The inverse matrix of left[begin{array}{lll}0 | vedclass

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Question

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

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$\left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}ight]$

Vedclass · Answer

(A) Let $A = \left[\begin{array}{lll}0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1\end{array}ight]$.First,we calculate the determinant $|A|$:$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.Since $|A| 
eq 0$,the inverse exists.Next,we find the matrix of cofactors $C_{ij}$:$C_{11} = +(2-3) = -1, C_{12} = -(1-9) = 8, C_{13} = +(1-6) = -5$.$C_{21} = -(1-2) = 1, C_{22} = +(0-6) = -6, C_{23} = -(0-3) = 3$.$C_{31} = +(3-4) = -1, C_{32} = -(0-2) = 2, C_{33} = +(0-1) = -1$.Thus,$\operatorname{adj}(A) = \left[\begin{array}{rrr}-1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1\end{array}ight]^T = \left[\begin{array}{rrr}-1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1\end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = \frac{1}{-2} \left[\begin{array}{rrr}-1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1\end{array}ight] = \left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ -4 & 3 & -1 \ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}ight]$.

The inverse matrix of $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ is

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