## College Algebra (10th Edition)

$(9,23)$
Solving $AX=B,\ \qquad X=A^{-1}B$. This problem is connected to exercise $32$, as the coefficient matrix is $A=\left[\begin{array}{rr}{3}&{-1}\\{-2}&{1}\end{array}\right]$, and $A^{-1}=\left[\begin{array}{ll}{1}&{1}\\{2}&{3}\end{array}\right]$ Writing the system \left\{\begin{aligned}3x-y&=4\\-2x+y&=5\end{aligned}\right. in matrix form, $AX=B,$ the solution is $X=A^{-1}B=\left[\begin{array}{rr} {1}&{1}\\ {2}&{3}\end{array}\right]\left[\begin{array}{l} 4\\ 5 \end{array}\right]$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{l} 4+5\\ 8+15 \end{array}\right]=\left[\begin{array}{l} 9\\ 23 \end{array}\right]$ Solution: $(x,y)=(9,23)$