## Precalculus (6th Edition) Blitzer

a) $A+B=\left[ \begin{matrix} 6 & 4 \\ 5 & 5 \\ \end{matrix} \right]$ b) $A-B=\left[ \begin{matrix} -10 & 2 \\ -5 & -3 \\ \end{matrix} \right]$ c) $\left( -4 \right)A=\left[ \begin{matrix} 8 & -12 \\ 0 & -4 \\ \end{matrix} \right]$ d) $3A+2B=\left[ \begin{matrix} 10 & 11 \\ 10 & 11 \\ \end{matrix} \right]$
(a) Perform the addition of the matrices $A$ and $B$ as follows: \begin{align} & A+B=\left[ \begin{matrix} -2 & 3 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 8 & 1 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2+8 & 3+1 \\ 0+5 & 1+4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 6 & 4 \\ 5 & 5 \\ \end{matrix} \right] \end{align} Hence, $A+B=\left[ \begin{matrix} 6 & 4 \\ 5 & 5 \\ \end{matrix} \right]$ (b) Perform the subtraction of the matrices $A$ and $B$ as follows: \begin{align} & A-B=\left[ \begin{matrix} -2 & 3 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 8 & 1 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2-8 & 3-1 \\ 0-5 & 1-4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -10 & 2 \\ -5 & -3 \\ \end{matrix} \right] \end{align} Hence, $A-B=\left[ \begin{matrix} -10 & 2 \\ -5 & -3 \\ \end{matrix} \right]$ (c) Perform the multiplication of the matrices $A$ with constant number. \begin{align} & \left( -4 \right)A=\left( -4 \right)\left[ \begin{matrix} -2 & 3 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -4\times \left( -2 \right) & -4\times \left( 3 \right) \\ -4\times 0 & -4\times \left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 8 & -12 \\ 0 & -4 \\ \end{matrix} \right] \end{align} Hence, $\left( -4 \right)A=\left[ \begin{matrix} 8 & -12 \\ 0 & -4 \\ \end{matrix} \right]$ (d) Perform the multiplication with a content number and addition of the matrices $A$ and $B$ as follows: \begin{align} & 3A+2B=3\left[ \begin{matrix} -2 & 3 \\ 0 & 1 \\ \end{matrix} \right]+2\left[ \begin{matrix} 8 & 1 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3\times \left( -2 \right) & 3\times 3 \\ 3\times 0 & 3\times 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 2\times 8 & 2\times 1 \\ 2\times 5 & 2\times 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & 9 \\ 0 & 3 \\ \end{matrix} \right]+\left[ \begin{matrix} 16 & 2 \\ 10 & 8 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 10 & 11 \\ 10 & 11 \\ \end{matrix} \right] \end{align} Hence, $3A+2B=\left[ \begin{matrix} 10 & 11 \\ 10 & 11 \\ \end{matrix} \right]$