## Precalculus (6th Edition) Blitzer

a. $\begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -2 & -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -6 \\ -10 \end{bmatrix}$ b. $(-1,4,2)$
a. Based on the given system, we have $\begin{bmatrix} 1 & -1 & 1 \\ 0 & -2 & 1 \\ -2 & -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -6 \\ -10 \end{bmatrix}$ b. As $X=A^{-1}B$, with the given $A^{-1}$ in the exercise, we have $X=\begin{bmatrix} 3 & -3 & 1 \\ -2 & 2 & -1 \\ -4 & 5 & -2 \end{bmatrix} \begin{bmatrix} -3 \\ -6 \\ -10 \end{bmatrix}=\begin{bmatrix} -9+18-10 \\ 6-12+10 \\ 12-30+20 \end{bmatrix}=\begin{bmatrix} -1 \\ 4 \\ 2 \end{bmatrix}$ Thus we have the solution as $(-1,4,2)$