \amp = -2</mrow>
      <mrow>-x-2y \amp = 1</mrow>
      </md>

      We apply the <em>standard</em> method again: multiply the second equation by <m>2</m>
      and add it to the first. The result is
      <me>0=0</me> 
      This is certainly a valid, although not very interesting, equation. In fact, if we
      multiply both sides of the second equation by <m>-2</m> the system becomes

      <md>
      <mrow>2x+4y \amp = -2</mrow>
      <mrow>2x+4y \amp = -2</mrow>
      </md>

      This means that any solution of the first equation is also a solution of the second one.
      Geometrically, if we plot the graph of the two equations, the same line results for
      each one. How do we find all solutions in this case? Let us assign a value to <m>y</m>.
      Let's call it <m>t</m> so <m>y=t</m>. Then, using either equation, we have
      <m>x=-2t-1</m>.
      This means that for <em>any</em> value of <m>t</m> we know that
      <m>(x,y)=(-2t-1,t)</m> is a solution to both equations. So, in fact we have an infinite
      number of solutions.</p>

      <theorem xml:id="TwoEquationaTwoUnknowns"><title>Two equations in two unknowns</title>
         <statement>
         <p>Two equations in two unknowns may have:
         <ul>
         <li><p>No solutions</p></li>
         <li><p>A single (unique)  solutions</p></li>
         <li><p>An infinite number of solutions</p></li>
         </ul>
         </p>
         </statement>
         <proof><p>
        The corresponding lines in the plane are parallel, intersect at
        a single point, or are identical.
         </p></proof>
      </theorem>
    </subsection>
  </section>
  
  <section><title>Important Definitions for Linear Equations</title> 
   
   <subsection><title>The Definition of a Linear Equation</title>
     
     <p>
     The equations given above are linear equations in two variables.
      A <em>linear equation in <m>n</m> variables</em> 
      <m>x_1, x_2, \dots, x_n</m> is an equation of the form

      \<me>
      a_1x_1+ a_2x_2+ \cdots +a_n x_n= b
      </me>

      where <m>a_1,\dots,a_n</m> and <m>b</m> are constants.
      The numbers <m>a_1,a_2,\dots,a_n</m> are called the <em>coefficients</em>
      of the equation, that is, <m>a_1</m> is the coefficient of <m>x_1</m>, 
      <m>a_2</m> is the coefficient of <m>x_2</m>, etc.
      This is also called a <em>linear equation in <m>n</m> unknowns</em>.</p>
      
      <p>The following are examples of linear equations:
      <ul>
        <li><p> <m>2x=4</m>   (one variable)</p></li> 
        <li><p> <m>4x-2y=7</m>   (two variables)</p></li> 
        <li><p><m>\tfrac12 x -\tfrac35 y +\pi z=0</m>   (three variables)</p></li> 
        <li><p> <m>x_1-3x_2+4x_3-7x_4=100</m>   (four variables)</p></li> 
        <li><p><m>3x_1+15x_5=-10</m>
            (five variables: the ones with a coefficient of zero are not written)</p>
        </li> 
      </ul>
      </p>
   </subsection>
 
   <subsection><title>The Definition of a System of Linear Equations</title>
      
      <p>A <em>system of <m>m</m> linear equations in <m>n</m>
      unknowns</em> is a list of <m>m</m> linear equations, each of which
      has the same set of <m>n</m> unknowns. They are usually presented
      with the purpose of finding all of their simultaneous solutions.</p>
      
      <p>The original example given above is a system of 2 equations in 2 unknowns.
      Here are some more examples:
      <ul>
        <li><p>A system of 2 equations in 3 unknowns:
            <md>
            <mrow>x-y+z \amp = 1</mrow>
            <mrow>2x+3y-z \amp = 2</mrow>
            </md>
            </p>
        </li>
        <li><p>A system of 3 equations in 2 unknowns:
            <md>
            <mrow>u+v \amp = 1</mrow>
            <mrow>2u+v \amp = 3</mrow>
            <mrow>u+2v \amp =3</mrow>
            </md>
            </p>
        </li>
        <li><p>A system of 4 equations in 5 unknowns:
            <md alignment="alignat">
            <mrow>x_1 \amp + \amp x_2 \amp \amp \amp \amp \amp \amp \amp =1</mrow>
            <mrow>\amp \amp x_2 \amp +\amp x_3 \amp \amp \amp \amp \amp =2</mrow>
            <mrow>\amp \amp \amp \amp x_3 \amp +\amp x_4 \amp \amp \amp = 3</mrow>
            <mrow>\amp \amp \amp \amp \amp \amp x_4 \amp +\amp x_5 \amp =4</mrow>
            </md>
            </p>
        </li>
        <li><p>A system of <m>m</m> equations in <m>n</m> unknowns <m>x_1, x_2, x_3\dots,x_n</m>:
            <md>
            <mrow>a_{1,1}x_1 + a_{1,2}x_2 + \cdots + a_{1,n}x_n  \amp = b_1</mrow>
            <mrow>a_{2,1}x_1 + a_{2,2}x_2 + \cdots + a_{2,n}x_n  \amp = b_2</mrow>
            <mrow>\amp \vdots</mrow>
            <mrow>a_{m,1}x_1 + a_{m,2}x_2 + \cdots + a_{m,n}x_n  \amp = b_m</mrow>
            </md>
            </p>
        </li>
      </ul>
      </p>
      
   </subsection>
 
   <subsection><title>The Coefficient and Augmented Matrix of a System of Linear Equations </title>
      
      <p>A <em>matrix</em> is a rectangular array of numbers (the plural is <em>matrices</em>). Here is a matrix:
      <me>
      \begin{bmatrix}
      1 \amp  2 \amp  3 \amp  4\\
      5 \amp  6 \amp  7 \amp  8\\
      9 \amp  10\amp 11\amp 12
      \end{bmatrix}
      </me>
      This matrix has 3 rows:
      <ul>
      <li><p>
            <m>
            \begin{bmatrix}
            1 \amp  2 \amp  3 \amp  4
            \end{bmatrix}
            </m> is row 1,
      </p></li>
      <li><p>
            <m>
            \begin{bmatrix}
            5 \amp  6 \amp  7 \amp  8
            \end{bmatrix}
            </m> is row 2 and
      </p></li>
      <li><p>
         <m>
         \begin{bmatrix}
         9 \amp  10\amp 11\amp 12
         \end{bmatrix}
         </m>
         is row 3.
      </p></li>
      </ul>
      </p>
      
      <p>Similarly, the matrix has 4 columns:
      <ul>
      <li><p>
         <m>
         \begin{bmatrix}
         1\\ 5\\ 9 
         \end{bmatrix}
         </m>
         is  column <m>1</m>
      </p></li>
      <li><p>
         <m>
         \begin{bmatrix}
         2\\ 6\\ 10
         \end{bmatrix}
         </m>
         is  column <m>2</m>
      </p></li>
      
      <li><p>
         <m>
         \begin{bmatrix}
         3 \\ 7\\ 11
         \end{bmatrix}
         </m>
         is  column <m>3</m>, and
      </p></li>
      <li><p>
         <m>
         \begin{bmatrix}
         4\\ 8\\ 12
         \end{bmatrix}
         </m>
         is  column <m>4</m>
      </p></li>
      </ul>
      </p>

      <p>We call a matrix with 3 rows and 4 columns a <m>3\times4</m>
      matrix.</p>
      
      <p>More generally, a matrix with <m>m</m> rows and <m>n</m> columns is called an
      <m>m\times n</m> <em>matrix</em>. An <m>m\times n</m> matrix <m>A</m> has the form
      <me>
      A=\begin{bmatrix}
      a_{1,1} \amp  a_{1,2} \amp  \cdots \amp  a_{1,n}\\
      a_{2,1} \amp  a_{2,2} \amp  \cdots \amp  a_{2,n}\\
      \amp \amp \vdots\\
      a_{m,1} \amp  a_{m,2} \amp  \cdots \amp  a_{m,n}
      \end{bmatrix}
     </me>
      This notation means that <m>a_{i,j}</m> is the number in both row
      <m>i</m> and column <m>j</m>. We will call the rows
      <m>R_1, R_2,\ldots,R_m</m> and the columns <m>C_1, C_2, \ldots,C_n</m>.
      In other words,
      <me>
      R_i =\begin{bmatrix} a_{i,1} \amp  a_{i,2} \amp  a_{i,3}, \amp \cdots \amp  a_{i,n}\end{bmatrix},
      </me>
      and
      <me>
      C_j=\begin{bmatrix}
      a_{1,j}\\ a_{2,j}\\ a_{3,j}\\ \vdots\\ a_{m,j}
      \end{bmatrix}
      </me>
      The notation for this matrix is similar to that used for
      a system of linear equations, and for good reason. Suppose
      we have a system of <m>m</m> linear equations in <m>n</m> unknowns:
      <me>
      \begin{array}{lcl}
      a_{1,1}x_1 + a_{1,2}x_2 +{} \amp \cdots \amp {}+ a_{1,n}x_n  = b_1\\
      a_{2,1}x_1 + a_{2,2}x_2 +{} \amp \cdots \amp {}+ a_{2,n}x_n  = b_2\\
      \amp \vdots\\
      a_{m,1}x_1 + a_{m,2}x_2 +{} \amp \cdots \amp {}+ a_{m,n}x_n  = b_m
      \end{array}
      </me>
      The <em>coefficient matrix</em> <m>A</m> is then the <m>m\times n</m> matrix
      <me>
      A=\begin{bmatrix}
      a_{1,1} \amp  a_{1,2} \amp  \cdots \amp  a_{1,n}\\
      a_{2,1} \amp  a_{2,2} \amp  \cdots \amp  a_{2,n}\\
      \amp \amp \vdots\\
      a_{m,1} \amp  a_{m,2} \amp  \cdots \amp  a_{m,n}
      \end{bmatrix}
      </me>
      and the <em>augmented matrix</em> of the system is the <m>m\times( n+1)</m> matrix
      <me>
      A=\begin{bmatrix}
      a_{1,1} \amp  a_{1,2} \amp  \cdots \amp  a_{1,n}\amp b_1\\
      a_{2,1} \amp  a_{2,2} \amp  \cdots \amp  a_{2,n}\amp b_2\\
      \amp \amp \vdots\amp \amp \\
      a_{m,1} \amp  a_{m,2} \amp  \cdots \amp  a_{m,n}\amp b_m
      \end{bmatrix}
      </me>
      Hence the augmented is the coefficient matrix with one column (the constants on the
      right side of the equations) added. To emphasize the extra column, the augmented
      matrix is sometimes written as
      <me>
      A=\left[\begin{array}{llcl|l}
      a_{1,1} \amp  a_{1,2} \amp  \cdots \amp  a_{1,n}\amp b_1\\
      a_{2,1} \amp  a_{2,2} \amp  \cdots \amp  a_{2,n}\amp b_2\\
      \amp \amp \vdots\amp \amp \\
      a_{m,1} \amp  a_{m,2} \amp  \cdots \amp  a_{m,n}\amp b_m
      \end{array}\right]
      </me></p>
      
      <p>Here are some examples of coefficient matrices and augmented matrices:
      <ul>
      <li><p>A system of 2 equations in 3 unknowns:
            <md>
            <mrow>x-y+z \amp = 1</mrow>
            <mrow>2x+3y-z \amp = 2</mrow>
            </md>
      Coefficient matrix: 
        <me>
        \begin{bmatrix}
        1\amp -1\amp 1\\
        2\amp 3\amp -1
        \end{bmatrix}
        </me>
      Augmented matrix: 
        <me>
        \begin{bmatrix}
        1\amp -1\amp 1\amp 1\\
        2\amp 3\amp -1\amp 2
        \end{bmatrix}
        </me>
      </p></li> 
     
     <li>
      <p>A system of 3 equations in 2 unknowns:
      <md>
      <mrow>u+v  \amp = 1</mrow>
      <mrow>2u+v \amp = 3</mrow>
      <mrow>u+2v \amp =3</mrow>
      </md>
      </p>
      
      <p>Coefficient matrix: <m>\begin{bmatrix}
      1\amp 1\\
      2\amp 1\\
      1\amp 2
      \end{bmatrix}</m></p>
      
      <p>Augmented matrix: <m>\left[\begin{array}{cc|c}
      1\amp 1\amp 1\\
      2\amp 1\amp 3\\
      1\amp 2\amp 3
      \end{array}\right]</m></p>
      </li> 
      
