Let A = begin{bmatrix} 5 & sin^2 theta & cos^ | vedclass

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(A) Given $A = \begin{bmatrix} 5 & \sin^2 	heta & \cos^2 	heta \ -\sin^2 	heta & -5 & 1 \ \cos^2 	heta & 1 & 5 \end{bmatrix}$.Expanding the dete

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$-125$

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(A) Given $A = \begin{bmatrix} 5 & \sin^2 	heta & \cos^2 	heta \ -\sin^2 	heta & -5 & 1 \ \cos^2 	heta & 1 & 5 \end{bmatrix}$.Expanding the determinant along the first row:$\det(A) = 5((-5)(5) - (1)(1)) - \sin^2 	heta((-\sin^2 	heta)(5) - (1)(\cos^2 	heta)) + \cos^2 	heta((-\sin^2 	heta)(1) - (-5)(\cos^2 	heta))$$= 5(-25 - 1) - \sin^2 	heta(-5\sin^2 	heta - \cos^2 	heta) + \cos^2 	heta(-\sin^2 	heta + 5\cos^2 	heta)$$= 5(-26) + 5\sin^4 	heta + \sin^2 	heta \cos^2 	heta - \sin^2 	heta \cos^2 	heta + 5\cos^4 	heta$$= -130 + 5(\sin^4 	heta + \cos^4 	heta)$Using the identity $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2\sin^2 	heta \cos^2 	heta = 1 - 2\sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2}\sin^2 2	heta$.$\det(A) = -130 + 5(1 - \frac{1}{2}\sin^2 2	heta) = -130 + 5 - \frac{5}{2}\sin^2 2	heta = -125 - \frac{5}{2}\sin^2 2	heta$.Since $\sin^2 2	heta \in [0, 1]$,the expression $-125 - \frac{5}{2}\sin^2 2	heta$ is maximized when $\sin^2 2	heta = 0$.Therefore,the maximum value is $-125 - 0 = -125$.

List-$I$	List-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

Let $A = \begin{bmatrix} 5 & \sin^2 \theta & \cos^2 \theta \\ -\sin^2 \theta & -5 & 1 \\ \cos^2 \theta & 1 & 5 \end{bmatrix}$. Then the maximum value of $\det(A)$ is

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