## College Algebra (6th Edition)

$x=\left[\begin{array}{ll} 5 & 17\\ -5 & 45/2\\ -11 & -2 \end{array}\right]$
$2X+5A=B\qquad$ ... subtract $5A$ from both sides $2X=B-5A\qquad$ ... multiply both sides with $\displaystyle \frac{1}{2}$ $X=\displaystyle \frac{1}{2}(B-5A)$ $X=\displaystyle \frac{1}{2}\cdot(\left[\begin{array}{ll} -5 & -1\\ 0 & 0\\ 3 & -4 \end{array}\right]-5\displaystyle \cdot\left[\begin{array}{ll} -3 & -7\\ 2 & -9\\ 5 & 0 \end{array}\right])$ $X=\displaystyle \frac{1}{2}\cdot\left[\begin{array}{ll} -5-5(-3) & -1-5(-7)\\ 0-5(2) & 0-2(-9)\\ 3-5(5) & -4-5(0) \end{array}\right]$ $X=\displaystyle \frac{1}{2}\cdot\left[\begin{array}{ll} 10 & 34\\ -10 & 45\\ -22 & -4 \end{array}\right]$ $x=\left[\begin{array}{ll} 5 & 17\\ -5 & 45/2\\ -11 & -2 \end{array}\right]$