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

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(C) First, we calculate the determinant of $A$: $|A| = 1(1-3) - 2(2-1) + 3(6-1) = 1(-2) - 2(1) + 3(5) = -2 - 2 + 15 = 11$. Next, we calculate the dete

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$198$

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(C) First, we calculate the determinant of $A$: $|A| = 1(1-3) - 2(2-1) + 3(6-1) = 1(-2) - 2(1) + 3(5) = -2 - 2 + 15 = 11$. Next, we calculate the determinant of $B$: $|B| = 2(4-8) - 3(6-4) + 4(12-4) = 2(-4) - 3(2) + 4(8) = -8 - 6 + 32 = 18$. Since $|AB| = |A| 	imes |B|$, we have $|AB| = 11 	imes 18 = 198$. Using the property of the adjoint matrix, $|\operatorname{Adj}(M)| = |M|^{n-1}$ where $n$ is the order of the matrix. Here $n=3$, so $|\operatorname{Adj}(AB)| = |AB|^{3-1} = |AB|^2$. Thus, $\sqrt{|\operatorname{Adj}(AB)|} = \sqrt{|AB|^2} = |AB| = 198$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{bmatrix}$, then $\sqrt{|\operatorname{Adj}(AB)|} = $

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