     <li><p>A system of 4 equations in 5 unknowns</p>
     
     <p>
      <md>
      <mrow>x_1\amp + \amp x_2 \amp \amp \amp \amp \amp \amp \amp =1</mrow>
      <mrow>\amp \amp x_2 \amp +\amp x_3 \amp \amp \amp \amp \amp =2</mrow>
      <mrow>\amp \amp \amp \amp x_3 \amp +\amp x_4 \amp \amp \amp = 3</mrow>
      <mrow>\amp \amp \amp \amp \amp \amp x_4 \amp +\amp x_5 \amp =4</mrow>
      </md>
      </p>
      
      <p>Coefficient matrix: <m>\begin{bmatrix}
     \begin{array}{rrrrr}
      1\amp 1\amp 0\amp 0\amp 0\\
      0\amp 1\amp 1\amp 0\amp 0\\
      0\amp 0\amp 1\amp 1\amp 0\\
      0\amp 0\amp 0\amp 1\amp 1
       \end{array}
      \end{bmatrix}</m></p>
      
      <p>Augmented matrix: <m>\left[\begin{array}{ccccc|c}
      1\amp 1\amp 0\amp 0\amp 0\amp 1\\
      0\amp 1\amp 1\amp 0\amp 0\amp 2\\
      0\amp 0\amp 1\amp 1\amp 0\amp 3\\
      0\amp 0\amp 0\amp 1\amp 1\amp 4
      \end{array}\right]</m></p></li> 
      </ul>
    </p>     
   </subsection>


   <subsection><title>Exercises</title>
   <!-- <exercises> -->
       <exercise>
       <statement>
       <p>Which of the following equations are linear?
       <ol>
       <li> <p><m>x+y-z^2=0</m></p></li>
       <li> <p><m>2x+yz=1</m></p></li>
       <li> <p><m>z=3</m></p></li>
       <li> <p><m>\pi^2 x-\sqrt[3]{3}y=8\sqrt{5}</m></p></li>
       <li> <p><m>3x-4y=\frac73 z-5+t</m></p></li>
       <li> <p><m>\cos{x}=3y+2z</m></p></li>
       </ol></p>
       </statement>
       <solution>
       <p>
       <ol>
       <li><p>Not linear because of <m>z^2</m></p></li>
       <li><p>Not linear because of <m>yz</m></p></li>
       <li><p>Linear</p></li>
       <li><p>Linear (the exponents involve the constants, but not the variables)</p></li>
       <li><p>Linear (Tricky because variables are on both sides of the equal sign.
       It is the same as <m>3x-4y-\frac73 z-t =-5 </m>.)</p></li>
       <li><p>Not linear because of the <m>\cos</m> function</p></li>
       </ol>
       </p>
       </solution>    
       </exercise>

       <exercise>
       <statement>
       <p>Consider the following system of linear equations:
       <me>
       2x-4y+z=-1\\
       3x-3y+z=1
       </me>
       Which of the following are solutions to this system
       <ul>
       <li><p><m> (x,y,z)=(2,1,-1) </m></p></li>
       <li><p><m> (x,y,z)=(1,1,1) </m></p></li>
       <li><p><m> (x,y,z)=(0,0,1) </m></p></li>
       </ul>
       </p>
       </statement>
       <solution>
       <p>
       <ul>
       <li><p>
       Substitute <m> (x,y,z)=(2,1,-1) </m> into the equations:
       <me>
         2x-4y+z=4-4-1=-1\\
         3x-3y+z=6-3-1=2 \not=1
       </me>
       and so <m>(2,1,-1)</m> is <em>not</em> a solution.
       </p></li>
       <li><p> Substitute <m> (x,y,z)=(1,1,1) </m> into the equations:
       <me>
         2x-4y+z=2-4+1=-1\\
         3x-3y+z=3-3+1=1
       </me>
       and so <m>(1,1,1)</m> <em>is</em> a solution.
       </p></li>
       <li><p> Substitute <m> (x,y,z)=(0,0,1) </m> into the equations:
       <me>
         2x-4y+z=0+0+1\not=-1\\
         3x-3y+z=0+0+1=1
       </me>
       and so <m>(0,0,1)</m> is <em>not</em> a solution.
       </p></li>
       </ul></p>
       </solution>
       </exercise>
   
       <exercise>
       <statement>
       <p>
       Give a system of two equations in two unknowns that has <m>(1,2)</m>
       as its unique solution.
       </p>
       </statement>
       <solution>
       <p>
       There are many solutions corresponding to two lines intersecting
       at <m>(1,2)</m>. Surely the easiest is
       <md>
       <mrow>x\amp=1</mrow>
       <mrow>y\amp=2</mrow>
       </md>
       </p>
       </solution>
       </exercise>
   
       <exercise>
       <statement>
           <p>
           Give the coefficient matrix and the augmented matrix this system of
           linear equations:
           <md>
           <mrow>x+y \amp= 2 </mrow>
           <mrow>x-y \amp= 0 </mrow>
           </md>
           </p>
       </statement>
       <solution>
         <p>
         <m>
          \begin{bmatrix}
          1\amp1\\1\amp-1
          \end{bmatrix}
         </m> 
      and
      <m>
         \left[\begin{array}{cc|c}
         1 \amp 1 \amp 2 \\  1 \amp -1 \amp 0
         \end{array}\right]
      </m>
      </p>
       </solution>    
       </exercise>
   
       <exercise>
       <statement>
           <p>
           Give the coefficient matrix and the augmented matrix this system of
           linear equations:
           <md>
           <mrow>x_1+x_2-x_3+x_5\amp= 4 </mrow>
           <mrow>2x_1-3x_2 +x_4\amp= 7 </mrow>
           </md>
           </p>
       </statement>
       <solution>
      <p>
      <m>
      \begin{bmatrix}
      1 \amp 1 \amp -1 \amp 0 \amp 1 \\
      2\amp-3 \amp 0 \amp 1 \amp 0 
      \end{bmatrix}
      </m> 
      and
      <m>
      \left[\begin{array}{ccccc|c}
      1 \amp 1 \amp -1 \amp 0 \amp 1 \amp 4\\
      2\amp-3 \amp 0 \amp 1 \amp 0 \amp7
      \end{array}\right]
      </m>
      </p>
       </solution>
       </exercise>
   
       <exercise>
       <statement>
           <p>
           Give the coefficient matrix and the augmented matrix this system of
           linear equations:
           <md>
           <mrow>x+y-z \amp=1</mrow>
           <mrow>3x+y\amp= 2</mrow>
           <mrow>y+z \amp=3</mrow>
           <mrow>x+z\amp=4</mrow>
           <mrow>x+y+z\amp=5</mrow>
           </md>
           </p>
       </statement>
       <solution>
      <p>
      <m>
      \begin{bmatrix}
      1 \amp 1 \amp -1 \\
      3 \amp 1\amp 0 \\
      0 \amp 1 \amp 1\\
      1 \amp 0 \amp 1 \\
      1 \amp 1 \amp 1
      \end{bmatrix}
      </m> 
      and
      <m>
      \left[\begin{array}{ccc|c}
      1 \amp 1 \amp -1 \amp 1\\
      3 \amp 1\amp 0 \amp 2\\
      0 \amp 1 \amp 1\amp 3\\
      1 \amp 0 \amp 1 \amp 4\\
      1 \amp 1 \amp 1\amp 5
      \end{array}\right]
      </m>
      </p>
       </solution>
       </exercise>
   
       <exercise>
       <statement>
       <p>
       Give the system of equations whose augmented matrix is
       <me>
      \left[\begin{array}{cc|c}
      1 \amp 4  \amp 1\\
      3 \amp -2\amp 0 \\
      0 \amp 1 \amp 1\\
      -1 \amp 1 \amp 3
      \end{array}\right]
       </me>
       </p>
       </statement>
       <solution>
       <p>
       <md>
       <mrow>x+4y\amp= 1</mrow>
       <mrow>3x-2y\amp=0</mrow>
       <mrow>y\amp=1</mrow>
       <mrow>-x+y\amp=3</mrow>
       </md>
       </p>
       </solution>
       </exercise>
   
   
       <exercise>
       <statement>
       <p>
       There is a special matrix <m>I_n</m> called the identity
       matrix. It has <m>n</m> rows and <m>n</m> columns. The entries <m>a_{i,j}</m>
       of the matrix satisfy
       <me>
       a_{i,j}=
       \begin{cases}
       1 \amp \text{if } i=j\\
       0 \amp \text{if } i\not=j\\
       \end{cases}
       </me>
       <ol>
       <li><p>
       Show the the matrix <m>I_n</m> has the value <m>1</m> along the (main) diagonal going from the upper
       left corner to the lower right corner of the matrix, and has the value <m>0</m> elsewhere.
       </p></li> 
       <li><p>Suppose that a system of linear equations has an augmented matrix of the form
       <me>
       \left[ I_n \mid B \right]
       </me>
       where
       <me>
       B= \begin{bmatrix}
       b_1\\b_2\\ \vdots\\b_n
       \end{bmatrix}
       </me>
       What does the say about the number of variables (unknowns)
       for the corresponding system of linear equations, and
       what does it say about the solutions to the system?
       </p>
       </li>
       </ol>
       </p>
       </statement>
   
       <solution>
       <p>
       <ol>
       <li>
       <p>
       When <m>i=j</m>, the row number and column number are identical, and so the entry
       <m>a_{i.j}</m> is on the diagonal. This implies that 
       <me>
       a_{i,j}=
       \begin{cases}
       1 \amp \text{ for entries on the main diagonal}\\
       0 \amp \text{ for entries not on the main diagonal} \\
       \end{cases}
       </me>
       </p>
       </li>
       <li>
       <p>
       The number of unknowns is <m>n</m>, the number of columns in the matrix.
       We may call them <m>x_1, x_2,\dots,x_n</m>.
       The augmented matrix now says that
       <md>
       <mrow>x_1\amp =b_1</mrow>
       <mrow>x_2\amp =b_2</mrow>
       <mrow>\amp \vdots</mrow>
       <mrow>x_n\amp =b_n</mrow>
       </md>
       </p>
       </li>
       </ol>
       </p>
       </solution>
       </exercise>
  
   <!-- </exercises> -->
   
   </subsection>
   
</section>


<section><title>Elementary row operations</title>
    <introduction>
      <p>
      Gaussian elimination, which we shall describe in detail
      presently, is an algorithm (a well-defined procedure for computation
      that eventually completes) that finds all solutions to any <m>m\times n</m> system
      of linear equations. It is defined via certain operations carried out
      on the augmented matrix. These operations change the matrix (and hence
      the system of linear equations associated with it), <em>but they leave the
      set of solutions unchanged</em>. There are three of them, which we now describe.
     </p></introduction>

      <subsection><title>Interchanging two rows</title>
      <p>
         Exchanging two rows in the augmented matrix is the same as writing the the equations 
         in a different order. The equations themselves are unchanged, as is the set of all 
         solutions. When we interchange row <m>i</m> and row <m>j</m>, we denote it by 
         <m>R_i\leftrightarrow R_j</m>. 
      </p>

        <example><title>Augmented matrix change (interchange rows)</title>
        <p>The system of equations
        <me>
        \begin{array}{rl}
        2x+3y-z \amp = 2\\
        -x+y+z \amp =4\\
        3x-y-z \amp = 3
        \end{array}
        </me> 
        corresponds to the augmented matrix
        <me>
        \left[\begin{array}{ccc|c}
        2 \amp 3 \amp -1 \amp 2\\
        -1 \amp 1 \amp 1 \amp 4\\
        3 \amp -1 \amp -1 \amp 3
        \end{array}\right]
        </me>
        Now we interchange rows 1 and 3 (<m>R_1\leftrightarrow R_3</m>) to get
        the matrix
        <me>
        \left[\begin{array}{ccc|c}
        3 \amp -1 \amp -1 \amp 3\\
        -1 \amp 1 \amp 1 \amp 4\\
        2 \amp 3 \amp -1 \amp 2
        \end{array}\right]
        </me>
        which corresponds to the equations
        <me>
        \begin{array}{rl}
        3x-y-z \amp = 3\\
        -x+y+z \amp =4\\
        2x+3y-z \amp = 2
        \end{array}
        </me>
        </p>
        </example>

      </subsection>
      
      <subsection><title>Multiplying a row by a nonzero constant</title>
       <p>
       If we have a linear equation and multiply both sides of the equation by a nonzero
        constant <m>\lambda</m> (the Greek letter lambda (<m>\lambda</m>) 
        is the traditional name for the constant) then the linear equation
        <me>
        a_1x_1 + a_2x_2 +a_3x_3 + \cdots + a_nx_n =b
        </me>
        becomes
        <me>
        \lambda(a_1x_1 + a_2x_2 +a_3x_3 + \cdots + a_nx_n) =\lambda b
        </me>
        and so
        <me> \lambda a_1x_1 + \lambda a_2x_2 +\lambda a_3x_3 + \cdots +\lambda a_nx_n =\lambda b.
        </me>
        This would apply to any equation in a system, of course, 
        and so when we apply this operation to row <m>i</m> we denote it by 
        <m>R_i \gets \lambda R_i</m> (The symbol <m>\gets</m> means <em>is replaced by</em>). 
        <em>As long as <m>\lambda</m> is nonzero</em>, this operation leaves the set of solutions unchanged.
        </p>

