Chapter 7, Problem 89RE

To determine

To use: the matrices to solve the system of equations using Gaussian elimination with back substitution.

Answer to Problem 89RE

The solution of the above system of equations is $x=1-2k,y=-2k,z=k$ .

Explanation of Solution

Given information :

  $\begin{array}{l}x+y+z=-3\\ 4x+y-2z=-12\end{array}$

Writing the above system of equation in the matrix form,

  $\begin{array}{l}AX=B\\ \left[\begin{array}{ccc}1& 1& 1\\ 4& 1& -2\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-3\\ -12\\ 0\end{array}\right]\end{array}$

Writing the matrix in the augmented form,

  $\left[\begin{array}{ccc}1& 1& 1\\ 4& 1& -2\\ 0& 0& 0\end{array}\begin{array}{c}:\\ :\\ :\end{array}\begin{array}{c}-3\\ -12\\ 0\end{array}\right]$

Applying Gauss-Jordan elimination,

  $\left[\begin{array}{ccc}1& 1& 1\\ 4& 1& -2\\ 0& 0& 0\end{array}\begin{array}{c}:\\ :\\ :\end{array}\begin{array}{c}-3\\ -12\\ 0\end{array}\right]\stackrel{R_2\to R_2-4R_1}{\to }\left[\begin{array}{ccc}1& 2& 6\\ 0& -3& -6\\ 0& 0& 0\end{array}\begin{array}{c}:\\ :\\ :\end{array}\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Now, from the last matrix, it is clear that the above system of equations have infinitely many solutions.

So, let $z=k$ ,

Then,

  $\begin{array}{c}-3y-6z=0\\ -y-2z=0\\ y=-2z\\ y=-2k\end{array}$

Further,

  $\begin{array}{c}x+2y+6z=1\\ x+2\left(-2k\right)+6k=1\\ x+2k=1\\ x=1-2k\end{array}$

The solution of the above system of equations is $x=1-2k,y=-2k,z=k$ .