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Problems on Inhomogeneous Systems

1.
Row reducing the augmented matrix with Octave gives
octave:5> A=[1,2,3,4;2,4,8,10;3,6,11,14];
octave:6> B=[1;6;7];
octave:7> rref([A,B])
ans =

   1.00000   2.00000   0.00000   1.00000  -5.00000
   0.00000   0.00000   1.00000   1.00000   2.00000
   0.00000   0.00000   0.00000   0.00000   0.00000
This leads to the system

\begin{eqnarray*}x_1 + 2 x_2 + x_4 &=& -5\\
x_3 + x_4 &=& 2
\end{eqnarray*}


Assigning the values s and t to the variables x2 and x4, and solving for the leading variables, gives

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\\ x_3\\ x_4\end{array}\right]...
...]
+ t \left[\begin{array}{r}-1\\ 0\\ -1\\ 1\end{array}\right]
\end{displaymath}

2.
Row reduce the augmented matrix:
octave:8> A=[1,3,3;2,6,9;-1,-3,3];
octave:9> B=[1;5;5];
octave:10> rref([A,B])
ans =

   1   3   0  -2
   0   0   1   1
   0   0   0   0
Solving the reduced system for x1 and x3 gives

\begin{eqnarray*}x_1 &=& -2 - 3x_2\\
x_3&=&1
\end{eqnarray*}


Assigning the value s to x2 gives

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\\ x_3\end{array}\right]
= \l...
...ight]
+ s \left[\begin{array}{r}-3\\ 1\\ 0\end{array}\right].
\end{displaymath}

3.
Row reducing the augmented matrix gives:
octave:11> A=[1,3,1,2;2,6,4,8;0,0,2,4];
octave:12> B=[1;3;1];
octave:13> rref([A,B])
ans =

  1.00000  3.00000  0.00000  0.00000  0.50000
  0.00000  0.00000  1.00000  2.00000  0.50000
  0.00000  0.00000  0.00000  0.00000  0.00000
Solving for x1 and x3 gives

\begin{eqnarray*}x_1 &=& 0.5 - 3 x_2\\
x_3 &=& 0.5 - 2 x_4
\end{eqnarray*}


Setting x2=s and x3=t gives

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\\ x_3\\ x_4\end{array}\right]...
...]
+ t \left[\begin{array}{r}0\\ 0\\ -2\\ 1\end{array}\right].
\end{displaymath}

4.
Row reducing the augmented matrix gives
octave:14> A=[1,2,2,4,6;1,2,3,6,9;0,0,1,2,3];
octave:15> B=[0;0;0];
octave:16> rref([A,B])
ans =

  1  2  0  0  0  0
  0  0  1  2  3  0
  0  0  0  0  0  0
Solving for x1 and x3 yields

\begin{eqnarray*}x_1 &=& -2 x_2\\
x_3 &=& -2 x_4 - 3 x_5
\end{eqnarray*}


Setting x2=s, x4=t, and x5=u gives

\begin{displaymath}\left[\begin{array}{c}x_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\...
...+ u \left[\begin{array}{r}0\\ 0\\ -3\\ 0\\ 1\end{array}\right]
\end{displaymath}

5.
Row reducing the augmented matrix gives
octave:17> A=[1,2,-2;2,5,-4;4,9,-8];
octave:18> B=[1;0;0];
octave:19> rref([A,B])
ans =

   1   0  -2   0
   0   1   0   0
   0   0   0   1
This gives the equivalent system of equations

\begin{eqnarray*}x_1 -2 x_3 &=&0\\
x_2 &=& 0\\
0 &=& 1
\end{eqnarray*}


Since no value values for x1, x2, and x3 can make the last equation true, the system has no solutions.


next up previous
Up: Solutions to Supplementary Homework Previous: Problems on Row Reduction
Frank Beatrous
1998-09-21