The element in the 1^{st} row and 2^{nd} colu | vedclass

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(C) Step $1$: Find the determinant $|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0) = 1(25) - 3(10) - 2(-15) = 25 - 30 + 30 = 25$.Step $2$: The element

Vedclass · Accepted Answer

$-\frac{2}{25}$

Vedclass · Answer

(C) Step $1$: Find the determinant $|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0) = 1(25) - 3(10) - 2(-15) = 25 - 30 + 30 = 25$.Step $2$: The element in the $1^{st}$ row and $2^{nd}$ column of $A^{-1}$ is given by $\frac{C_{21}}{|A|}$, where $C_{21}$ is the cofactor of the element at the $2^{nd}$ row and $1^{st}$ column of $A$.Step $3$: $C_{21} = (-1)^{2+1} 	imes 	ext{minor}_{21} = -1 	imes \begin{vmatrix} 3 & -2 \ 5 & 0 \end{vmatrix} = -1 	imes (0 - (-10)) = -10$.Step $4$: The required element is $\frac{C_{21}}{|A|} = \frac{-10}{25} = -\frac{2}{5}$. Note: The provided options do not contain the correct value. Re-evaluating the cofactor: $C_{21} = -10$, $|A|=25$, result is $-2/5$.

The element in the $1^{st}$ row and $2^{nd}$ column of the inverse of the matrix $A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$ is ...

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