        <example><title>Augmented matrix change (multiply row by <m>\lambda</m>)</title>
        <p>The system of equations
        <me>
        \begin{array}{rl}
        2x+3y-z \amp = 2\\
        -x+y+z \amp =4\\
        3x-y-z \amp = 3
        \end{array}
        </me> 
        corresponds to the augmented matrix
        <me>
        \begin{bmatrix}
        2 \amp 3 \amp -1 \amp 2\\
        -1 \amp 1 \amp 1 \amp 4\\
        3 \amp -1 \amp -1 \amp 3
        \end{bmatrix}
        </me>
        Now we multiply row <m>2</m> by <m>\lambda=-3</m> (<m>R_2\gets -3R_2</m>) to get
        the matrix
        <me>
        \begin{bmatrix}
        2 \amp 3 \amp -1 \amp 2\\
        3 \amp -3 \amp -3 \amp -12\\
        3 \amp -1 \amp -1 \amp 3
        \end{bmatrix}
        </me>
        which corresponds to the system of equations
        <me>
        \begin{array}{rl}
        2x+3y-z \amp = 2\\
        3x-3y-3z \amp =-12\\
        3x-y-z \amp = 3
        \end{array}
        </me>
        </p>
        </example>
      
      </subsection>

      <subsection><title>Adding a multiple of one row to another</title>
        <p>Recall that one of our basic techniques for solving a
        system of equations is to pick a variable,
        make the coefficients of that variable equal in two equations
        and then take the difference of the equations
        to create a new one with the variable eliminated. In terms of
        the associated augmented matrix, this means we get a new row
        formed by subtracting the corresponding entries in the rows of
        the two equations.  In other words, we replace one row by the
        difference of two rows. We use the notation <m>R_i\gets R_i-R_j</m>
        to indicate that row <m>i</m> is replaced by the difference of
        row <m>i</m> and row <m>j</m>.  The set of solutions is unchanged. We
        have already seen that multiplying a row by a nonzero constants
        leaves the set of solutions unchanged; it is often convenient to
        combine these two steps as one: <m>R_i\gets R_i+\lambda R_j.</m>
        Note that <m>\lambda=-1</m> is the case of subtracting one row
        from another.</p>

        <example><title>Augmented matrix change (add multiple of one
        row to another)</title> 
        <p>Eliminating <m>x</m> from the second
        equation: 
        <me> 
        \begin{array}{rl}
        -x+y+z \amp =4\\
        2x+3y-z \amp = 2\\
        3x-y-z \amp = 3
        \end{array}
        </me> 
        corresponds to the
        augmented matrix 
        <me> 
        \begin{bmatrix} 
        -1 \amp 1 \amp 1 \amp 4\\
        2 \amp 3 \amp -1 \amp 2\\ 
        3 \amp -1 \amp -1 \amp 3 
        \end{bmatrix}
        </me> 
        Now we replace row 2 by the sum of row 2 and twice row 1
        <m>(R_2\gets R_2+2R_1)</m> to get the matrix 
        <me> 
        \begin{bmatrix}
        -1 \amp 1 \amp 1 \amp 4\\ 
        0 \amp 5 \amp 1 \amp 10\\ 
        3 \amp -1 \amp -1 \amp 3 
        \end{bmatrix} 
        </me> 
        which corresponds to the system of linear equations 
        <me> 
        \begin{array}{rl} -
        x+y+z
        \amp =4\\ 
        5y+z \amp =10\\ 
        3x-y-z \amp = 3 
        \end{array} 
        </me>
        </p> 
        </example>
      </subsection>

      <subsection><title>Summary of the three elementary row operations</title>
        <p> In summary, here is a table of the three elementary row
        operations: 
        </p>

        <table xml:id="TableOfElemantaryRowOperations">
        <title>Elementary Row Operations</title>
        <tabular bottom="minor" top="minor" left="minor" right="minor" halign="center">
        <row>
            <cell><em>Elementary Operation</em></cell>
            <cell><em>Notation</em></cell>
        </row>
        <row>
            <cell>Interchange rows</cell>
            <cell><m>R_i\leftrightarrow R_j</m></cell>
        </row>
        <row>
            <cell>Multiply row by constant</cell>
            <cell><m>R_i\gets\lambda R_i</m>, <m>\lambda\not=0</m></cell>
        </row>
        <row>
            <cell>Add multiple of one row to another</cell>
            <cell><m>R_i\gets R_i+\lambda R_j</m></cell>
        </row>
        </tabular>
        </table>
        
        <p>
        Each operation changes
        the matrix and the associated system of linear equations, but
        it leaves the set of solutions unchanged.  </p>
      </subsection>

      <subsection><title>Changing a specific entry of a matrix using elementary row
      operations</title>

      <p>
      We want to show that it is possible to change specific entries
      in a matrix in an advantageous way using elementary row 
      operations. Specifically, we will show two things:
      <ul>
      <li><p> Any <m>a_{i,k}\not=0</m> may be changed to <m>1</m> with one elementary row
         operation. </p>
      </li>
      <li>
        <p>If <m>a_{i,k}\not=0</m>, then any other entry in the same column
        may be changed to <m>0</m> with one elementary row operation.
        </p>
      </li>
      </ul>
      </p>

      <theorem xml:id="ChangeTo01"><title>Change a matrix entry to <m>0</m> or <m>1</m> 
      with elementary row operations</title>
      <statement>
      <p>
      <ol>
      <li>
        <p>If a matrix has an entry <m>a_{i,k}\not=0</m>, then, with one elementary
        row operation, it may be changed to <m>1</m>.</p>
      </li>
      <li>
        <p>If some column <m>k</m> of a matrix has two non-zero entries
        <m>a_{i,k}</m> and <m>a_{j,k}</m> then, with one elementary row
        operation, (either) one of them 
        may be changed to <m>0</m>. </p>
      </li>
      </ol>
      </p>
      </statement>
      <proof>
    <p>
    <ol>
      <li>
      <p>If a matrix has an entry <m>a_{i,k}\not=0</m>, then multiply
      the row <m>R_i</m> by <m>\frac 1{a_{i,k}}</m>, that is, carry out
      the elementary row operation <m>R_i\gets \frac 1{a_{i,k}}R_i</m>.
      In the resulting matrix, the <m>i</m>-<m>k</m> entry is
      <m>\frac {a_{i,k}}{a_{i,k}}=1</m>.
      </p>
      </li>
      <li>
      <p>Suppose a matrix has two non-zero entries <m>a_{i,k}</m> and <m>a_{j,k}</m>.
      First, do the elementary row operation 
      <m>R_i\gets \frac 1{a_{i,k}}R_i</m> to change <m>a_{i,k}</m> to <m>1</m>.
      Then, do the elementary row operation 
      <m>R_j\gets R_j-a_{j,k}R_i</m>. The matrix  now has a new value
      in the <m>j</m>-<m>k</m> position which is <m>a_{j,k}-a_{j,k}=0</m>, and we
      have accomplished our goal using two elementary operations.
      Now we note that these two operations can be carried out with
      the single operation 
      <m>
      R_j\gets R_j-\frac {a_{j,k}}{a_{i,k}}R_i
      </m>.
      </p>
      </li>
      </ol>
      </p>
      </proof>
      </theorem>
   </subsection>


   <subsection><title>Exercises</title>
   <p>
   For these exercises, let
   <m>A=
   \begin{bmatrix}
   1\amp2\amp3\amp4\\  5\amp6\amp7\amp-6\\ -5\amp-4\amp-3\amp-2
   \end{bmatrix}
   </m>
   </p>
<!--   <exercises> -->
   <exercise>
   <statement>
   <p>Find the elementary row operation that
   <ul>
   <li><p>changes the <m>2</m> to a <m>1</m></p></li>
   <li><p>changes the <m>7</m> to a <m>1</m></p></li>
   <li><p>changes the <m>-4</m> to a <m>1</m></p></li>
   </ul>
   In each case give the matrix that results from the 
   application of your elementary row operation.</p>
   </statement>
   <solution>
   
   <p>
   <ul>
   <li><p><m>R_1\gets \frac12 R_1</m>
   <m>\begin{bmatrix}
   \frac12\amp1\amp\frac32\amp\frac12\\  5\amp6\amp7\amp-6\\ -5\amp-4\amp-3\amp-2
   \end{bmatrix}</m>
   </p></li>
   <li><p><m>R_2\gets \frac17 R_2</m>
   <m>
   \begin{bmatrix}
   1\amp2\amp3\amp4\\  \frac57\amp\frac67\amp1\amp-\frac67\\ -5\amp-4\amp-3\amp-2
   \end{bmatrix}
   </m> </p></li>
   <li><p><m>R_3\gets -\frac14 R_3</m>
   <m>\begin{bmatrix}
   1\amp2\amp3\amp4\\  5\amp6\amp7\amp-6\\ \frac54\amp1\amp\frac34\amp\frac12
   \end{bmatrix}</m>
   </p></li>
   </ul>
   </p>
   </solution>
   </exercise>
   
   
   <exercise>
   <statement>
   <p>Find an elementary row operation that
   <ul>
       <li><p>changes the <m>5</m> to a <m>0</m></p></li>
       <li><p>changes the <m>6</m> to a <m>0</m></p></li>
       <li><p>changes the <m>-4</m> to a <m>0</m></p></li>
       <li><p>changes the <m>7</m> to a <m>0</m></p></li>
   </ul>
   In each case give the matrix that results from the 
   application of your elementary row operation.</p>
   </statement>
   <solution>
   <p>
   <ul>
       <li><p><m>R_2\gets R_2+R_3</m>
       <m>\begin{bmatrix}
       1\amp2\amp3\amp4\\  0\amp2\amp4\amp-8\\ -5\amp-4\amp-3\amp-2
       \end{bmatrix}</m>
       </p></li>
   
       <li><p><m>R_2\gets R_2-3R_1</m>
       \begin{bmatrix}
       1\amp2\amp3\amp4\\  2\amp0\amp-2\amp-18\\ -5\amp-4\amp-3\amp-2
       \end{bmatrix}
       </p></li>
   
       <li><p><m>R_3\gets R_3+2R_1</m>
       <m>\begin{bmatrix}
       1\amp2\amp3\amp4\\  5\amp6\amp7\amp-6\\ -3\amp0\amp3\amp6
       \end{bmatrix}</m>
       </p></li>
   
       <li><p><m>R_2\gets R_2-\frac73 R_1</m>
       <m>\begin{bmatrix}
       1\amp2\amp3\amp4\\  \frac83\amp\frac43\amp0\amp-\frac{46}3\\ -5\amp-4\amp-3\amp-2
       \end{bmatrix}</m>
       </p></li>
   </ul>
   </p>
   </solution>
   </exercise>
   
   
    <exercise>
    <statement>
    <p>Use two elementary rwo operations to make the first column
    of the resulting matrix
    <m>
    \begin{bmatrix}
    1\\0\\0
    \end{bmatrix}
    </m></p>
    </statement>
    <solution>
    <p><m>R_2\gets R_2-5R_1</m> and  <m>R_3\gets R_3+5R_1</m> give
    <me>
    \begin{bmatrix}
    1\amp2\amp3\amp4\\
    0\amp-4\amp-8\amp-26\\
    0\amp6\amp12\amp18
    \end{bmatrix}
    </me>
    </p>
    </solution>
    </exercise>

<!--   </exercises> -->

   </subsection>



    </section>

    
    <section><title>The Gaussian elimination algorithm</title>
    
        <subsection><title> Some matrices whose associated system of equations
        are easy to solve</title>
        <p>The elementary row operations allow us to change matrices
        and their associated system of linear equations without
        changing the solutions of those equations. The goal is
        to end up with matrices which make these common solutions
        obvious.  Here are some examples.

