Answer
(2,3)- element in the inverse of the given matrix is $\frac{9}{16}$
Work Step by Step
$ \det(A)=1\begin{bmatrix}
-1 & 1 & 3\\
0& 1 & 2 \\
-1 & 1 & 0
\end{bmatrix}+1\begin{bmatrix}
2 & -1 & 3\\
0& 1 & 2 \\
-1 & 1 & 0
\end{bmatrix}=15+1=16$
To find (2,3)- element in the inverse of the given matrix A:
$(A^{-1})_{23}=\frac{1}{\det(A)}adj.A_{23}=\frac{1}{\det(A)}.C_{32}$
Now find the cofactor $C_{11}$ of the matrix A:
$C_{32}=(-1)^{3+2}.M_{32}=1.\begin{vmatrix}
1 & 1 &0\\
2& 1&3\\
-1 & 2 & 0
\end{vmatrix}=(-1).(-9)=9$
Hence, $(A^{-1})_{23}=\frac{1}{\det(A)}.C_{32}=\frac{1}{16}.(9)=\frac{9}{16}$
(2,3)- element in the inverse of the given matrix is $\frac{9}{16}$
