The inverse of the matrix A = begin{bmatrix} | vedclass

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(D) Step $1$: Calculate the determinant of matrix $A$.$|A| = 2((-3)(1) - (1)(2)) - (-1)((4)(1) - (1)(1)) + 4((4)(2) - (-3)(1))$$|A| = 2(-3 - 2) + 1(4

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$-2$

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(D) Step $1$: Calculate the determinant of matrix $A$.$|A| = 2((-3)(1) - (1)(2)) - (-1)((4)(1) - (1)(1)) + 4((4)(2) - (-3)(1))$$|A| = 2(-3 - 2) + 1(4 - 1) + 4(8 + 3)$$|A| = 2(-5) + 1(3) + 4(11) = -10 + 3 + 44 = 37$.Step $2$: The inverse $A^{-1}$ is given by $\frac{1}{|A|} 	ext{adj}(A)$.Step $3$: The element at position $(3,3)$ of the adjoint matrix is the cofactor $C_{33}$.$C_{33} = (-1)^{3+3} \begin{vmatrix} 2 & -1 \ 4 & -3 \end{vmatrix} = (1)((2)(-3) - (-1)(4)) = -6 + 4 = -2$.Step $4$: Since $B = A^{-1} = \frac{1}{37} 	ext{adj}(A)$, the element at $(3,3)$ in $B$ is $\frac{C_{33}}{37} = \frac{-2}{37}$.Comparing this with the given matrix $B$, we get $k = -2$.

The inverse of the matrix $A = \begin{bmatrix} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{bmatrix}$ is $B = \frac{1}{37} \begin{bmatrix} -5 & 9 & 11 \\ -3 & -2 & 14 \\ 11 & -5 & k \end{bmatrix}$, then the value of $k$ is...

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