Chapter 8.2, Problem 94E

To determine

To find: The solution for the system of equation by the gauss Jordan method.

Answer to Problem 94E

The solution of the matrix is $x=2$ , $y=-6$ and $z=3$ .

Explanation of Solution

Given:

The given system of equation is $\{\begin{array}{l}-x+4y+10z=4\\ 5x-3y+z=31\\ 8x+2y-3z=-5\end{array}$

Calculation:

Re-write the system of equation in the matrix form of $AX=B$ as,

  $\begin{array}{l}A=\left[\begin{array}{ccc}-1& 4& 10\\ 5& -3& 1\\ 8& 2& -3\end{array}\right]\\ X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ B=\left[\begin{array}{c}4\\ 31\\ -5\end{array}\right]\end{array}$

Then, the augmented matrix of the given system is,

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 4& 10& 4\\ 5& -3& 1& 31\\ 8& 2& -3& -5\end{array}\right]$

Apply elementary row operation in the given matrix as.

  $\begin{array}{l}{R}_2\to R_2+5R_1\\ R_3\to R_3+8R_1\end{array}$

Then,

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 4& 10& 4\\ 0& 1& 3& 3\\ 0& 34& 77& 27\end{array}\right]$

Apply $R_2\to \left(\frac{1}{17}\right)R_2$

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 4& 10& -8\\ 0& 1& 3& 3\\ 0& 34& 77& 27\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to R_1-4R_2\\ R_3\to R_3-34R_2\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 0& -2& -8\\ 0& 1& 3& 3\\ 0& 0& -25& -75\end{array}\right]$

Apply $R_3\to \left(-\frac{1}{25}\right)R_3$

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 0& -2& -8\\ 0& 1& 3& 3\\ 0& 0& 1& 3\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to R_1+2R_3\\ R_2\to R_2-3R_3\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}-1& 0& 0& 2\\ 0& 1& 0& -6\\ 0& 0& 1& 3\end{array}\right]$

Apply $R_1\to \left(-1\right)R_1$

  $\left[A:B\right]=\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& -6\\ 0& 0& 1& 3\end{array}\right]$

Compare the above matrix with the Gauss Jordan matrix and the unique values are obtained as,

  $\begin{array}{l}\left[\begin{array}{cccc}1& 0& 0& x\\ 0& 1& 0& y\\ 0& 0& 1& z\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& -6\\ 0& 0& 1& 3\end{array}\right]\\ x=2\\ y=-6\\ z=3\end{array}$

Thus, the solution of the matrix is $x=2$ , $y=-6$ and $z=3$ .