        <ul> <li><p>The augmented matrix 
        <me>
        \left[\begin{array}{cccc|c} 
        1\amp 0\amp 0\amp 0\amp 1\\ 
        0\amp 1\amp 0\amp 0\amp 2\\
        0\amp 0\amp 1\amp 0\amp 5\\ 
        0\amp 0\amp 0\amp 1\amp -1 
        \end{array}\right] 
        </me> 
        corresponds to the system of linear equations 
        <me>\begin{array}{cccccc} 
        x_1\amp
        \amp \amp \amp =\amp 1\\ 
        \amp x_2\amp \amp \amp =\amp 2\\ 
        \amp \amp x_3\amp \amp =\amp 5\\ 
        \amp \amp \amp x_4\amp =\amp -1
        \end{array} </me> 
        The equations themselves actually describe
        the unique solution! Notice the structure of the coefficient
        matrix that makes this possible. There is only one nonzero
        entry in each row, its value is 1, and as you proceed down
        through from one row to the next, the nonzero entry moves
        one column to the right.</p></li>

        <li><p> The augmented matrix 
        <me> 
        \left[\begin{array}{cc|c}
        1\amp 1\amp 4\\ 0\amp 2\amp 6 
        \end{array}\right]
        </me> 
        corresponds to the system of linear equations
        <me>
        \begin{array}{rcc}
        x_1+x_2\amp =\amp 4\\ 
        2x_2\amp =\amp 6   
        \end{array} 
        </me> 
        The last row is easy to solve:
        we get <m>x_2=3</m>. Using this value, it is also easy to solve
        <m>x_1+x_2=x_1+3=4</m>, or <m>x_1=1</m>.  </p></li>

       <li><p>
        The augmented matrix <me> \left[\begin{array}{ccc|c} 1\amp
        1\amp 1\amp 4\\ 0\amp 0\amp 2\amp 6 \end{array}\right]
        </me> corresponds to the system of linear equations <me>
        \begin{array}{rcc} x_1+x_2+x_3\amp =\amp 4\\ 2x_3\amp =\amp 6
        \end{array} </me> As before, we get <m>x_3=3</m>. We still have two
        variables undefined: we assign a parameter to the second one:
        <m>x_2=t</m>. Using this value, we have <m>x_1+x_2+x_3=x_1+t+3=4</m>,
        or <m>x_1=1-t</m>. We now know the solutions: <m>x_1=1-t</m>, <m>x_2=t</m>,
        <m>x_3=3</m> where <m>t</m> can be any real number. 
        In fact we can check this result with the first
        equation: <m>x_1+x_2+x_3=(1-t) + t + 3=4</m>. The compact way
        of writing this solution is <m>(x_1,x_2,x_3)=(1-t,t,3)</m>.
        </p></li> </ul>

         It's clear from the examples given that having lots of
        zeros in the coefficient matrix is helpful for computing the
        solutions. It's also clear that leading ones can also make
        the computation easier. Our plan is to use elementary row
        operations to change a given coefficient matrix into one with
        these properties, and then to describe all the solutions.
        Here are some observations that will help us:
        <ol>
       <li><p>If a column has some nonzero entry, we can always 
           make the top entry nonzero by interchanging rows, if
            necessary using <m>R_1\leftrightarrow R_i</m>
            for some <m>i>1</m>.</p></li> 
        <li><p>If the first nonzero entry of
            a row <m>R_i</m> is <m>\lambda</m>, we can turn it into a 1 by using
            <m>R_i\leftarrow \tfrac 1\lambda R_i</m>.</p></li> 
        <li><p> If two rows <m>R_i</m> and <m>R_j</m> have nonzero entries 
            in some column <m>k</m>, we can turn the <m>j,k</m> entry into 
            a zero using <m>R_j \leftarrow R_j - \frac{a_{j,k}}{a_{i,k}} R_i</m>.
            </p></li> 
        </ol></p> 

       </subsection>


       <subsection><title>Row echelon form</title>
       <p>
        We now want to define a general class of matrices whose
        corresponding system of linear equations have solutions
        that are easy to find. These matrices have a special pattern
        of zeros and ones, and are said to be in
        <em>row echelon form.</em>
        </p>

        
        <figure>
        <caption/>
        <image source="images/gausselim.png" width="50%">
        <description>A matrix in row echelon form</description>
        </image>
        </figure>
        
        <p>
        The matrix above gives an idea of what we want.
        Notice the staircase line drawn through the matrix
        has all entries below it equal to zero.
        The entries marked with a <m>*</m> can take on any value.
        </p>
        <p>
        The first nonzero entry in a row (if there is one)
        is called the <em>leading entry.</em>
        If it equals <m>1</m>, then it is called a <em>leading one.</em>
        </p>
        
        <definition><title>Row echelon form</title>
        <statement>
        <p>A matrix is in  <term> row echelon form</term> if
        <ul>
        <li><p>Every leading entry is a leading one. </p></li>
        <li><p>Every entry below a leading one is <m>0</m>. </p></li>
        <li><p>As you go down the matrix, the leading ones move to the right.</p></li>
        <li><p>Any all zero rows are at the bottom.</p></li>
        </ul></p>
        </statement>
        </definition>
        
        <exercise>
        <statement>
        <p>Which of the following matrices are in row echelon form?
        <ul>
       
        <li><p><m>
        \begin{bmatrix}
        1\amp2\amp3\amp4\\
        0\amp5\amp6\amp7\\
        0\amp0\amp8\amp9\\
        0\amp0\amp0\amp10
        \end{bmatrix}
        </m></p></li>
        
        <li><p><m>
        \begin{bmatrix}
        1\amp2\amp3\amp4\\
        0\amp1\amp2\amp3\\
        0\amp0\amp0\amp0\\
        0\amp0\amp0\amp1
        \end{bmatrix}
        </m></p></li>
        
        <li><p><m>
        \begin{bmatrix}
        1\amp2\amp3\amp4\\
        0\amp1\amp2\amp3\\
        0\amp0\amp0\amp1\\
        0\amp0\amp0\amp0
        \end{bmatrix}
        </m></p></li>
        
        <li><p><m>
        \begin{bmatrix}
        1\amp2\amp3\amp4\\
        0\amp1\amp6\amp7\\
        0\amp0\amp1\amp1\\
        0\amp0\amp1\amp1
        \end{bmatrix}
        </m></p></li>
        
        <li><p><m>
        \begin{bmatrix}
        1\amp0\amp0\amp0\\
        0\amp0\amp0\amp0\\
        0\amp0\amp0\amp0\\
        0\amp0\amp0\amp0
        \end{bmatrix}
        </m></p></li>
        
        
        </ul></p>
        </statement>
        <solution>
        <p>
        <ul>
        <li><p>Not in row echelon form because not every leading entry is a <m>1</m>. </p></li>
        <li><p>Not in row echelon form because the zero row is not at the bottom. </p></li>
        <li><p>It is row echelon form.</p></li>
        <li><p>Not row echelon form because a leading entry has a 
        nonzero entry below it.</p></li>
        <li><p>It is row echelon form.</p></li>
        </ul></p>
        </solution>
        </exercise>

       </subsection>
    

            <subsection><title>The Gaussian elimination algorithm </title> 
             <p>The plan is now start with the augmented matrix and, by using
             a sequence of elementary row operations, change it to a
             new matrix where it is easy to identify the solutions of
             the associated system of linear
            equations. Since any elementary row operation leaves the
            solutions unchanged, the solutions to the final system of
            linear equations will be identical to the solutions of the
            original one. </p>

            <p>We work on the columns of the matrix from left to
            right and change the matrix in the following way:
            <ol> 
            <li><p>Start with the first column. If it has all
            entries equal to zero, move on to the next column to the
            right.
            </p></li> 
            <li><p>If the column has nonzero entries,
            interchange rows, if necessary, to get a nonzero entry on
            top.
            </p></li> 
            <li>
            <p>
            Change the top entry, if necessary, to make it a <m>1</m>.
            </p>
            </li>
            <li><p>For any nonzero entry below the top one,
            use an elementary row operation to change it to zero.
            </p></li>
            <li><p>Now consider the part of the matrix below the top
            row and to the right of the column under consideration: if
            there are no such rows or columns, stop since the procedure
            is finished.Otherwise, carry out the same procedure on the
            new matrix.
            </p></li> 
            </ol></p> 
            
            <p>Here is a first example:
            <me>
            \begin{array}{r}
            3x_1-2x_2-x_3+x_4=1\\
            6x_1-8x_2+x_3+2x_4=4
            \end{array}
            </me>
            has augmented matrix
            <me>
            \begin{bmatrix}
            3 \amp  -2 \amp  -1 \amp 1 \amp 1 \\
            6 \amp  -8 \amp  1 \amp  2 \amp  4
            \end{bmatrix}.
           </me>
           We don't need to exchange rows to make the top entry of the
           first column nonzero, so we proceed to make the top entry <m>1</m> using
           the elementary row operation
           <m>R_1\gets \frac13R_1</m>. The matrix becomes
            <me>
            \begin{bmatrix}
            1 \amp  -\frac23 \amp  -\frac13 \amp \frac13 \amp \frac13 \\
            6 \amp  -8 \amp  1 \amp  2 \amp  4
            \end{bmatrix}.
           </me>


           Now we must make all entries below the top
           one in the column equal to zero. There is, of course, only one such entry,
           and so,  using <m>R_2\leftarrow R_2-6R_1</m>, we get
           <me>
           \begin{bmatrix}
            1 \amp  -\frac23 \amp  -\frac13 \amp \frac13 \amp \frac13 \\
            0 \amp  -4 \amp  3 \amp  0 \amp  2
           \end{bmatrix}
           </me>.

           We're now done with the first column, so we 
           continue with the same procedure on the matrix obtained by deleting the first
           row and first column:

           <me>
           \begin{bmatrix}
             -4 \amp  3 \amp  0 \amp  2
           \end{bmatrix}.
           </me>

           Since there is only one row, we need only change the top entry to <m>1</m>
           using division by <m>-4</m>, that is, <m>R_1\gets -\frac14R_1</m>. We then get
           <me>
           \begin{bmatrix}
             1 \amp  -\frac34 \amp  0 \amp  -\frac12
           \end{bmatrix}.
           </me>
           and, putting it back into the original matrix, we get
           <me>
           \begin{bmatrix}
            1 \amp  -\frac23 \amp  -\frac13 \amp \frac13 \amp \frac13 \\
              0 \amp  1 \amp  -\frac34 \amp  0 \amp  -\frac12
          \end{bmatrix}
          </me></p>
        
          <p>The matrix is in row echelon form. Now we can determine all of the solutions
          to the original system of linear equations. 
          The first nonzero
          entry in the first row is in the first column, the column associated
          with <m>x_1</m>. The first nonzero entry in the second row, similarly, is
          associated with <m>x_2</m>. We assign parameters to the other variables:
          <m>x_3=s</m> and <m>x_4=t</m>. The second row then tells us that
          <m>x_2-\frac34s =-\frac12 </m>, or <m>x_2=\frac34s-\frac12</m>. Now
          that we know <m>x_2</m>, we can use the first row to find <m>x_1</m>: we get
          <m>x_1-\frac23x_2-\frac13s+\frac13t=\frac13</m>. We substitute in our known
          value for <m>x_2</m> into this equation, and after some
          simplification, we get <m>x_1=\tfrac56s-\tfrac13t</m>.  In summary, we have:
          All solutions to
          <me>
          \begin{array}{r}
          3x_1-2x_2-x_3+x_4=1\\
          6x_1-8x_2+x_3+2x_4=4
          \end{array}
          </me>
          are of the form
          <me>
          \begin{array}{rcl}
          x_1\amp =\amp \tfrac56s-\tfrac13t\\
          x_2\amp =\amp -\tfrac12 +\tfrac34 s\\
          x_3\amp =\amp s\\
          x_4\amp =\amp t
          \end{array}
          </me>
          where <m>s</m> and <m>t</m> are any real numbers. 
          More compactly, we write this as
          <m>(x_1,x_2,x_3,x_4)= \tfrac56s-\tfrac13t,-\tfrac12 +\tfrac34 s,s,t)</m>.
          In other words, for any assignment of real numbers to
          <m>s</m> and <m>t</m>, we get a solution to the system of linear equations.</p>
          
          <p>It is easy (and worthwhile) to check that substituting the values of <m>x_1,\dots,x_4</m>
          into the two equations does indeed give a solution.</p>
        
