Chapter 7, Problem 96RE

To determine

To solve the system of equations using Gaussian elimination with back substitution.

Answer to Problem 96RE

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

Explanation of Solution

Given information :

  $\begin{array}{l}x-3y+z=2\\ 3x-y-z=-6\\ -x-5y+3z=10\end{array}$

Writing the above system of equation in the matrix form,

  $\begin{array}{l}AX=B\\ \left[\begin{array}{ccc}1& -3& 1\\ 3& -1& -1\\ -1& -5& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -6\\ 10\end{array}\right]\end{array}$

Writing the matrix in the augmented form,

  $\left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 3& -1& -1& :& -6\\ -1& -5& 3& :& 10\end{array}\right]$

Applying Gauss-Jordan elimination,

  $\begin{array}{l}\left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 3& -1& -1& :& -6\\ -1& -5& 3& :& 10\end{array}\right]\stackrel{R_2\to R_2-3R_1}{\to }\left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 0& 8& -4& :& -12\\ -1& -5& 3& :& 10\end{array}\right]\\ \left[\begin{array}{ccccc}1& -3& 1& :& 2\\ 0& 8& -4& :& -12\\ -1& -5& 3& :& 10\end{array}\right]\stackrel{R_3\to R_3+R_1}{\to }\left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 0& 8& -4& :& -12\\ 0& -8& 4& :& 12\end{array}\right]\\ \left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 0& 8& -4& :& -12\\ 0& -8& 4& :& 12\end{array}\right]\stackrel{R_3\to R_3+R_2}{\to }\left[\begin{array}{ccccc}1& 2& -1& :& 3\\ 0& 8& -4& :& -12\\ 0& 0& 0& :& 0\end{array}\right]\end{array}$

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

So, let

  $z=k$

Further,

  $\begin{array}{l}8y-4z=-12\\ \Rightarrow 8y-4k=-12\\ \Rightarrow 8y=-12+4k\\ \Rightarrow y=\frac{-12+4k}{8}\\ \Rightarrow y=\frac{-3+k}{2}\end{array}$

Further,

  $\begin{array}{l}x-3y-z=2\\ x-3\left(\frac{-3+k}{2}\right)-k=2\\ \Rightarrow x-\left(\frac{-9+3k}{2}\right)-k=2\\ \Rightarrow 2x+9-3k-2k=2\\ \Rightarrow 2x-5k=-7\\ \Rightarrow 2x=-7+5k\\ \Rightarrow x=\frac{-7+5k}{2}\end{array}$

So, the solution of the above system of equations is $x=\frac{-3+k}{2},y=\frac{-7+5k}{2},z=k$ .