# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 893: 22

$(1,-1,1)$

#### Work Step by Step

Step 1. Writing a matrix based on the given system of equations, we have $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 2 & -1 & 1 & | & 4 \\ -1 & 1 & -2 & | & -4 \end{bmatrix} \begin{array} ..\\-2R1+R2\to R2\\ R1+R3\to R3 \end{array}$ Step 2. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 0 & 3 & 3 & | & 0 \\ 0 & -1 & -3 & | & -2 \end{bmatrix} \begin{array} ..\\..\\ R2+3R3\to R3 \end{array}$ Step 3. Perform row operations given to the right of the matrix: $\begin{bmatrix} 1 & -2 & -1 & | & 2 \\ 0 & 3 & 2 & | & 0 \\ 0 & 0 & -6 & | & -6 \end{bmatrix} \begin{array} ..\\..\\ .. \end{array}$ Step 4. The last equation gives $-6z=-6$ or $z=1$ Step 5. Use back-substitution to get $3y+3z=0$ and $y=-1$, $x-2y-z=2$ and $x=2+2(-1)+1=1$ Step 6. The solution set for the system is $(1,-1,1)$

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