          <!-- continue here -->
          <p>Now consider another example with the equations
          <me>
          \begin{array}{lrcc}
          x_1  + 2x_2 \amp -2x_3  \amp =\amp  -1\\
          3x_1 -2x_2  \amp -4x_3 \amp =\amp  3\\
          4x_1        \amp -2x_3 \amp =\amp  -2\\
          -x_1 -x_2   \amp +2x_3 \amp =\amp 0
          \end{array}
          </me>
          and its corresponding augmented matrix as it is changed by Gaussian elimination:
          <me>
          \begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          3\amp -2\amp -4\amp 3\\
          4\amp 0\amp -2\amp -2\\
          -1\amp -1\amp 2\amp 0
          \end{bmatrix}
          \begin{array}{c}
          \\R_2 \leftarrow R_2-3R_1\\ R_3\gets R_3-4R_1\\ R_4\gets R_4+R_1
          \end{array}
          </me>
          <me>\begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          0\amp -8\amp 2\amp 6\\
          0\amp -8\amp 6\amp 2\\
          0\amp 1\amp 0\amp -1
          \end{bmatrix}
          \begin{array}{c}
          R_2\gets-\frac18 R_2\\ R_3\gets R_3+8R_2\\R_4\gets R_4-R_2
          \end{array}
          </me>
          <me>
          \begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          0\amp 1\amp -\frac14\amp -\frac34\\
          0\amp 0\amp 4\amp -4\\
          0\amp 0\amp \tfrac14 \amp -\tfrac14
          \end{bmatrix}
          \begin{array}{c}
           R_3 \gets \frac14 R_3\\R_4\gets R_4-\frac14 R_3
          \end{array}
          </me>
          <me>
          \begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          0\amp 1\amp -\frac14\amp -\frac34\\
          0\amp 0\amp 1\amp -1\\
          0\amp 0\amp 0\amp 0
          \end{bmatrix}
          </me>
          The last row, for any choice of <m>x_1, x_2, x_3</m>, reduces to
          <m>0=0,</m> so any solution of the associated first three equations will also be
          a solution to the last one. In other words, the last row of the matrix has no effect
          on the solution set and may be dropped from the matrix. The third row gives
          <m>x_3=-1.</m> The second row gives <m>x_2=-1</m> and the first row gives <m>x_1=-1.</m>
          Hence there is one solution: <m>(x_1,x_2,x_3)=(-1,-1,-1).</m>
          </p>
        
          <p>
          Now change the equations from the last example very slightly 
          (the right-hand side of the last equation is changed from <m>0</m> to <m>1</m>):
          <me>
          \begin{array}{rcc}
          x_1 + 2x_2 -2x_3  \amp =\amp  -1\\
          3x_1-2x_2-4x_3 \amp =\amp  3\\
          4x_1 -2x_3 \amp =\amp  -2\\
          -x_1-x_2+2x_3 \amp =\amp  1
          \end{array}
          </me>
          The Gaussian elimination is almost identical as
          <me>
          \begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          3\amp -2\amp -4\amp 3\\
          4\amp 0\amp -2\amp -2\\
          -1\amp -1\amp 2\amp 1
          \end{bmatrix}
          </me>
          is reduced to
          <me>
          \begin{bmatrix}
          1\amp 2\amp -2\amp -1 \\
          0\amp 1\amp -\frac14\amp -\frac34\\
          0\amp 0\amp 1\amp -1\\
          0\amp 0\amp 0\amp 1

          \end{bmatrix}
          </me>
          </p>
        
          <p>Now the last row says <m>0x_1+0x_2+0x_3=1</m>, which, for any choice of
          <m>x_1</m>, <m>x_2</m> and <m>x_3</m>, reduces to <m>0=1</m>
          which is never true. This means the original system of equations
          have no solutions, that is, the system is inconsistent.
          </p>
        
          <p>We can make a useful observation here: 
          If a row of the augmented matrix is of the form
          <me>
          \begin{bmatrix}
          0 \amp 0 \amp 0 \amp \cdots \amp 0 \amp *
          \end{bmatrix}
          </me>
          where <m>*</m> is either zero or nonzero, then one of two things happens:
          <ol>
          <li><p><m>*=0</m> in which case the row may be dropped from the matrix</p></li>
          <li><p><m>*\not=0</m> in which case there is no solution.</p></li>
          </ol>
          </p>
        
          <example>
          <p>Consider the system of linear equations:
          <me>
          \begin{array}{rcl}
          x+y+z\amp =\amp  1\\
          2x+y+z \amp =\amp  2\\
          3x+ay+bz\amp =\amp c
          \end{array}
          </me>
          We wish to know the values of <m>a,</m> <m>b</m> and <m>c</m> for which there 
          are there no solutions, one solution or more than one solution.
          To solve this problem, we apply Gaussian elimination to the augmented matrix: 
          
          <me>
          \begin{bmatrix} 1\amp 1\amp 1\amp 1\\2\amp 1\amp 1\amp 2\\3\amp a\amp b\amp c \end{bmatrix}
          \begin{matrix}R_2\gets R_2-2R_1\\ R_3\gets R_3-3 R_1\end{matrix}
          </me> 
          <me>
          \begin{bmatrix} 1\amp 1\amp 1\amp 1\\0\amp -1\amp -1\amp 0\\0\amp a-3\amp b-3\amp c-3 \end{bmatrix}
          \begin{matrix}R_2\gets -R_2\\R_3\gets R_3-(a-3)R_2\end{matrix}
          </me>
          <me>
          \begin{bmatrix} 1\amp 1\amp 1\amp 1\\0\amp 1\amp 1\amp 0\\0\amp 0\amp b-a\amp c-3 \end{bmatrix}
          </me>

          An analysis of the last row tells us everything: If <m>b-a\not=0,</m>
          then there is exactly one solution. If <m>b-a=0,</m> and <m>c-3\not=0,</m> 
          then there are no solutions. Otherwise (when
          <m>b=a</m> and <m>c=3</m>) there are an infinite number of solutions. 
          </p>
          </example>

          <exercise>
          <statement>
          <p>
            Find all solutions to the system of equations
            <md>
            <mrow>x+2y-z \amp=2</mrow>
            <mrow>x+y-z\amp=0</mrow>
            <mrow>2x-y+z\amp=3</mrow>
            </md>
          </p>
          </statement>
          <solution>
          <p>We put the augemented matrix into row echelon form:
            <me>
            \begin{bmatrix}
            1 \amp 2 \amp -1 \amp 2\\
            1 \amp 1 \amp -1 \amp 0\\
            2 \amp -1 \amp 1 \amp 3\\
            \end{bmatrix}
            \\
            \begin{array}{l}
            R_2\gets R_2 - R_1\\
            R_3\gets R_3-2R_1
            \end{array}
            \\
            \begin{bmatrix}
            1 \amp 2 \amp -1 \amp 2\\
            0 \amp -1 \amp 0 \amp -2\\
            0 \amp -5 \amp 3 \amp -1\\
            \end{bmatrix}
            \\
            \begin{array}{l}
            R_2\gets -R_2\\
            \end{array}
            \\
            \begin{bmatrix}
            1 \amp 2 \amp -1 \amp 2\\
            0 \amp 1 \amp 0 \amp 2\\
            0 \amp -5 \amp 3 \amp -1\\
            \end{bmatrix}
            \\
            R_3\gets R_3+5R_2
            \\
            \begin{bmatrix}
            1 \amp 2 \amp -1 \amp 2\\
            0 \amp 1 \amp 0 \amp 2\\
            0 \amp 0 \amp 3 \amp 9\\
            \end{bmatrix}
            \\
            R_3\gets \frac13 R_3
            \\
            \begin{bmatrix}
            1 \amp 2 \amp -1 \amp 2\\
            0 \amp 1 \amp 0 \amp 2\\
            0 \amp 0 \amp 1 \amp 3\\
            \end{bmatrix}
            </me>
            The last row gives <m>z=3</m>. The second row gives <m>y=2</m>. The first row gives
            <m>x=2-2y+z=2-4+3=1</m>, and so there is a unique solution <m>(x,y,z)=(1,2,3)</m>.
          </p>
          </solution>
          </exercise>
    
    
      </subsection>
    </section>
    
    
    <section><title>Gauss-Jordan reduction</title>
    <introduction>
     <p>Gauss-Jordan reduction is an extension of the Gaussian elimination algorithm.
      It produces a matrix, called <em>the reduced row echelon form</em> in the following
      way: after carrying out Gaussian elimination, continue by changing
      all nonzero entries <em>above</em> the leading ones to a zero. The resulting
      matrix looks something like:
      </p>
      
        
        <figure>
        <caption/>
        <image source="images/rref.png" width="50%">
        <description>A matrix in row echelon form</description>
        </image>
        </figure>
        
        <p>
        The matrix above gives an idea of what we want.
        Notice the staircase line drawn through the matrix
        has all entries below it equal to zero.
        The entries marked with a <m>*</m> can take on any value.
        </p>

        <definition><title>Reduced row echelon form</title>
        <statement>
        <p>A matrix is in <term>reduced row echelon form</term>  if
        <ul>
        <li><p>Every leading entry is a leading one. </p></li>
        <li><p>Every entry below <em>and above</em> a leading one is <m>0</m>. </p></li>
        <li><p>As you go down the matrix, the leading ones move to the right.</p></li>
        <li><p>Any all zero rows are at the bottom.</p></li>
        </ul></p>
        </statement>
        </definition>

      <p>

      For completeness, we'll describe Gauss-Jordan reduction algorithm in detail. As with Gaussian 
      elimination, the columns of the matrix are processed from left to right.
      
      <ol>
      <li><p>Start with the first column. If it has all entries equal to zero, 
      move on to the next column to the right.</p></li>
      <li><p>If, to the contrary, the column has nonzero entries, interchange rows, 
      if necessary, to get a nonzero entry on top.</p></li>
      <li><p>Multiply the top row by a constant to change the nonzero entry to a (leading) one.</p></li>
      <li><p>If there are nonzero entries above or below this (leading) one, 
      use an elementary row operation on each one to change it to a zero.</p></li>
      <li><p>Now consider the part of the matrix below the top row and to the right of the column 
      under consideration: if there are no such rows or columns, 
      stop and the algorithm is finished. Otherwise, carry out the same procedure on the 
      new matrix.</p></li>
      </ol>
      </p>
      
      <example><title>Putting a matrix in reduced row echelon form</title>
      <p>
      <m>
      \begin{align}
      \amp \begin{bmatrix}
      1 \amp  2 \amp  3 \\
      4 \amp  5 \amp  6 \\
      7 \amp  8 \amp  9 \\
      10 \amp  12 \amp  15 
      \end{bmatrix} \amp \amp 
      \begin{array}{l}
      R_2\gets R_2-4R_1\\
      R_3\gets R_3-7R_1\\
      R_4\gets R_4-10R_1
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  3 \\
      0 \amp  -3 \amp  -6 \\
      0 \amp  -6 \amp  -12 \\
      0 \amp  -8 \amp  -15
      \end{bmatrix} \amp \amp 
      \begin{array}{l}
      R_2\gets -\tfrac13 R_2
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  3 \\
      0 \amp  1 \amp  2 \\
      0 \amp  -6 \amp  -12 \\
      0 \amp  -8 \amp  -15
      \end{bmatrix} \amp \amp 
      \begin{array}{l}
      R_1\gets R_1-2R_2\\
      R_3\gets R_3+6R_2\\
      R_4\gets R_4+8R_2
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  0 \amp  -1 \\
      0 \amp  1 \amp  2 \\
      0 \amp  0 \amp  0 \\
      0 \amp  0 \amp  1
      \end{bmatrix} \amp \amp 
      \begin{array}{l}
      R_3\leftrightarrow R_4 \\
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  0 \amp  -1 \\
      0 \amp  1 \amp  2 \\
      0 \amp  0 \amp  1 \\
      0 \amp  0 \amp  0
      \end{bmatrix} \amp \amp 
      \begin{array}{l}
      R_1 \gets R_1+R_3 \\
      R_2\gets R_2-2R_3
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  0 \amp  0 \\
      0 \amp  1 \amp  0 \\
      0 \amp  0 \amp  1 \\
      0 \amp  0 \amp  0
      \end{bmatrix}
      \end{align} 
      </m>
      </p>
      </example>
     
     <example><title>Putting another matrix in reduced row echelon form</title>
     <p>
     <m>
      \begin{align}
      \amp \begin{bmatrix}
      1 \amp  2 \amp  6 \amp  1 \amp  4 \amp  6 \\
      2 \amp  4 \amp  9 \amp  2 \amp  8 \amp  9 \\
      1 \amp  2 \amp  9 \amp  2 \amp  10 \amp  9
      \end{bmatrix}\amp \amp 
      \begin{array}{c}
      R_2 \gets R_2-2R_1 \\
      R_3 \gets R_3-R_1
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  6 \amp  1 \amp  4 \amp  6 \\
      0 \amp  0 \amp  -3 \amp  0 \amp  0 \amp  -3 \\
      0 \amp  0 \amp  3 \amp  1 \amp  6 \amp  3
      \end{bmatrix}\amp \amp 
      \begin{array}{c}
      R_2 \gets -\tfrac13R_2
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  6 \amp  1 \amp  4 \amp  6 \\
      0 \amp  0 \amp  1 \amp  0 \amp  0 \amp  1 \\
      0 \amp  0 \amp  3 \amp  1 \amp  6 \amp  3
      \end{bmatrix}\amp \amp 
      \begin{array}{c}
      R_1 \gets R_1 -6R_2\\
      R_3 \gets R_3 -3R_2
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  0 \amp  1 \amp  4 \amp  0 \\
      0 \amp  0 \amp  1 \amp  0 \amp  0 \amp  1 \\
      0 \amp  0 \amp  0 \amp  1 \amp  6 \amp  0
      \end{bmatrix}\amp \amp 
      \begin{array}{c}
      R_1 \gets R_1 - R_3
      \end{array}\\[6pt]

