## Precalculus (6th Edition) Blitzer

$\left[ \begin{matrix} -4 & -\frac{1}{2} \\ 3 & 3 \\ \end{matrix} \right]$
To solve the provided expression, follow the below methods: Matrix addition: Consider two matrices $X,Y$ $X=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]$ and $Y=\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & X+Y=\left[ \begin{matrix} {{x}_{1}} & {{x}_{2}} \\ {{x}_{3}} & {{x}_{4}} \\ \end{matrix} \right]+\left[ \begin{matrix} {{y}_{1}} & {{y}_{2}} \\ {{y}_{3}} & {{y}_{4}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} {{x}_{1}}+{{y}_{1}} & {{x}_{2}}+{{y}_{2}} \\ {{x}_{3}}+{{y}_{3}} & {{x}_{4}}+{{y}_{4}} \\ \end{matrix} \right] \end{align} Matrix multiplication with a scalar: Consider a matrix $Z$ and a scalar value a: $Z=\left[ \begin{matrix} {{z}_{1}} & {{z}_{2}} \\ {{z}_{3}} & {{z}_{4}} \\ \end{matrix} \right]$ Now, \begin{align} & aZ=a\left[ \begin{matrix} {{z}_{1}} & {{z}_{2}} \\ {{z}_{4}} & {{z}_{3}} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} a{{z}_{1}} & a{{z}_{2}} \\ a{{z}_{4}} & a{{z}_{3}} \\ \end{matrix} \right] \end{align} Substitute $C=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right]$ in $2C-\frac{1}{2}B$ \begin{align} & 2C-\frac{1}{2}B=2\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]-\frac{1}{2}\left[ \begin{matrix} 4 & 1 \\ -6 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2 & 0 \\ 0 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 2 & \frac{1}{2} \\ -3 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2-2 & 0-\frac{1}{2} \\ 0+3 & 2+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -4 & -\frac{1}{2} \\ 3 & 3 \\ \end{matrix} \right] \end{align} Hence, $2C-\frac{1}{2}B=\left[ \begin{matrix} -4 & -\frac{1}{2} \\ 3 & 3 \\ \end{matrix} \right]$