If A = begin{bmatrix} 2 & 1 & 3 -1 & 2 & 0 | vedclass

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(D) We know that $\operatorname{det}(kM) = k^n \operatorname{det}(M)$ for an $n 	imes n$ matrix $M$,and $\operatorname{det}(XY) = \operatorname{det}(

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$\frac{-1}{6}$

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(D) We know that $\operatorname{det}(kM) = k^n \operatorname{det}(M)$ for an $n 	imes n$ matrix $M$,and $\operatorname{det}(XY) = \operatorname{det}(X) \operatorname{det}(Y)$.Given $A = \begin{bmatrix} 2 & 1 & 3 \ -1 & 2 & 0 \ 4 & 1 & 3 \end{bmatrix}$.$|A| = 2(6 - 0) - 1(-3 - 0) + 3(-1 - 8) = 12 + 3 - 27 = -12$.Given $B = \begin{bmatrix} 3 & 2 & 1 \ 1 & 2 & 3 \ 3 & 1 & 0 \end{bmatrix}$.$|B| = 3(0 - 3) - 2(0 - 9) + 1(1 - 6) = -9 + 18 - 5 = 4$.We need to find $\operatorname{det}(2 B^{-1} A^{-1})$.Since $A$ and $B$ are $3 	imes 3$ matrices,$\operatorname{det}(2 B^{-1} A^{-1}) = 2^3 \operatorname{det}(B^{-1}) \operatorname{det}(A^{-1})$.Using the property $\operatorname{det}(M^{-1}) = \frac{1}{\operatorname{det}(M)}$,we have:$\operatorname{det}(2 B^{-1} A^{-1}) = 8 	imes \frac{1}{\operatorname{det}(B)} 	imes \frac{1}{\operatorname{det}(A)}$.Substituting the values:$= 8 	imes \frac{1}{4} 	imes \frac{1}{-12} = 2 	imes \left(-\frac{1}{12}ight) = -\frac{1}{6}$.

If $A = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$,then $\operatorname{det}(2 B^{-1} A^{-1})$ is equal to

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