      \amp \begin{bmatrix}
      1 \amp  2 \amp  0 \amp  0 \amp  -2 \amp  0 \\
      0 \amp  0 \amp  1 \amp  0 \amp  0 \amp  1 \\
      0 \amp  0 \amp  0 \amp  1 \amp  6 \amp  0
      \end{bmatrix}
      \end{align}
     </m>
     </p> 
     </example>
      
      <p>
      Notice the pattern of zero, 1 and nonzero entries after Gauss-Jordan reduction
      (with the leading ones in red):
      <me>
      \left[\begin{array}{ccccccc}
      {{\color{red}1}} \amp  0 \amp  0 \amp  * \amp  0 \amp  \cdots \amp  *\\
      0 \amp  {\color{red}1} \amp  0 \amp  * \amp  0 \amp  \cdots \amp  *\\
      0 \amp  0 \amp  {\color{red}1}\amp  * \amp  0 \amp  \cdots \amp  *\\
      0 \amp  0 \amp  0 \amp  0 \amp  {\color{red}1} \amp  \cdots \amp  *\\
      \end{array}\right]
      </me>

      Now that we can put a matrix in reduced row echelon form, let's see what this implies for 
      finding all solutions to the associated system of linear equations. Remember that the 
      first <m>n</m> columns correspond to coefficients of variables 
      <m>x_1,x_2,\dots,x_n</m>, and the last column corresponds to the constants on
      the right side of the equations. Each column either contains a leading one or it does not.
      If it does, the corresponding variable is called <term>constrained</term> or
      <term>basic</term>. 
      If not the variable is called <term>free.</term> 
      Each free variable is assigned a parameter which may take on any number
      when finding solutions. The values of the constrained variables are then determined by
      the reduced row echelon form.
      </p>
      
      <p>
     As an example, suppose we have a system of linear equations whose augmented matrix has
      the following reduced row echelon form:
      
      <me>
      \begin{bmatrix}
      {\color{red}1} \amp  2 \amp  0 \amp  1 \amp  0 \amp  1\\
      0 \amp  0 \amp   {\color{red}1}\amp  2 \amp  0 \amp  2\\
      0 \amp  0 \amp  0 \amp  0 \amp  {\color{red}1} \amp  3
      \end{bmatrix}
      </me>
      
      Notice that this means that our system has three equations and five unknowns.
      The leading ones are in columns one, three and five, so <m>x_1</m>, 
      <m>x_3</m> and <m>x_5</m> are constrained variables. This leaves
      <m>x_2</m> and <m>x_4</m> as free variables. We assign parameters
      to the free variables: <m>x_2=s</m> and <m>x_4=t</m>. The rows of
      the matrix then determine the constrained variables: 
      
     <ul>
     <li><p><m>x_5=3</m> from the bottom row</p></li> 
     <li><p><m>x_3 = 2-2t</m> from the middle row</p></li> 
     <li><p><m>x_1=1-2s-t</m> from the top row</p></li> 
     </ul>
     </p>
      
     <p>
      The compact way of writing this is 
      <m>(x_1,x_2,x_3,x_4,x_5) = (1-2s-t,s,2-2t,t,3).</m>
     </p>
      
     <p>
     Notice the role of the zeros above and below each leading one: the evaluation of a
      constrained variable involves only free variables.
     </p>
      
     <p>
      In summary, we can say the following:
     <ol>
     <li><p>
         If the reduced row echelon form has a row of the form <m>[0,0,\dots,0,1]</m>, 
        then the system of linear equations has no solution.</p></li>
     <li><p>
        If the reduced row echelon form has no free variables, then it looks like:
        <me>
        \begin{bmatrix}
        1 \amp  0 \amp  0 \amp \cdots \amp 0 \amp 0 \amp c_1\\
        0 \amp  1 \amp  0 \amp \cdots \amp 0 \amp 0 \amp c_2\\
        0 \amp  0 \amp  1 \amp \cdots \amp 0 \amp 0 \amp c_3\\
            \amp    \amp    \amp \vdots \\
        0 \amp  0 \amp  0 \amp \cdots \amp 1 \amp 0 \amp c_{n-1}\\
        0 \amp  0 \amp  0 \amp \cdots \amp 0 \amp 1 \amp c_n
        \end{bmatrix}
        </me>
        and there is a unique solution, namely, <m>x_1=c_1, x_2=c_2,\dots x_n=c_n</m>.
     </p></li>
     <li><p>
        If the reduced row echelon form has free variables, then there are an infinite 
        number of solutions. Indeed, the parameter assigned to any one free variable can take 
        on an infinite number of values.
     </p></li>
     </ol> 
      
      Consider the following system of linear equations:
      <me>
      \begin{array}{rl}
      x_1-x_2+2x_3-x_4 \amp =  -1 \\
      2x_1+x_2-2x_3-2x_4 \amp =  -2 \\
      -x_1+2x_2-4x_3+x_4 \amp =  1 \\
      3x_1-3x_4 \amp =  -3
      \end{array}
      </me>


    The augmented matrix is then put into reduced row echelon form:

      <me>\begin{bmatrix}
      1 \amp  -1 \amp  2 \amp  -1 \amp  -1\\
      2 \amp  1 \amp -2 \amp  -2 \amp  -2\\
      -1 \amp  2 \amp  -4 \amp  1 \amp  1\\
      3 \amp  0 \amp  0 \amp  -3 \amp  -3
      \end{bmatrix}</me>

      <me>\begin{array}{rl}
      R_2 \amp \gets R_2-2R_1\\
      R_3 \amp \gets R_3+R_1\\
      R_4 \amp \gets R_4-3R_1
      \end{array}</me>
      </p>
      
      <p>
     <me>\begin{bmatrix}
      1 \amp  -1 \amp  2 \amp  -1 \amp  -1\\
      0 \amp  3 \amp  -6 \amp  0 \amp  0\\
      0 \amp  1 \amp  -2 \amp  0 \amp  0\\
      0 \amp  3 \amp  -6 \amp  0 \amp  0
      \end{bmatrix}</me>

      <me>\begin{array}{rl}
      R_2 \amp \gets \tfrac13 R_2\\
      \end{array}</me>

      <me>\begin{bmatrix}
      1 \amp  -1 \amp  2 \amp  -1 \amp  -1\\
      0 \amp  1 \amp  -2 \amp  0 \amp  0\\
      0 \amp  1 \amp  -2 \amp  0 \amp  0\\
      0 \amp  3 \amp  -6 \amp  0 \amp  0
      \end{bmatrix}</me>
      
      <me>\begin{array}{rl}
      R_1 \amp \gets R_1 + R_2\\
      R_3 \amp \gets R_3- R_2\\
      R_4 \amp \gets R_4- 3R_2
      \end{array}</me> 
      
      <me>\begin{bmatrix}
      {\color{red} 1} \amp  0 \amp  0 \amp  -1 \amp  -1\\
      0 \amp  {\color{red} 1} \amp  -2 \amp  0 \amp  0\\
      0 \amp  0 \amp  0 \amp  0 \amp  0\\
      0 \amp  0 \amp  0 \amp  0 \amp  0
      \end{bmatrix}</me>
      </p>
      
      <p>
     Since  <m>x_3</m> and <m>x_4</m> are free variables,
      we assign them parameters: <m>x_3=s</m> and <m>x_4=t</m>.  We can 
      now evaluate the constrained variables:
      <m>x_1=-1+t</m> (from the first row) and <m>x_2=2s</m> (from the second row).
      In short, <m>(x_1,x_2,x_3,x_4)=(-1+t,2s,s,t)</m> for any choice of <m>s</m> or <m>t</m>. 
      </p>
      
     <p>
       Now, as a check, let's put these solutions back into the original equations.

      <m>\begin{array}{rlll}
      x_1-x_2+2x_3-x_4 \amp =\amp  (-1+t) - (2s) + 2(s) - (t) \amp= -1\\
      2x_1+x_2-2x_3-2x_4 \amp =\amp  2(-1+t) + 2s -2s-2t \amp= -2\\
      -x_1+2x_2-4x_3+x_4 \amp =\amp  -(-1+t) +2(2s) -4s+t \amp= 1 \\
      3x_1-3x_4 \amp =\amp  3(-1+t) -3t \amp= -3
      \end{array}</m> 
     </p>

    <example>
    <title>A matrix row-reducing itself</title>
        <p>
        The following example is a matrix row reducing itself. Notice how it
        proceeds left to right in columns, and each column with a leading one
        has zeros above and below when the algorithm completes.
        </p>
    </example>

    <sidebyside>
    <image width="100%" source="images/rref.gif" />
    </sidebyside>
    
</introduction>

    <subsection><title>The rank of a matrix</title>
    <definition>
    <statement>
    <p>
        The <term>rank</term> of a matrix is the number of leading ones in the reduced row
        echelon form.
    </p>
    </statement>
    </definition>

    <p>
    Since the matrix
    <me>A=
    \begin{bmatrix}
    1\amp 2\amp 3\amp 4\amp 5 \\
    6\amp 7\amp 8\amp 9\amp 10 \\
    11\amp 12\amp 13\amp 14\amp 15 \\
    \end{bmatrix}
    </me>
    has a reduced row echelon form of
    <me>
    \begin{bmatrix}
    1\amp 0\amp -1\amp -2\amp -3 \\
    0\amp 1\amp 2\amp 3\amp 4 \\
    0\amp 0\amp 0\amp 0\amp 0 \\
    \end{bmatrix}
    </me>
    the rank of <m>A</m> is <m>2</m>.
    </p>

    </subsection>

      <exercises>

      <exercisegroup>
      <introduction><p>Put the following matrices into reduced row echelon
      form.</p></introduction>
       
            <exercise> <statement> <p>
            <me>
            \begin{bmatrix}
            2\amp4\amp6\amp8\\
            3\amp6\amp9\amp10
            \end{bmatrix}
            </me>
            </p> </statement> <solution> <p> 
            <me>
            \begin{bmatrix}
            2\amp4\amp6\amp8\\
            3\amp6\amp9\amp10
            \end{bmatrix}\\
            \rowmul1{\frac12}\\
            \begin{bmatrix}
            1\amp2\amp3\amp4\\
            3\amp6\amp9\amp10
            \end{bmatrix}\\
            \rowsub231\\
            \begin{bmatrix}
            1\amp2\amp3\amp4\\
            0\amp0\amp0\amp-2
            \end{bmatrix}\\
            \rowmul 2{-\frac12}\\
            \begin{bmatrix}
            1\amp2\amp3\amp4\\
            0\amp0\amp0\amp1
            \end{bmatrix}\\
            \rowsub142\\
            \begin{bmatrix}
            1\amp2\amp3\amp0\\
            0\amp0\amp0\amp1
            \end{bmatrix}
            </me>
            </p></solution> </exercise>
        
            <exercise> <statement> <p>
            <me>
            \begin{bmatrix}
            0 \amp  1 \amp -1\\
            1 \amp  0 \amp -1\\
            1 \amp -1 \amp  0
            \end{bmatrix}
            </me>
            </p> </statement> <solution> <p> 
            <me>
            \begin{bmatrix}
            0 \amp  1 \amp -1\\
            1 \amp  0 \amp -1\\
            1 \amp -1 \amp  0
            \end{bmatrix}\\
            \rowint 12\\
            \begin{bmatrix}
            1 \amp  0 \amp -1\\
            0 \amp  1 \amp -1\\
            1 \amp -1 \amp  0
            \end{bmatrix}\\
            \rowsub 3{}1\\
            \begin{bmatrix}
            1 \amp  0 \amp -1\\
            0 \amp  1 \amp -1\\
            0 \amp -1 \amp  1
            \end{bmatrix}\\
            \rowadd3{}2\\
            \begin{bmatrix}
            1 \amp  0 \amp -1\\
            0 \amp  1 \amp -1\\
            0 \amp  0 \amp  0
            \end{bmatrix}\\
            </me>
            </p></solution> </exercise>
            
