Given A=(2132)A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}A=(2312) with det(A)=1\det(A) = 1det(A)=1, use the formula A−1=1det(A)adj(A)A^{-1} = \frac{1}{\det(A)}\text{adj}(A)A−1=det(A)1adj(A) to find the (1,2)(1,2)(1,2) entry of A−1A^{-1}A−1.
−1-1−1
111
222
333