Let M=begin{bmatrix} sin^4 theta & -1-sin^2 t | vedclass

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(B) Given $M = \alpha I + \beta M^{-1}$. Multiplying by $M$,we get $M^2 = \alpha M + \beta I$,or $M^2 - \alpha M - \beta I = O$. By the Cayley-Hamilto

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$-\frac{29}{16}$

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(B) Given $M = \alpha I + \beta M^{-1}$. Multiplying by $M$,we get $M^2 = \alpha M + \beta I$,or $M^2 - \alpha M - \beta I = O$. By the Cayley-Hamilton theorem,$M^2 - 	ext{tr}(M)M + \det(M)I = O$. Comparing coefficients,$\alpha = 	ext{tr}(M) = \sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2\sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2}\sin^2(2	heta)$. Since $\sin^2(2	heta) \in [0, 1]$,the minimum value $\alpha^* = 1 - \frac{1}{2} = \frac{1}{2}$. Also,$-\beta = \det(M) = \sin^4 	heta \cos^4 	heta + (1+\cos^2 	heta)(1+\sin^2 	heta) = \frac{\sin^4(2	heta)}{16} + 1 + \sin^2 	heta + \cos^2 	heta + \sin^2 	heta \cos^2 	heta = \frac{\sin^4(2	heta)}{16} + 2 + \frac{\sin^2(2	heta)}{4}$. Let $t = \sin^2(2	heta) \in [0, 1]$. Then $-\beta(t) = \frac{t^2}{16} + \frac{t}{4} + 2$. To find the minimum of $\beta$,we find the maximum of $-\beta$. Since $f(t) = \frac{t^2}{16} + \frac{t}{4} + 2$ is increasing on $[0, 1]$,its maximum is at $t=1$,which is $\frac{1}{16} + \frac{4}{16} + \frac{32}{16} = \frac{37}{16}$. Thus,$\beta^* = -\frac{37}{16}$. Therefore,$\alpha^* + \beta^* = \frac{1}{2} - \frac{37}{16} = \frac{8-37}{16} = -\frac{29}{16}$.

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