            <exercise> <statement> <p>
            <me>\begin{bmatrix}
            1\amp 2\amp3\\
            4\amp5\amp6\\
            7\amp8\amp9
            \end{bmatrix}</me>
            </p> </statement> <solution> <p> 
            <me>\begin{bmatrix}
            1\amp 2\amp3\\
            4\amp5\amp6\\
            7\amp8\amp9
            \end{bmatrix}\\
            \rowsub241\\
            \rowsub371\\
            \begin{bmatrix}
            1\amp 2\amp3\\
            0\amp-3\amp-6\\
            0\amp-6\amp-12
            \end{bmatrix}\\
            \rowmul2{-\frac13}\\
            \begin{bmatrix}
            1\amp 2\amp3\\
            0\amp1\amp2\\
            0\amp-6\amp-12
            \end{bmatrix}\\
            \rowsub 122\\
            \begin{bmatrix}
            1\amp 0\amp-1\\
            0\amp1\amp2\\
            0\amp-6\amp-12
            \end{bmatrix}\\
            \rowadd 362\\
            \begin{bmatrix}
            1\amp 0\amp-1\\
            0\amp1\amp2\\
            0\amp0\amp0
            \end{bmatrix}</me>
            </p></solution> </exercise>
        
            <exercise> <statement> <p>
            <me>
            \begin{bmatrix}
            1\amp2\\ 1\amp3 \\1\amp4
            \end{bmatrix}
            </me>
            </p> </statement> <solution> <p> 
            <me>
            \begin{bmatrix}
            1\amp2\\ 1\amp3 \\1\amp4
            \end{bmatrix}\\
            \rowsub 2 {} 1\\
            \rowsub 3 {} 1\\
            \begin{bmatrix}
            1\amp2\\ 0\amp1 \\0\amp2
            \end{bmatrix}\\
            \rowsub 1 2 2\\
            \rowsub 3 2 2\\
            \begin{bmatrix}
            1\amp0\\ 0\amp1 \\0\amp0
            \end{bmatrix}\\
            </me>
            </p></solution> </exercise>
        
            <exercise> <statement> <p>
            <me>
            \begin{bmatrix}
            0\amp 1\amp 2\amp 3\amp 4 \\
            4\amp 3\amp 2\amp 1\amp 0 
            \end{bmatrix}
            </me>
            </p> </statement> <solution> <p> 
            <me>
            \begin{bmatrix}
            0\amp 1\amp 2\amp 3\amp 4 \\
            4\amp 3\amp 2\amp 1\amp 0 
            \end{bmatrix}\\
            \rowint12\\
            \begin{bmatrix}
            4\amp 3\amp 2\amp 1\amp 0 \\
            0\amp 1\amp 2\amp 3\amp 4 
            \end{bmatrix}\\
            \rowmul1{\frac14}\\
            \begin{bmatrix}
            1\amp \frac34\amp \frac12\amp \frac14\amp 0 \\
            0\amp 1\amp 2\amp 3\amp 4 
            \end{bmatrix}\\
            \rowsub 1{\frac34}2\\
            \begin{bmatrix}
            1\amp 0\amp -1\amp -2\amp -3 \\
            0\amp 1\amp 2\amp 3\amp 4 
            \end{bmatrix}
            </me>
            </p></solution> </exercise>
            </exercisegroup>

            <exercisegroup>
                <introduction>
                <p>Consider the system of four linear equations in four unknowns:
                <md>
                <mrow>a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1</mrow>
                <mrow>a_{2,1}x_1+a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2</mrow>
                <mrow>a_{3,1}x_1+a_{2,2}x_2+a_{3,3}x_3+a_{3,4}x_4=b_3</mrow>
                <mrow>a_{4,1}x_1+a_{4,2}x_2+a_{4,3}x_3+a_{4,4}x_4=b_4</mrow>
                </md>
                In each case, suppose that the given matrix is the reduced
                row echelon form of the augmented matrix. Find all of the solutions.
                to the original system of linear equations.
                </p>
                </introduction>
            <exercise>
            <statement>
            <p>
            <m>
            \begin{bmatrix}
            1\amp 0 \amp 0 \amp 0 \amp 1 \\
            0\amp 1 \amp 0 \amp 0 \amp -1 \\
            0\amp 0 \amp 1 \amp 0 \amp 2 \\
            0\amp 0 \amp 0 \amp 1 \amp 3 \\
            \end{bmatrix}
            </m>
            </p></statement>
            <solution><p> 
            <m>x_1=1</m>, <m>x_2=-1</m>, <m>x_3=2</m> and <m>x_4=3</m> is the unique solution. 
            </p></solution>
            </exercise>
            <exercise>
            <statement>
            <p>
            <m>
            \begin{bmatrix}
            1\amp 0 \amp 0 \amp 0 \amp 1 \\
            0\amp 1 \amp 0 \amp 0 \amp -1 \\
            0\amp 0 \amp 1 \amp 3 \amp 2 \\
            0\amp 0 \amp 0 \amp 0 \amp 1 \\
            \end{bmatrix}
            </m>
            </p>
            </statement>
            <solution>
            <p>
            The last line satisfies the condition for there to be no solution.
            </p>
            </solution>
            </exercise>
        
            <exercise> <statement> <p>
            <m>
            \begin{bmatrix}
            1\amp 0 \amp 0 \amp 1 \amp 1 \\
            0\amp 1 \amp 0 \amp -1 \amp -1 \\
            0\amp 0 \amp 1 \amp 3 \amp 2 \\
            0\amp 0 \amp 0 \amp 0 \amp 0 \\
            \end{bmatrix}
            </m>
            </p> </statement> 
            <solution> 
            <p><m>x_4=t</m> since it is a free variable.</p> 
            <p>From row 1, we have <m>x_1=1-t</m></p> 
            <p>From row 2, we have <m>x_2=-1+t</m></p> 
            <p>From row 3, we have <m>x_3=2-3t</m></p> 
            <p>
            In other words, <m>(x_1,x_2,x_3,x_4)=(1-t,-1+t,2-3t,t)</m>.
            </p>
            </solution></exercise>
        
            <exercise> <statement> <p>
            <m>
            \begin{bmatrix}
            1\amp -2 \amp 0 \amp 3 \amp -1 \\
            0\amp 0 \amp 1 \amp 1 \amp 2 \\
            0\amp 0 \amp 0 \amp 0 \amp 0 \\
            0\amp 0 \amp 0 \amp 0 \amp 0 \\
            \end{bmatrix}
            </m>
            </p> </statement> 
            <solution> 
            <p><m>x_2=t</m> and <m>x_4=u</m> since they are free variables.</p> 
            <p>From row 1, we have <m>x_1=-1+2t-3u</m></p> 
            <p>From row 2, we have <m>x_3=2-u</m></p> 
            <p>
            In other words, <m>(x_1,x_2,x_3,x_4)=(-1+2t-3u,t,2-u,u)</m>.
            </p>
            </solution></exercise>
            </exercisegroup>

       <exercisegroup>
       <introduction>
        <p>Find <em>all</em> solutions to the following 
        systems of linear equations
        </p>
       </introduction>
      <exercise>
      <statement>
      <p>
        <m>
        \begin{align}
        x+2y-z \amp=2\\
        x+y-z\amp=0\\
        2x-y+z\amp=3
        \end{align}
        </m>
      </p>
      </statement>
      <solution>
      <p>We put the augemented matrix into reduced row echelon form:
        <m>
        \begin{bmatrix}
        1 \amp 2 \amp -1 \amp 2\\
        1 \amp 1 \amp -1 \amp 0\\
        2 \amp -1 \amp 1 \amp 3\\
        \end{bmatrix}
        \\
        \begin{array}{l}
        R_2\gets R_2 - R_1\\
        R_3\gets R_3-2R_1
        \end{array}
        \\
        \begin{bmatrix}
        1 \amp 2 \amp -1 \amp 2\\
        0 \amp -1 \amp 0 \amp -2\\
        0 \amp -5 \amp 3 \amp -1\\
        \end{bmatrix}
        \\
        \begin{array}{l}
        R_2\gets -R_2\\
        \end{array}
        \\
        \begin{bmatrix}
        1 \amp 2 \amp -1 \amp 2\\
        0 \amp 1 \amp 0 \amp 2\\
        0 \amp -5 \amp 3 \amp -1\\
        \end{bmatrix}
        \\
        R_3\gets R_3-5R_2
        \\
        \begin{bmatrix}
        1 \amp 2 \amp -1 \amp 2\\
        0 \amp 1 \amp 0 \amp 2\\
        0 \amp 0 \amp 3 \amp 9\\
        \end{bmatrix}
        \\
        R_3\gets \frac13 R_3
        \\
        \begin{bmatrix}
        1 \amp 2 \amp -1 \amp 2\\
        0 \amp 1 \amp 0 \amp 2\\
        0 \amp 0 \amp 1 \amp 3\\
        \end{bmatrix} \\
        R_1 \gets R_1-2R_2
        \\
        \begin{bmatrix}
        1 \amp 0 \amp -1 \amp-2\\
        0 \amp 1 \amp 0 \amp 2\\
        0 \amp 0 \amp 1 \amp 3\\
        \end{bmatrix} \\
        R_1 \gets R_1+R_3
        \\
        \begin{bmatrix}
        1 \amp 0 \amp 0 \amp1\\
        0 \amp 1 \amp 0 \amp 2\\
        0 \amp 0 \amp 1 \amp 3\\
        \end{bmatrix} \\
        </m>
        The first, second and third rows give us, respectively,
        <me>
        \begin{matrix}
        x=1\\y=2\\z=3
        \end{matrix}
        </me>
      </p>
      </solution>
      </exercise>

      <exercise> <statement> <p>
      <m>
      \begin{align*}
      x+2y+z=0\\
      x+y+2z=5\\
      -x+y+z=0
      \end{align*}
      </m>
      </p> </statement> <solution> <p> 
      <m>
      \begin{bmatrix}
      1 \amp 2 \amp 1 \amp 0\\
      1 \amp 1 \amp 2 \amp 5\\
      -1 \amp 1 \amp 1 \amp 0
      \end{bmatrix}\\
      \rowsub 211\\
      \rowadd 3{}1\\
      \begin{bmatrix}
      1 \amp 2 \amp 1 \amp 0\\
      0 \amp -1 \amp 1 \amp 5\\
      0 \amp 3 \amp 2 \amp 0
      \end{bmatrix}\\
      \rowmul 2{-}\\
      \begin{bmatrix}
      1 \amp 2 \amp 1 \amp 0\\
      0 \amp 1 \amp -1 \amp -5\\
      0 \amp 3 \amp 2 \amp 0
      \end{bmatrix}\\
      \rowsub122\\
      \rowsub332\\
      \begin{bmatrix}
      1 \amp 0 \amp 3 \amp 10\\
      0 \amp 1 \amp -1 \amp -5\\
      0 \amp 0 \amp 5 \amp 15
      \end{bmatrix}\\
      \rowmul3{\frac15}\\
      \begin{bmatrix}
      1 \amp 0 \amp 3 \amp 10\\
      0 \amp 1 \amp -1 \amp -5\\
      0 \amp 0 \amp 1 \amp 3
      \end{bmatrix}\\
      \rowsub 133\\
      \rowadd 3{}2\\
      \begin{bmatrix}
      1 \amp 0 \amp 0 \amp 1\\
      0 \amp 1 \amp 0 \amp -2\\
      0 \amp 0 \amp 1 \amp 3
      \end{bmatrix}\\
      </m>
      and so the unique solution is <m>(x,y,z)=(1,-2,3)</m>. 
      </p>
      </solution> 
      </exercise>

