If A = begin{bmatrix} 2 & 0 & -1 5 & 1 & 0 | vedclass

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(C) Step $1$: Calculate the determinant of $A$. $|A| = 2(3-0) - 0(15-0) + (-1)(5-0) = 6 - 5 = 1$.Step $2$: Use the formula $A^{-1} = \frac{1}{|A|} 	e

Vedclass · Accepted Answer

$225$

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(C) Step $1$: Calculate the determinant of $A$. $|A| = 2(3-0) - 0(15-0) + (-1)(5-0) = 6 - 5 = 1$.Step $2$: Use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. Since $|A| = 1$, $A^{-1} = 	ext{adj}(A)$.Step $3$: Find the cofactor matrix $C_{ij}$.$C_{11} = +(3-0) = 3$, $C_{12} = -(15-0) = -15$, $C_{13} = +(5-0) = 5$.$C_{21} = -(0 - (-1)) = -1$, $C_{22} = +(6-0) = 6$, $C_{23} = -(2-0) = -2$.$C_{31} = +(0 - (-1)) = 1$, $C_{32} = -(0 - (-5)) = -5$, $C_{33} = +(2-0) = 2$.Step $4$: The adjoint is the transpose of the cofactor matrix: $	ext{adj}(A) = \begin{bmatrix} 3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2 \end{bmatrix}$.Step $5$: Comparing with $A^{-1}$, we get $\gamma = 3$, $\alpha = -15$, $\beta = 5$.Step $6$: Calculate $|\alpha \cdot \beta \cdot \gamma| = |(-15) \cdot 5 \cdot 3| = |-225| = 225$.

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ (where $| \cdot |$ denotes the absolute value)

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