Answer
(3,2)- element in the inverse of the given matrix is $-1$
Work Step by Step
$ \det(A)=\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 2 \\
1 & 2 & 3
\end{bmatrix}=1.2+(-1).1+1.0=1$
To find (3,2)- element in the inverse of the given matrix A:
$(A^{-1})_{32}=\frac{1}{\det(A)}adj.A_{32}=\frac{1}{\det(A)}.C_{32}$
Now find the cofactor $C_{11}$ of the matrix A:
$C_{32}=(-1)^{3+2}.M_{32}=-1.\begin{vmatrix}
1 & 1\\
1 & 2
\end{vmatrix}=(-1).1=-1$
Hence, $(A^{-1})_{32}=\frac{1}{\det(A)}.C_{32}=\frac{1}{1}.(-1)=-1$
(3,2)- element in the inverse of the given matrix is $-1$
