The element in the third row and second colum | vedclass

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Question

(B) Let $A = \begin{bmatrix} 3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5 \end{bmatrix}$. The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} 	ext

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$-2$

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(B) Let $A = \begin{bmatrix} 3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5 \end{bmatrix}$. The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,we calculate the determinant $|A|$:$|A| = 3(1 	imes 5 - 2 	imes 2) - 2(1 	imes 5 - 2 	imes 2) + 6(1 	imes 2 - 1 	imes 2)$$|A| = 3(5 - 4) - 2(5 - 4) + 6(2 - 2) = 3(1) - 2(1) + 6(0) = 3 - 2 = 1$.The element in the third row and second column of $A^{-1}$ is the entry $(A^{-1})_{32} = \frac{C_{23}}{|A|}$,where $C_{23}$ is the cofactor of the element at the second row and third column of matrix $A$.The cofactor $C_{23} = (-1)^{2+3} 	imes M_{23}$,where $M_{23}$ is the minor of the element at $(2,3)$.$M_{23} = \det \begin{bmatrix} 3 & 2 \ 2 & 2 \end{bmatrix} = (3 	imes 2) - (2 	imes 2) = 6 - 4 = 2$.Thus,$C_{23} = (-1)^5 	imes 2 = -2$.Therefore,$(A^{-1})_{32} = \frac{-2}{1} = -2$.

The element in the third row and second column of the inverse of the matrix $\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ is

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