College Algebra 7th Edition

We find the determinant $(ad-bc)$ $2-5/3=1/3$ Use the formula $M=\frac{1}{ad-bc} \left[ {\begin{array}{cc} d &-b \\ -c & a \\ \end{array} } \right]$ to get $M= 3 \left[ {\begin{array}{cc} 4 & -1/3 \\ -5 & 1/2 \\ \end{array} } \right]$ or $M= \left[ {\begin{array}{cc} 12 & -1 \\ -15 & 3/2 \\ \end{array} } \right]$