## Algebra and Trigonometry 10th Edition

X = $\begin{bmatrix} 3 & -\frac{1}{2} & -\frac{13}{2}\\ 3 & 0 & -\frac{11}{2}\\ \end{bmatrix}$
2X + 3A = B 2X = B - 3A X = $\frac{1}{2}$(B - 3A) First multiply matrix A by the scalar multiple 3: $\begin{bmatrix} -2(3) & 1(3) & 3(3)\\ -1(3) & 0(3) & 4(3)\\ \end{bmatrix}$ = $\begin{bmatrix} -6 & 3 & 9\\ -3 & 0 & 12\\ \end{bmatrix}$ Then perform B - 3A to get matrix Y: Y = $\begin{bmatrix} 0+6 & 2-3 & -4-9\\ 3+3 & 0-0 & 1-12\\ \end{bmatrix}$ = $\begin{bmatrix} 6 & -1 & -13\\ 6 & 0 & -11\\ \end{bmatrix}$ Then multiply matrix Y by the scalar $\frac{1}{2}$ to get matrix X: X = $\begin{bmatrix} 6(\frac{1}{2}) & -1(\frac{1}{2}) & -13(\frac{1}{2})\\ 6(\frac{1}{2}) & 0(\frac{1}{2}) & -11(\frac{1}{2})\\ \end{bmatrix}$ = $\begin{bmatrix} 3 & -\frac{1}{2} & -\frac{13}{2}\\ 3 & 0 & -\frac{11}{2}\\ \end{bmatrix}$