## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} -\frac{11}{3} & -\frac{31}{3}\\ 1 & \frac{3}{2}\\ -8 & -1\\ \end{bmatrix}$
First add the two matrices in the parentheses to create matrix X: X = $\begin{bmatrix} -5+7 & -1+5\\ 3-9 & 4-1\\ 0+6 & 13-1\\ \end{bmatrix}$ Then multiply matrix X by the scalar multiple $\frac{1}{6}$ to create matrix A: A = $\begin{bmatrix} 2(\frac{1}{6}) & 4(\frac{1}{6})\\ -6(\frac{1}{6}) & 3(\frac{1}{6})\\ 6(\frac{1}{6}) & 12(\frac{1}{6})\\ \end{bmatrix}$ = $\begin{bmatrix} \frac{1}{3} & \frac{2}{3}\\ -1 & \frac{1}{2}\\ 1 & 2\\ \end{bmatrix}$ Then multiply the matrix not in parentheses by the scalar multiple -1 to create matrix B: B = $\begin{bmatrix} 4(-1) & 11(-1)\\ -2(-1) & -1(-1)\\ 9(-1) & 3(-1)\\ \end{bmatrix}$ = $\begin{bmatrix} -4 & -11\\ 2 & 1\\ -9 & -3\\ \end{bmatrix}$ Then perform A + B to get the final answer: $\begin{bmatrix} -4+\frac{1}{3} & -11+\frac{2}{3}\\ 2-1 & 1+\frac{1}{2}\\ -9+1 & -3+2\\ \end{bmatrix}$ = $\begin{bmatrix} -\frac{11}{3} & -\frac{31}{3}\\ 1 & \frac{3}{2}\\ -8 & -1\\ \end{bmatrix}$