## Precalculus (10th Edition)

$-8$
We know that for a matrix $\left[\begin{array}{rr} a & b \\ c &d \\ \end{array} \right]$ the determinant, $D=ad-bc.$ Hence here $D=xb-ay=8$ is given. Our determinant is: $ay-xb=-(xb-ay)=-(8)=-8$