If A = begin{bmatrix} 3 & 1 -1 & 2 end{bmatr | vedclass

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(B) Given $AB = C$. Multiplying by $A^{-1}$ on the left, we get $B = A^{-1}C$. Taking the inverse of both sides, $B^{-1} = (A^{-1}C)^{-1} = C^{-1}A$.F

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$\frac{1}{6} \begin{bmatrix} 3 & 0 \ -1 & 2 \end{bmatrix}$

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(B) Given $AB = C$. Multiplying by $A^{-1}$ on the left, we get $B = A^{-1}C$. Taking the inverse of both sides, $B^{-1} = (A^{-1}C)^{-1} = C^{-1}A$.First, find $C^{-1}$. The determinant $|C| = (7 	imes 6) - (3 	imes 0) = 42$. The adjoint $adj(C) = \begin{bmatrix} 6 & -3 \ 0 & 7 \end{bmatrix}$. Thus, $C^{-1} = \frac{1}{42} \begin{bmatrix} 6 & -3 \ 0 & 7 \end{bmatrix}$.Now, $B^{-1} = C^{-1}A = \frac{1}{42} \begin{bmatrix} 6 & -3 \ 0 & 7 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.$B^{-1} = \frac{1}{42} \begin{bmatrix} (6 	imes 3) + (-3 	imes -1) & (6 	imes 1) + (-3 	imes 2) \ (0 	imes 3) + (7 	imes -1) & (0 	imes 1) + (7 	imes 2) \end{bmatrix} = \frac{1}{42} \begin{bmatrix} 21 & 0 \ -7 & 14 \end{bmatrix}$.This does not match the options directly. Re-evaluating: $B = A^{-1}C \implies B^{-1} = C^{-1}A$. Calculation: $C^{-1}A = \frac{1}{42} \begin{bmatrix} 21 & 0 \ -7 & 14 \end{bmatrix} = \frac{7}{42} \begin{bmatrix} 3 & 0 \ -1 & 2 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 3 & 0 \ -1 & 2 \end{bmatrix}$.Thus, the correct option is $B$.

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, $C = \begin{bmatrix} 7 & 3 \\ 0 & 6 \end{bmatrix}$ and $AB = C$, then the inverse of matrix $B$ is

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