If A = begin{bmatrix} 1 & -3 & -5 -2 & 4 & - | vedclass

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(A) Given matrix $A = \begin{bmatrix} 1 & -3 & -5 \ -2 & 4 & -6 \ 7 & -11 & 13 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 1(4

Vedclass · Accepted Answer

$64$

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(A) Given matrix $A = \begin{bmatrix} 1 & -3 & -5 \ -2 & 4 & -6 \ 7 & -11 & 13 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 1(4 	imes 13 - (-6) 	imes (-11)) - (-3)((-2) 	imes 13 - (-6) 	imes 7) + (-5)((-2) 	imes (-11) - 4 	imes 7)$$|A| = 1(52 - 66) + 3(-26 + 42) - 5(22 - 28)$$|A| = 1(-14) + 3(16) - 5(-6)$$|A| = -14 + 48 + 30 = 64$.We know that $|\operatorname{Adj} A| = |A|^{n-1}$,where $n$ is the order of the matrix.Here,$n = 3$,so $|\operatorname{Adj} A| = |A|^{3-1} = |A|^2$.$|\operatorname{Adj} A| = (64)^2$.Therefore,$\sqrt{|\operatorname{Adj} A|} = \sqrt{(64)^2} = 64$.

If $A = \begin{bmatrix} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{bmatrix}$,then $\sqrt{|\operatorname{Adj} A|} = $

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