Chapter 8.2, Problem 93E

To determine

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

Answer to Problem 93E

The solution of the matrix is $x=1$ , $y=5$ and $z=-1$ .

Explanation of Solution

Given:

The given system of equation is $\{\begin{array}{l}2x-y+7z=-10\\ 3x+2y-4z=17\\ 6x-5y+z=-20\end{array}$ .

Calculation:

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

  $\begin{array}{l}A=\left[\begin{array}{ccc}2& -1& 7\\ 3& 2& -4\\ 6& -5& 1\end{array}\right]\\ X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ B=\left[\begin{array}{c}-10\\ 17\\ -20\end{array}\right]\end{array}$

Then, the augmented matrix of the given system is,

  $\left[A:B\right]=\left[\begin{array}{cccc}2& -1& 7& -10\\ 3& 2& -4& 17\\ 6& -5& 1& -20\end{array}\right]$

Apply elementary row operation in the given matrix as.

  $\begin{array}{l}{R}_2\to 2R_2-3R_1\\ R_3\to R_3-3R_1\end{array}$

Then,

  $\left[A:B\right]=\left[\begin{array}{cccc}2& -1& 7& -10\\ 0& 7& -29& 64\\ 0& -2& -20& 10\end{array}\right]$

Apply $R_3\to \left(-\frac{1}{2}\right)R_3$

  $\left[A:B\right]=\left[\begin{array}{cccc}2& -1& 7& -10\\ 0& 7& -29& 64\\ 0& 1& -10& -5\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to 7R_1+R_2\\ R_3\to 7R_3-R_2\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}14& 0& 20& -6\\ 0& 7& -29& 64\\ 0& 0& 99& -99\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to 7R_1+R_2\\ R_3\to 7R_3-R_2\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}7& 0& 10& -3\\ 0& 7& -29& 64\\ 0& 0& 1& -1\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to \frac{R_1}{2}\\ R_3\to \frac{R_3}{99}\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}7& 0& 10& -3\\ 0& 7& 0& 35\\ 0& 0& 1& -1\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to R_1-10R_3\\ R_2\to R_2+29R_3\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}7& 0& 0& 7\\ 0& 7& 0& 35\\ 0& 0& 1& -1\end{array}\right]$

Apply $\begin{array}{l}{R}_1\to \frac{1}{7}\left(R_1\right)\\ R_2\to \frac{1}{7}\left(R_3\right)\end{array}$

  $\left[A:B\right]=\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 5\\ 0& 0& 1& -1\end{array}\right]$

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

  $\begin{array}{l}\left[\begin{array}{cccc}1& 0& 0& x\\ 0& 1& 0& y\\ 0& 0& 1& z\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 5\\ 0& 0& 1& -1\end{array}\right]\\ x=1\\ y=5\\ z=-1\end{array}$

Thus, the solution of the matrix is $x=1$ , $y=5$ and $z=-1$ .