The inverse of the matrix left[begin{array}{c | vedclass

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(D) Let $A = \left[\begin{array}{ccc}7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1\end{array}ight]$.First,calculate the determinant $|A|$:$|A| = 7(1 - 0)

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$\left[\begin{array}{lll}1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4\end{array}ight]$

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(D) Let $A = \left[\begin{array}{ccc}7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1\end{array}ight]$.First,calculate the determinant $|A|$:$|A| = 7(1 - 0) - (-3)(-1 - 0) + (-3)(0 - (-1)) = 7(1) + 3(-1) - 3(1) = 7 - 3 - 3 = 1$.Since $|A| 
eq 0$,the inverse exists.Next,find the cofactors $C_{ij}$ of matrix $A$:$C_{11} = +\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1, C_{12} = -\begin{vmatrix} -1 & 0 \ -1 & 1 \end{vmatrix} = 1, C_{13} = +\begin{vmatrix} -1 & 1 \ -1 & 0 \end{vmatrix} = 1$.$C_{21} = -\begin{vmatrix} -3 & -3 \ 0 & 1 \end{vmatrix} = 3, C_{22} = +\begin{vmatrix} 7 & -3 \ -1 & 1 \end{vmatrix} = 4, C_{23} = -\begin{vmatrix} 7 & -3 \ -1 & 0 \end{vmatrix} = 3$.$C_{31} = +\begin{vmatrix} -3 & -3 \ 1 & 0 \end{vmatrix} = 3, C_{32} = -\begin{vmatrix} 7 & -3 \ -1 & 0 \end{vmatrix} = 3, C_{33} = +\begin{vmatrix} 7 & -3 \ -1 & 1 \end{vmatrix} = 4$.The adjoint matrix $\operatorname{adj}(A)$ is the transpose of the cofactor matrix:$\operatorname{adj}(A) = \left[\begin{array}{ccc} 1 & 1 & 1 \ 3 & 4 & 3 \ 3 & 3 & 4 \end{array}ight]^T = \left[\begin{array}{ccc} 1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4 \end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = \frac{1}{1} \left[\begin{array}{ccc} 1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4 \end{array}ight] = \left[\begin{array}{ccc} 1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4 \end{array}ight]$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

The inverse of the matrix $\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is

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