Chapter 9, Problem 12CLT

To determine

Solve the system using the matrix.

Answer to Problem 12CLT

  $x=-2,y=3,z=1$

Explanation of Solution

Calculation:

Let us attempt to solve the system by Gauss-Jordan elimination

Let us consider the following system of equations

  $System:\{\begin{array}{l}-x+2y-z=9\\ 2x-y+2z=-9\\ 3x+3y-4z=7\end{array}$

Augmented matrix is

  $\begin{array}{ccccc}{R}_1& -1& 2& -1:& 9\\ R_2& 2& -1& 2:& -9\\ R_3& 3& 3& -4:& 7\end{array}$

Multiply $R_1$ by $2$ and adding to $R_2$ . Multiply by $3$ and adding to $R_3$ ,we get

  $\begin{array}{ccccc}{R}_1& 1& -2& 1:& -9\\ 2R_1+R_2& 0& 3& 0:& 9\\ 3R_1+R_3& 0& 0& -7:& 7\end{array}$

Multiply $R_1$ by $-1$ . Multiply $R_2$ by $-3$ and add to $R_3$

  $\begin{array}{ccccc}{R}_1& 1& -2& 1:& -9\\ & 0& 3& 0:& 9\\ 3R_2+R_3& 0& 0& -7:& 7\end{array}$

Multiply $R_2$ by $\frac{1}{3}$ and $R_3$ by $-\frac{1}{7}$

  $\begin{array}{ccccc}& 1& -2& 1:& -9\\ 1/3R_2& 0& 1& 0:& 3\\ -1/7R_3& 0& 0& 1:& -1\end{array}$

Multiply $R_2$ by 2 and add to $R_1$

  $\begin{array}{ccccc}2R_2+R_1& 1& 0& 1:& -3\\ & 0& 1& 0:& 3\\ & 0& 0& 1:& -1\end{array}$

Multiply $R_3$ by $-1$ and add to $R_1$

  $\begin{array}{ccccc}-R_3+R_1& 1& 0& 1:& -2\\ & 0& 1& 0:& 3\\ & 0& 0& 1:& -1\end{array}$

Hence, solving the system yields $x=-2,y=3,z=1$