If for a matrix A,|A|=6 and text{adj } A = be | vedclass

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(C) We know that for any square matrix $A$ of order $n$,the property $	ext{adj } A \cdot A = |A| I$ holds,and $|	ext{adj } A| = |A|^{n-1}$.Given $|A

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$2$

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(C) We know that for any square matrix $A$ of order $n$,the property $	ext{adj } A \cdot A = |A| I$ holds,and $|	ext{adj } A| = |A|^{n-1}$.Given $|A| = 6$ and $n = 3$,we have $|	ext{adj } A| = |A|^{3-1} = |A|^2 = 6^2 = 36$.Now,calculate the determinant of $	ext{adj } A$:$|	ext{adj } A| = \begin{vmatrix} 1 & -2 & 4 \ 4 & 1 & 1 \ -1 & k & 0 \end{vmatrix} = 1(0 - k) - (-2)(0 - (-1)) + 4(4k - (-1))$$= 1(-k) + 2(1) + 4(4k + 1)$$= -k + 2 + 16k + 4$$= 15k + 6$.Equating the two values:$15k + 6 = 36$$15k = 30$$k = 2$.

If for a matrix $A$,$|A|=6$ and $\text{adj } A = \begin{bmatrix} 1 & -2 & 4 \\ 4 & 1 & 1 \\ -1 & k & 0 \end{bmatrix}$,then $k$ is equal to

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