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

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(C) Let $A = \left[ {\begin{array}{*{20}{c}}3&{ - 2}&{ - 1}\{ - 4}&1&{ - 1}\2&0&1\end{array}} ight]$.First,we calculate the determinant $|A|$:$|A|

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$\left[ {\begin{array}{*{20}{c}}1&2&3\2&5&7\{ - 2}&{ - 4}&{ - 5}\end{array}} ight]$

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(C) Let $A = \left[ {\begin{array}{*{20}{c}}3&{ - 2}&{ - 1}\{ - 4}&1&{ - 1}\2&0&1\end{array}} ight]$.First,we calculate the determinant $|A|$:$|A| = 3(1(1) - (-1)(0)) - (-2)((-4)(1) - (-1)(2)) + (-1)((-4)(0) - 1(2))$$|A| = 3(1) + 2(-4 + 2) - 1(0 - 2) = 3 - 4 + 2 = 1$.Next,we find the matrix of cofactors $C_{ij}$:$C_{11} = +((1)(1) - (-1)(0)) = 1$$C_{12} = -((-4)(1) - (-1)(2)) = -(-4 + 2) = 2$$C_{13} = +((-4)(0) - (1)(2)) = -2$$C_{21} = -((-2)(1) - (-1)(0)) = -(-2) = 2$$C_{22} = +((3)(1) - (-1)(2)) = 3 + 2 = 5$$C_{23} = -((3)(0) - (-2)(2)) = -(4) = -4$$C_{31} = +((-2)(-1) - (1)(-1)) = 2 + 1 = 3$$C_{32} = -((3)(-1) - (-4)(-1)) = -(-3 - 4) = 7$$C_{33} = +((3)(1) - (-2)(-4)) = 3 - 8 = -5$The cofactor matrix is $\left[ {\begin{array}{*{20}{c}}1&2&{-2}\2&5&{-4}\3&7&{-5}\end{array}} ight]$.The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:$adj(A) = \left[ {\begin{array}{*{20}{c}}1&2&3\2&5&7\{-2}&{-4}&{-5}\end{array}} ight]$.Since $A^{-1} = \frac{1}{|A|} adj(A)$ and $|A| = 1$,we have:$A^{-1} = \left[ {\begin{array}{*{20}{c}}1&2&3\2&5&7\{-2}&{-4}&{-5}\end{array}} ight]$.Thus,the correct option is $C$.

The inverse of the matrix $\left[ {\begin{array}{*{20}{c}}3&{ - 2}&{ - 1}\\{ - 4}&1&{ - 1}\\2&0&1\end{array}} \right]$ is

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