Find the inverse of each of the matrices (if | vedclass

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(B) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}ight]$.First,we find the determinant $|A|$:$|A| = 1((-3)(1) - (0

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$-\frac{1}{3}\left[\begin{array}{ccc}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}ight]$

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(B) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}ight]$.First,we find the determinant $|A|$:$|A| = 1((-3)(1) - (0)(2)) - 0 + 0 = -3$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the cofactor matrix $C_{ij}$:$C_{11} = +((-3) - 0) = -3$$C_{12} = -((-3) - 0) = 3$$C_{13} = +((6) - (15)) = -9$$C_{21} = -(0 - 0) = 0$$C_{22} = +((-1) - 0) = -1$$C_{23} = -(2 - 0) = -2$$C_{31} = +(0 - 0) = 0$$C_{32} = -(0 - 0) = 0$$C_{33} = +(3 - 0) = 3$The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:$adj(A) = \left[\begin{array}{ccc}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-3} \left[\begin{array}{ccc}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}ight] = -\frac{1}{3} \left[\begin{array}{ccc}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}ight]$.

Find the inverse of each of the matrices (if it exists). $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$

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