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Mathematics Question on Sets

Let M=[sin4θ−1−sin2θ 1+cos2θcos4θ]=αI+βM−1M=\begin{bmatrix}sin^{4}\theta &-1-sin^{2}\theta\\\ 1+cos^{2}\theta&cos^{4}\theta\end{bmatrix}=\alpha I+\beta M^{-1} where α=α(θ)\alpha=\alpha\left(\theta\right) and β=β(θ)\beta=\beta\left(\theta\right) are real numbers, and II is the 2 x 2 identity matrix. If α∗\alpha^* is the minimum of the set \left\\{\alpha\left(\theta\right):\theta\,\in[\,0,2\pi)\right\\} and β∗\beta^* is the minimum of the set \left\\{\beta\left(\theta\right):\theta\,\in[\,0,2\pi)\right\\}, then the value of α∗+β∗\alpha^*+\beta^* is

A

−3716-\frac{37}{16}

B

−3116-\frac{31}{16}

C

−2916-\frac{29}{16}

D

−1716-\frac{17}{16}

Answer

−2916-\frac{29}{16}

Explanation

Solution

M=[sin4 θ−1−sin2 θ 1+cos2 θcos4 θ]=αl+βM−1M=\begin{bmatrix}sin^{4}\,\theta &-1-sin^{2}\,\theta\\\ 1+cos^{2}\,\theta &cos^{4}\,\theta\end{bmatrix}=\alpha l+\beta M^{-1}
∴det(M)=∣M∣=sin4θ⋅cos4θ+sin2θcos2θ+2\therefore det \left(M\right) = |M| = sin^{4}\theta\cdot cos^{4}\theta+sin^{2}\theta cos^{2}\theta+2
=\left\\{\left(sin^{2}\theta cos^{2}\theta+\frac{1}{2}\right)^{2}+\frac{7}{4}\right\\}
[sin4θ−1−sin2θ 1+cos2θcos4θ]=[α0 0α]+β∣M∣[cos4θ1+sin2θ −1−cos2θsin4θ]\begin{bmatrix}sin^{4}\theta&-1-sin^{2}\theta\\\ 1+cos^{2}\theta&cos^{4}\theta\end{bmatrix}=\begin{bmatrix}\alpha&0\\\ 0&\alpha\end{bmatrix}+\frac{\beta}{\left|M\right|}\begin{bmatrix}cos^{4}\theta&1+sin^{2}\theta\\\ -1-cos^{2}\theta&sin^{4}\theta\end{bmatrix}
∴α=cos4θ+sin4θ=1−12(sin2 2θ)\therefore \alpha=cos^{4}\theta+sin^{4}\theta=1-\frac{1}{2}\left(sin^{2}\,2\theta\right)
and β=−∣M∣\beta=-\left|M\right|
=-\left\\{\left(sin^{2}\,\theta\cdot cos^{2}\,\theta+\frac{1}{2}\right)^{2}+\frac{7}{4}\right\\}
∴αmin=12\therefore \alpha_{min}=\frac{1}{2} and βmin=−3716\beta_{min}=-\frac{37}{16}
∴α∗+β∗=12−3716=−2916\therefore \alpha^*+\beta^*=\frac{1}{2}-\frac{37}{16}=-\frac{29}{16}