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

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(A) Given $A = \begin{bmatrix} k & 5 & 2 \ 2 & -k & 5 \ 5 & 2 & -k \end{bmatrix}$ and $\det A = 190$.Expanding the determinant along the first row:$

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$\begin{bmatrix} -1 & 19 & 31 \ 31 & -19 & -11 \ 19 & 19 & -19 \end{bmatrix}$

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(A) Given $A = \begin{bmatrix} k & 5 & 2 \ 2 & -k & 5 \ 5 & 2 & -k \end{bmatrix}$ and $\det A = 190$.Expanding the determinant along the first row:$\det A = k(k^2 - 10) - 5(-2k - 25) + 2(4 + 5k) = 190$$k^3 - 10k + 10k + 125 + 8 + 10k = 190$$k^3 + 10k - 57 = 0$By testing values, for $k=3$: $27 + 30 - 57 = 0$. Thus, $k=3$.Substituting $k=3$, $A = \begin{bmatrix} 3 & 5 & 2 \ 2 & -3 & 5 \ 5 & 2 & -3 \end{bmatrix}$.The cofactor matrix $C$ is calculated as:$C_{11} = ((-3)(-3) - (5)(2)) = 9 - 10 = -1$$C_{12} = -((2)(-3) - (5)(5)) = -(-6 - 25) = 31$$C_{13} = ((2)(2) - (-3)(5)) = 4 + 15 = 19$$C_{21} = -((5)(-3) - (2)(2)) = -(-15 - 4) = 19$$C_{22} = ((3)(-3) - (2)(5)) = -9 - 10 = -19$$C_{23} = -((3)(2) - (5)(5)) = -(6 - 25) = 19$$C_{31} = ((5)(5) - (-3)(2)) = 25 + 6 = 31$$C_{32} = -((3)(5) - (2)(2)) = -(15 - 4) = -11$$C_{33} = ((3)(-3) - (5)(2)) = -9 - 10 = -19$The adjoint matrix is the transpose of the cofactor matrix:$\operatorname{Adj} A = C^T = \begin{bmatrix} -1 & 19 & 31 \ 31 & -19 & -11 \ 19 & 19 & -19 \end{bmatrix}$.

If $A = \begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix}$ and $\det A = 190$, then $\operatorname{Adj} A = $

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