The element of the second row and third colum | vedclass

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(B) Let $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant $|A| = 1(1-0) - 2(2-0) + 1(0 - (-1

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(B) Let $A = \begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$.First,we calculate the determinant $|A| = 1(1-0) - 2(2-0) + 1(0 - (-1)) = 1 - 4 + 1 = -2$.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.The element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is $\frac{1}{|A|} 	imes C_{32}$,where $C_{32}$ is the cofactor of the element in the $3^{rd}$ row and $2^{nd}$ column of $A$.The cofactor $C_{32} = (-1)^{3+2} M_{32} = -1 	imes \begin{vmatrix} 1 & 1 \ 2 & 0 \end{vmatrix} = -1 	imes (0 - 2) = 2$.Thus,the required element is $\frac{C_{32}}{|A|} = \frac{2}{-2} = -1$.

The element of the second row and third column in the inverse of $\begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$ is

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