      <exercise> <statement> <p>
      <m>
      \begin{align}
      x+y-z-w=2\\
      x-y-z+w=4\\
      2x+y-z+w=1
      \end{align}
      </m>
      </p> </statement> <solution> <p> 
      <m>
      \begin{bmatrix} 
      1 \amp 1 \amp -1 \amp -1 \amp 2\\
      1 \amp -1 \amp -1 \amp 1 \amp 4\\
      2 \amp 1 \amp -1 \amp 1 \amp 1
      \end{bmatrix} \\
      \rowsub 2{}1\\
      \rowsub 321\\
      \begin{bmatrix} 
      1 \amp 1 \amp -1 \amp -1 \amp 2\\
      0 \amp -2 \amp 0 \amp 2 \amp 2\\
      0 \amp -1 \amp 1 \amp 3 \amp -3
      \end{bmatrix} \\
      \rowmul2{-\frac12}\\
      \begin{bmatrix} 
      1 \amp 1 \amp -1 \amp -1 \amp 2\\
      0 \amp 1 \amp 0 \amp -1 \amp -1\\
      0 \amp -1 \amp 1 \amp 3 \amp -3
      \end{bmatrix} \\
      \rowsub1{}2\\
      \rowadd3{}2\\
      \begin{bmatrix} 
      1 \amp 0 \amp -1 \amp 0 \amp 3\\
      0 \amp 1 \amp 0 \amp -1 \amp -1\\
      0 \amp 0 \amp 1 \amp 2 \amp -4
      \end{bmatrix} \\
      \rowadd1{}3\\
      \begin{bmatrix} 
      1 \amp 0 \amp 0 \amp 2 \amp -1\\
      0 \amp 1 \amp 0 \amp -1 \amp -1\\
      0 \amp 0 \amp 1 \amp 2 \amp -4
      \end{bmatrix} 
      </m>
      and so <m>w</m> is a free variable that can be set equal to <m>t</m>. This gives
      <m>(x,y,z,w)=(-1-2t, -1+t, -4-2t,t)</m>.
      </p></solution> </exercise>
  
      <exercise> <statement> <p>
      <m>
      \begin{align*}
      x-y=0\\
      y-z=0\\
      x-z=0\\
      \end{align*}
      </m>
      Clearly, if <m>x=y</m> and <m>y=z</m>, then we have a solution
      to this system. Show that every solution is of this form.
      </p> </statement> <solution> <p> 
      We put the augmented matrix in reduced row echelon form:
      <m>
      \begin{bmatrix}
      1 \amp -1 \amp 0 \amp 0\\
      0 \amp  1 \amp -1 \amp 0\\
      1 \amp  0 \amp -1 \amp 0\\
      \end{bmatrix}\\
      \rowsub3{}1\\
      \begin{bmatrix}
      1 \amp -1 \amp 0 \amp 0\\
      0 \amp  1 \amp -1 \amp 0\\
      0 \amp  1\amp -1 \amp 0\\
      \end{bmatrix}\\
      \rowadd1{}2\\
      \begin{bmatrix}
      1 \amp 0\amp -1 \amp 0\\
      0 \amp  1 \amp -1 \amp 0\\
      0 \amp  1\amp -1 \amp 0\\
      \end{bmatrix}\\
      \rowsub3{}2\\
      \begin{bmatrix}
      1 \amp  0 \amp -1 \amp 0\\
      0 \amp  1 \amp -1 \amp 0\\
      0 \amp  0\amp 0 \amp 0\\
      \end{bmatrix}
      </m>
      We then have that <m>z</m> is a free variable that may be set to <m>t</m>.
      All solutions are then of the form <m>(x,y,z)=(t,t,t)</m>, which
      implies that <m>x=y</m> and <m>y=z</m>.
      </p></solution> </exercise>
        
       </exercisegroup> 
  
      <exercise> <statement> <p>
      Consider the two sets of linear equations:
      </p>
      <p>
      <m>
      \begin{array}{rl}
      3x+y-2z=2\\
      -x+3y-z=1
      \end{array}
      </m>
      and 
      <m>
      \begin{array}{rl}
     x+y-z=1\\
     -3x+y+z=-1
      \end{array}
      </m>
      </p>
      <p>
      Show that every solution of the first set is a solution of
      the second set, and vice-versa.
      </p> </statement> 
      <hint><p>Compare the RREF of the corresponding augmented matrices. </p></hint>  
      
      <solution> <p> 
      Each system has a RREF of
      \begin{bmatrix}
      1 \amp 0 \amp -\frac12 \amp \frac12\\
      0 \amp 1 \amp -\frac12 \amp \frac12\\
      \end{bmatrix}
      and so all the solutions for either system consist of
      <m>(x,y,z)=(\frac12+\frac12t,\frac12+\frac12t,t)</m> where <m>t</m> is any
      real number.
      </p></solution> </exercise>
  
        </exercises>

    </section>


    
    <section><title>Geometric examples of three equations in three unknowns</title>
      
      <p>
      A single equation in three unknowns can be interpreted geometrically
      in 3-dimensional space. An equation
      <me>
      Ax+By+Cz=D
      </me>
      has solutions that form a plane as long as at least one of <m>A, B, C</m> is
      nonzero.
      If we look at a system of such linear equations, the solution set is
      the set of all points lying in all of the corresponding planes. 
      </p>
      
      
      <example><title>Three planes intersecting at a point</title>
      <p>
      Consider the system of linear equations
      <me>
      \begin{array}{rl}
      x+y+z \amp= 2\\
      x+2y+z\amp=3\\
      x+2y-3z \amp=2
      \end{array}
      </me>
      The augmented matrix is
      
      <me>
      \begin{bmatrix}
      1\amp1\amp1\amp2\\
      1\amp2\amp1\amp3\\
      1\amp2\amp-3\amp2
      \end{bmatrix}
      </me>
      whose reduced row echelon form is 
      <me>
      \begin{bmatrix}
      1\amp0\amp0\amp\frac34\\
      0\amp1\amp0\amp1\\
      0\amp0\amp1\amp\frac14
      \end{bmatrix}
      </me>
      
      This means that there is a single solution: <m>(x,y,z)=(\frac34,1,\frac14)</m>.
      Here are the three planes corresponding to the three equations:
      </p>
     <sidebyside>
     <image width="90%" source="images/ThreePlanesPoint.png" />
     </sidebyside>
      <!--
      <image width="100%" xml:id="ThreePlanesPoint">
      <asymptote>
      import graph3;
      size(500);
      currentprojection=orthographic(4,4,1);
      currentlight=White;
      
      real A=1, B=1, C=1, D=2;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(0,0),(2,2)),lightblue,nolight);
      label("\(x+y+z=2\)",(1,4.2,0.5),blue);
      
      real A=1, B=2, C=1, D=3;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(0,0),(2,2)),lightgrey,nolight);
      label("\(x+2y+z=3\)",(1,3.1,1.8),grey);
      
      real A=1, B=2, C=-3, D=2;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(0,0),(2,2)),lightgreen,nolight);
      label("\(x+2y-3z=2\)",(1,4.2,1.3),green);
      
      dot ((3/4,1,1/4),red);
      axes3("\(x\)","\(y\)","\(z\)",(-2,-2,-2),(3,3,3));
      </asymptote>
      </image>
      -->
      </example>
      
      <example><title>Three planes intersecting in a single line</title>
      <p>
      Consider the system of linear equations
      <me>
      \begin{array}{rl}
      x+y-z \amp= 3\\
      x+y+z\amp=1\\
      2x+2y \amp=4
      \end{array}
      </me>
      The augmented matrix is
      
      <me>
      \begin{bmatrix}
      1\amp1\amp-1\amp3\\
      1\amp1\amp1\amp1\\
      2\amp2\amp0\amp4
      \end{bmatrix}
      </me>
      whose reduced row echelon form is
      <me>
      \begin{bmatrix}
      1\amp1\amp0\amp2\\
      0\amp0\amp1\amp-1\\
      0\amp0\amp0\amp0
      \end{bmatrix}
      </me>
      This means <m>y=t</m>, <m>x=2-t</m> and <m>z=-1</m> 
      so that <m>(x,y,z)=(2-t,t,-1)</m> is a solution
      for any real number <m>t</m>.
      </p>
      <!-- 
      <image width="70%" xml:id="ThreePlanesLine">
      <asymptote>
      import graph3;
      
      size(400);
      
      currentprojection=orthographic(-3,6,1);
      currentlight=White;
      
      
      real A=1, B=1, C=-1, D=3;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(-5,-5),(5,5)),lightblue,nolight);
      label("\(x+y-z=3\)",(1,12,9),blue);
      
      real A=1, B=1, C=1, D=1;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(-5,-5),(5,5)),lightgrey,nolight);
      label("\(x+y+z=1\)",(-2,0,11),grey);
      
      real A=2, B=2, C=0, D=4;
      path3 g=((-3,5,-6)-/-(3,-1,-6)-/-(3,-1,4)-/-(-3,5,4)-/-cycle);
      draw(surface(g),lightgreen,nolight);
      label("\(2x+2y=4\)",(-3,7,-7),green);
      
      
      draw((-6,8,-1)-/-(8,-6,-1),red+linewidth(1));
      axes3("\(x\)","\(y\)","\(z\)",(-2,-2,-2),(8,8,8));
      </asymptote>
      </image>
     -->
     <sidebyside>
     <image width="70%" source="images/ThreePlanesLine.png" />
     </sidebyside>
      </example>
      
      <example><title>Three intersecting planes with no solutions</title>
      <p>
      Consider the system of linear equations
      <me>
      \begin{array}{rl}
      2x-y+z \amp= 1\\
      x+y+z\amp=2\\
      4x+y+3z \amp=3
      \end{array}
      </me>
      The augmented matrix is
      <me>
      \begin{bmatrix}
      2\amp-1\amp1\amp1\\
      1\amp1\amp1\amp2\\
      4\amp1\amp3\amp3
      \end{bmatrix}
      </me>
      whose reduced row echelon form is
      <me>
      \begin{bmatrix}
      1\amp0\amp\frac23\amp0\\
      0\amp1\amp\frac13\amp0\\
      0\amp0\amp0\amp1
      \end{bmatrix}
      </me>
      The last row indicates that there is no solution.
      </p>
      <!-- png image inserted until 3d problem resolved
      <image width="100%" xml:id="ThreePlanesNoSolution">
      <asymptote>
      import graph3;
      size(500);
      currentprojection=orthographic(-2,-0.8,3.0);
      currentlight=White;
      
      real A=2, B=-1, C=1, D=1;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(-2,-2),(2,2)),lightblue,nolight);
      label("\(x-y+z=1\)",(1,2,3),blue);
      
      real A=1, B=1, C=1, D=2;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(-2,-2),(2,2)),lightgrey,nolight);
      label("\(x+y+z=2\)",(1,-3,3),grey);
      
      real A=4, B=1, C=3, D=3;
      real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
      draw(surface(f,(-2,-2),(2,2)),lightgreen,nolight);
      label("\(4x+y+3z=3\)",(-0.5,-4,3),green);
      
      axes3("\(x\)","\(y\)","\(z\)",(-2,-2,-2),(3,3,3));
      </asymptote>
      </image>
      -->
     <sidebyside>
     <image width="90%" source="images/ThreePlanesNoSolution.png" />
     </sidebyside>
      </example>
      
      <example><title>Three nonintersecting planes with no solutions</title>
      <p>
      Consider the system of linear equations
      <me>
      \begin{array}{rl}
      2x-y+z \amp= 1\\
      2x-y+z \amp= 3\\
      2x-y+z \amp=5
      \end{array}
      </me>
      The augmented matrix is
      
      <me>
      \begin{bmatrix}
      2\amp-1\amp1\amp1\\
      2\amp-1\amp1\amp3\\
      2\amp-1\amp1\amp5
      \end{bmatrix}
      </me>
      whose reduced row echelon form is
      <me>
      \begin{bmatrix}
      1\amp -\frac12\amp\frac12\amp0\\
      0\amp0\amp0\amp1\\
      0\amp0\amp0\amp0\\
      \end{bmatrix}
      </me>
      The middle row indicates that there is no solution.
      </p>
      <!-- png image inserted until 3d problem resolved
      <image width="100%" xml:id="ThreeParallelPlanes">
      <asymptote>
        import graph3;
        size(500);
        currentprojection=perspective(-7/3,-2.4, 5.3);
        currentlight=White;
         
        real A=2, B=-1, C=1, D=1;
        real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
        draw(surface(f,(-2.0,-1.5),(2,2)),lightblue,nolight);
        label("\(x-y+z=1\)",(1,2,2.5),blue);
         
        real A=1, B=1, C=1, D=2;
        real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
        draw(surface(f,(-2,-2),(2,2)),lightgrey,nolight);
        label("\(x+y+z=2\)",(1,-2,3.4),grey);
         
        real A=4, B=1, C=3, D=3;
        real f(pair z) { return (D-A*z.x-B*z.y)/C;} // equation of plane
        draw(surface(f,(-2,-2),(2,2)),lightgreen,nolight);
        label("\(4x+y+3z=3\)",(-0.5,-3.4,3),green);
         
        axes3("\(x\)","\(y\)","\(z\)",(-1.5,-2,-2),(3,3.2,3.5));
      </asymptote>
      </image>
      -->
     <sidebyside>
     <image width="80%" source="images/ThreeParallelPlanes.png" />
     </sidebyside>
      </example>

      </section>
      
</chapter>
</book>
</pretext>