Question

Let $M=\begin{bmatrix}sin^{4}\theta &-1-sin^{2}\theta\\ 1+cos^{2}\theta&cos^{4}\theta\end{bmatrix}=\alpha I+\beta M^{-1}$ where $\alpha=\alpha\left(\theta\right)$ and $\beta=\beta\left(\theta\right)$ are real numbers, and $I$ is the 2 x 2 identity matrix. If $\alpha^*$ is the minimum of the set $\left\{\alpha\left(\theta\right):\theta\,\in[\,0,2\pi)\right\}$ and $\beta^*$ is the minimum of the set $\left\{\beta\left(\theta\right):\theta\,\in[\,0,2\pi)\right\}$, then the value of $\alpha^*+\beta^*$ is

• A. $-\frac{37}{16}$
• B. $-\frac{31}{16}$
• C. $-\frac{29}{16}$
• D. $-\frac{17}{16}$

Answer 1

The Correct Option is C

$M=\begin{bmatrix}sin^{4}\,\theta &-1-sin^{2}\,\theta\\ 1+cos^{2}\,\theta &cos^{4}\,\theta\end{bmatrix}=\alpha l+\beta M^{-1}$
$\therefore det \left(M\right) = |M| = sin^{4}\theta\cdot cos^{4}\theta+sin^{2}\theta cos^{2}\theta+2$
$=\left\{\left(sin^{2}\theta cos^{2}\theta+\frac{1}{2}\right)^{2}+\frac{7}{4}\right\}$
$\begin{bmatrix}sin^{4}\theta&-1-sin^{2}\theta\\ 1+cos^{2}\theta&cos^{4}\theta\end{bmatrix}=\begin{bmatrix}\alpha&0\\ 0&\alpha\end{bmatrix}+\frac{\beta}{\left|M\right|}\begin{bmatrix}cos^{4}\theta&1+sin^{2}\theta\\ -1-cos^{2}\theta&sin^{4}\theta\end{bmatrix}$
$\therefore \alpha=cos^{4}\theta+sin^{4}\theta=1-\frac{1}{2}\left(sin^{2}\,2\theta\right)$
and $\beta=-\left|M\right|$
$=-\left\{\left(sin^{2}\,\theta\cdot cos^{2}\,\theta+\frac{1}{2}\right)^{2}+\frac{7}{4}\right\}$
$\therefore \alpha_{min}=\frac{1}{2}$ and $\beta_{min}=-\frac{37}{16}$
$\therefore \alpha^*+\beta^*=\frac{1}{2}-\frac{37}{16}=-\frac{29}{16}$