## College Algebra 7th Edition

$X=\left[\begin{array}{rr} -1 & -1/2\\ 1 & 2 \end{array}\right]$
Observing the dimensions, B is a 2$\times$2 matrix, A is 2$\times$2 matrix. For the LHS to be defined, X must be a 2$\times$2 matrix. So, there is a solution. $2X+A=B\qquad$subtract A from both sides $2X=B-A\qquad$... muliply with the scalar $\displaystyle \frac{1}{2}$ $X=\displaystyle \frac{1}{2}(B-A)=\frac{1}{2}\left[\begin{array}{ll} 2-4 & 5-6\\ 3-1 & 7-3 \end{array}\right]$ $X=\left[\begin{array}{ll} -2/2 & -1/2\\ 2/2 & 4/2 \end{array}\right]$ $X=\left[\begin{array}{ll} -1 & -1/2\\ 1 & 2 \end{array}\right]$