If A and B are non-singular matrices of order | vedclass

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(C) We know that $(AB)^{-1} = B^{-1} A^{-1}$.Given $(AB)^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$ and $A^{-1} = \frac{1}{3}

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

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(C) We know that $(AB)^{-1} = B^{-1} A^{-1}$.Given $(AB)^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$ and $A^{-1} = \frac{1}{3} \begin{bmatrix} 4 & 3 \ -1 & 0 \end{bmatrix}$.Substituting these into the formula,we get:$B^{-1} A^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$$B^{-1} \left( \frac{1}{3} \begin{bmatrix} 4 & 3 \ -1 & 0 \end{bmatrix} ight) = \frac{1}{6} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$$B^{-1} \begin{bmatrix} 4 & 3 \ -1 & 0 \end{bmatrix} = \frac{3}{6} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$Let $M = \begin{bmatrix} 4 & 3 \ -1 & 0 \end{bmatrix}$. Then $B^{-1} M = \frac{1}{2} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix}$.$B^{-1} = \frac{1}{2} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix} M^{-1}$.First,find $M^{-1} = \frac{1}{|M|} 	ext{adj}(M)$.$|M| = (4)(0) - (3)(-1) = 0 + 3 = 3$.$	ext{adj}(M) = \begin{bmatrix} 0 & -3 \ 1 & 4 \end{bmatrix}$.$M^{-1} = \frac{1}{3} \begin{bmatrix} 0 & -3 \ 1 & 4 \end{bmatrix}$.Now,$B^{-1} = \frac{1}{2} \begin{bmatrix} -7 & -3 \ 2 & 3 \end{bmatrix} \left( \frac{1}{3} \begin{bmatrix} 0 & -3 \ 1 & 4 \end{bmatrix} ight) = \frac{1}{6} \begin{bmatrix} (-7)(0) + (-3)(1) & (-7)(-3) + (-3)(4) \ (2)(0) + (3)(1) & (2)(-3) + (3)(4) \end{bmatrix}$$B^{-1} = \frac{1}{6} \begin{bmatrix} -3 & 21 - 12 \ 3 & -6 + 12 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} -3 & 9 \ 3 & 6 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1 & 3 \ 1 & 2 \end{bmatrix}$.

If $A$ and $B$ are non-singular matrices of order $2$ such that $(AB)^{-1} = \frac{1}{6} \begin{bmatrix} -7 & -3 \\ 2 & 3 \end{bmatrix}$ and $A^{-1} = \frac{1}{3} \begin{bmatrix} 4 & 3 \\ -1 & 0 \end{bmatrix}$,then $B^{-1} = $

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