## Algebra and Trigonometry 10th Edition

X = $\begin{bmatrix} -\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\ \frac{2}{3} & 0 & 6\\ \end{bmatrix}$
3X - 4A = 2B 3X = 2B + 4A X = $\frac{1}{3}$(2B + 4A) First multiply matrix A by the scalar multiple 4: $\begin{bmatrix} -2(4) & 1(4) & 3(4)\\ -1(4) & 0(4) & 4(4)\\ \end{bmatrix}$ = $\begin{bmatrix} -8 & 4 & 12\\ -4 & 0 & 16\\ \end{bmatrix}$ Then multiply matrix B by the scalar multiple 2: $\begin{bmatrix} 0(2) & 2(2) & -4(2)\\ 3(2) & 0(2) & 1(2)\\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 4 & -8\\ 6 & 0 & 2\\ \end{bmatrix}$ Then perform 2B + 4A to get matrix Y: Y = $\begin{bmatrix} 0-8 & 4+4 & -8+12\\ 6-4 & 0+0 & 2+16\\ \end{bmatrix}$ = $\begin{bmatrix} -8 & 8 & 4\\ 2 & 0 & 18\\ \end{bmatrix}$ Then multiply Y by the scalar $\frac{1}{3}$ to get X: X = $\begin{bmatrix} -8(\frac{1}{3}) & 8(\frac{1}{3}) & 4(\frac{1}{3})\\ 2(\frac{1}{3}) & 0(\frac{1}{3}) & 18(\frac{1}{3})\\ \end{bmatrix}$ = $\begin{bmatrix} -\frac{8}{3} & \frac{8}{3} & \frac{4}{3}\\ \frac{2}{3} & 0 & 6\\ \end{bmatrix}$