Find the inverse of the matrix (if it exists) | vedclass

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(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$.First,calculate the determinant $|A|$ by expanding alon

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$\left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$

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(D) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$.First,calculate the determinant $|A|$ by expanding along the first column:$|A| = 1(8 - 6) - 0( -4 + 4) + 3(3 - 4) = 1(2) - 0 + 3(-1) = 2 - 3 = -1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,find the cofactor matrix $C_{ij}$:$C_{11} = +(8-6) = 2, C_{12} = -(0+9) = -9, C_{13} = +(0-6) = -6$$C_{21} = -(-4+4) = 0, C_{22} = +(4-6) = -2, C_{23} = -(-2+3) = -1$$C_{31} = +(3-4) = -1, C_{32} = -(-3-0) = 3, C_{33} = +(2-0) = 2$The adjoint matrix $	ext{adj } A$ is the transpose of the cofactor matrix:$	ext{adj } A = \left[\begin{array}{ccc}2 & 0 & -1 \ -9 & -2 & 3 \ -6 & -1 & 2\end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} 	ext{adj } A = \frac{1}{-1} \left[\begin{array}{ccc}2 & 0 & -1 \ -9 & -2 & 3 \ -6 & -1 & 2\end{array}ight] = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$.

Find the inverse of the matrix (if it exists): $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$

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