Chapter 6, Problem 1MC

To determine

To solve: The linear system using matrices.

Answer to Problem 1MC

The solution of the given system of equations is $x=1,y=-1\text{and}z=2$.

Explanation of Solution

Given:

The system of linear equations is, $\{\begin{array}{c}x+2y-3z=-7\\ 3x-y+2z=8\\ 2x-y+z=5\end{array}$.

Calculation:

Use Gauss-Jordan elimination method to solve the augmented matrix without the back substitution.

The augmented matrix for the linear equations is,

$A=\left[\begin{array}{cccc}1& 2& -3& -7\\ 3& -1& 2& 8\\ 2& -1& 1& 5\end{array}\right]$

Apply $\left({\text{R}}_2\to {\text{R}}_2-3{\text{R}}_1\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& -7& 11& 29\\ 2& -1& 1& 5\end{array}\right]$

Perform $\left({\text{R}}_3\to {\text{R}}_3-2{\text{R}}_1\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& -7& 11& 29\\ 0& -5& 7& 19\end{array}\right]$

Use $\left({\text{R}}_2\to {\text{R}}_2-{\text{R}}_3\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& -2& 4& 10\\ 0& -5& 7& 19\end{array}\right]$

Perform $\left({\text{R}}_2\to -\frac{1}{2}{\text{R}}_2\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& 1& -2& -5\\ 0& -5& 7& 19\end{array}\right]$

Apply $\left({\text{R}}_3\to {\text{R}}_3{\text{+5R}}_2\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& 1& -2& -5\\ 0& 0& -3& -6\end{array}\right]$

Use $\left({\text{R}}_3\to -\frac{1}{3}{\text{R}}_3\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& 1& -2& -5\\ 0& 0& 1& 2\end{array}\right]$

Apply $\left({\text{R}}_2\to {\text{R}}_2+2{\text{R}}_3\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& -3& -7\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{array}\right]$

Perform $\left({\text{R}}_1\to {\text{R}}_1{\text{+3R}}_3\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 2& 0& -1\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{array}\right]$

Apply $\left({\text{R}}_1\to {\text{R}}_1-2{\text{R}}_2\right)$ in the above matrix.

$\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{array}\right]$

The system of linear equations that corresponds to the above matrix is, $\{\begin{array}{c}x=1\\ y=-1\\ z=2\end{array}$.

Hence, the solution of the given system of equations is $x=1,y=-1\text{and}z=2$.