If A=left[begin{array}{ccc}1 & 2 & 1 -1 & 1 | vedclass

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(C) First,calculate the product $AB$: $AB = \left[\begin{array}{ccc}1 & 2 & 1 \ -1 & 1 & 3\end{array}ight] \left[\begin{array}{cc}1 & 2 \ -3 & 1 \

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$\left[\begin{array}{ll}-5 & 6 \ -4 & 5\end{array}ight]$

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(C) First,calculate the product $AB$: $AB = \left[\begin{array}{ccc}1 & 2 & 1 \ -1 & 1 & 3\end{array}ight] \left[\begin{array}{cc}1 & 2 \ -3 & 1 \ 0 & 2\end{array}ight]$ $AB = \left[\begin{array}{cc} (1)(1) + (2)(-3) + (1)(0) & (1)(2) + (2)(1) + (1)(2) \ (-1)(1) + (1)(-3) + (3)(0) & (-1)(2) + (1)(1) + (3)(2) \end{array}ight]$ $AB = \left[\begin{array}{cc} 1 - 6 + 0 & 2 + 2 + 2 \ -1 - 3 + 0 & -2 + 1 + 6 \end{array}ight] = \left[\begin{array}{cc} -5 & 6 \ -4 & 5 \end{array}ight]$ Now,find the inverse of the matrix $M = \left[\begin{array}{cc} -5 & 6 \ -4 & 5 \end{array}ight]$. The determinant $|M| = (-5)(5) - (6)(-4) = -25 + 24 = -1$. The inverse is given by $M^{-1} = \frac{1}{|M|} 	ext{adj}(M)$. For a $2 	imes 2$ matrix $\left[\begin{array}{cc} a & b \ c & d \end{array}ight]$,the adjoint is $\left[\begin{array}{cc} d & -b \ -c & a \end{array}ight]$. So,$	ext{adj}(M) = \left[\begin{array}{cc} 5 & -6 \ 4 & -5 \end{array}ight]$. $(AB)^{-1} = \frac{1}{-1} \left[\begin{array}{cc} 5 & -6 \ 4 & -5 \end{array}ight] = \left[\begin{array}{cc} -5 & 6 \ -4 & 5 \end{array}ight]$.

If $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -3 & 1 \\ 0 & 2\end{array}\right]$,then find $(AB)^{-1}$.

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