Answer
(1,1)- element in the inverse of the given matrix is $0$
Work Step by Step
$ \det(A)=\begin{bmatrix}
-1 & -2 & 4\\
0 & 2 & -1 \\
3 & -2 & 1
\end{bmatrix}=-1.0+2.3+4(-6)=-18$
To find (1,1)- element in the inverse of the given matrix A:
$(A^{-1})_{11}=\frac{1}{\det(A)}adj.A_{11}=\frac{1}{\det(A)}.C_{11}$
Now find the cofactor $C_{11}$ of the matrix A:
$C_{11}=(-1)^{1+1}.M_{11}=1.\begin{vmatrix}
2 & -1\\
-1 & 1
\end{vmatrix}=2-2=0$
Hence, $(A^{-1})_{11}=\frac{1}{\det(A)}.C_{11}=\frac{1}{-18}.0=0$
(1,1)- element in the inverse of the given matrix is $0$
