## Precalculus (6th Edition) Blitzer

$X=\left[ \begin{matrix} 0.5 & 0.5 \\ -3 & 0.5 \\ \end{matrix} \right]$
Given $C=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]$. Since both matrices are $2\times 2$, the matrix operations are valid. Now \begin{align} & 2X-3C=B \\ & 2X=B+3C \\ & X=\frac{B+3C}{2} \end{align} Substitute the corresponding values to get: \begin{align} & X=\frac{\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]+3\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]}{2}=\frac{\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]+\left[ \begin{matrix} -3 & 0 \\ 0 & 3 \\ \end{matrix} \right]}{2} \\ & =\frac{\left[ \begin{matrix} 4-3 & 1+0 \\ -6+0 & -2+3 \\ \end{matrix} \right]}{2}=\frac{\left[ \begin{matrix} 1 & 1 \\ -6 & 1 \\ \end{matrix} \right]}{2} \\ & =\frac{1}{2}\left[ \begin{matrix} 1 & 1 \\ -6 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{-6}{2} & \frac{1}{2} \\ \end{matrix} \right] \end{align} Therefore, $X=\left[ \begin{matrix} 0.5 & 0.5 \\ -3 & 0.5 \\ \end{matrix} \right]$