If AB = begin{bmatrix} -6 & 26 -1 & 19 end{b | vedclass

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(C) Given $AB = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.We know that $A = (AB

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$\begin{bmatrix} 2 & 4 \ 3 & 2 \end{bmatrix}$

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(C) Given $AB = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.We know that $A = (AB)B^{-1}$.From the second equation,$B^{-1} = \frac{1}{11} \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$.Substituting these into the expression for $A$:$A = \begin{bmatrix} -6 & 26 \ -1 & 19 \end{bmatrix} 	imes \frac{1}{11} \begin{bmatrix} 5 & -3 \ 2 & 1 \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} (-6)(5) + (26)(2) & (-6)(-3) + (26)(1) \ (-1)(5) + (19)(2) & (-1)(-3) + (19)(1) \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} -30 + 52 & 18 + 26 \ -5 + 38 & 3 + 19 \end{bmatrix}$$A = \frac{1}{11} \begin{bmatrix} 22 & 44 \ 33 & 22 \end{bmatrix}$$A = \begin{bmatrix} 2 & 4 \ 3 & 2 \end{bmatrix}$.Thus,the correct option is $C$.

If $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$,then $A = $ . . . . . . .

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