Question

$$LetM=\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\alpha I+\beta M^{-1},$$
Where $\alpha =\alpha \left(\theta \right)and\beta =\beta \left(\theta \right)$ are real numbers, and $I$ is the $2\times 2$ identity matrix. If ${\alpha }^{\ast }$ is the minimum of set $\left\{\alpha \right(\theta ):\theta \in [0,2\pi \left)\right\}$ and
${\beta }^{\ast }$ is the minimum of set $\left\{\beta \right(\theta ):\theta \in [0,2\pi \left]\right\}$, then the value of ${\alpha }^{\ast }$+${\beta }^{\ast }$ is

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

Answer 1

The correct option is C $-\frac{29}{16}$

$$M=\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\alpha I+\beta M^{-1},$$$M^2=\alpha M+\beta I$
$\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\alpha \left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]+\beta \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
On comparing both sides we get,
$\Rightarrow {\sin }^8\theta -1-{\sin }^2\theta -{\cos }^2\theta -{\sin }^2\theta \cdot {\cos }^2\theta =\alpha {\sin }^4\theta +\beta$
$\Rightarrow {\sin }^8\theta -{\sin }^2\theta \cdot {\cos }^2\theta -2=\alpha {\sin }^4\theta +\beta ...\left(i\right)$
$Also{\sin }^4\theta (1+{\cos }^2\theta )+{\cos }^4\theta (1+{\cos }^2\theta )=\alpha (1+{\cos }^2\theta )$
$\Rightarrow \frac{{\sin }^4\theta (1+{\cos }^2\theta )+{\cos }^4\theta (1+{\cos }^2\theta )}{(1+{\cos }^2\theta )}=\alpha$
$\alpha ={\sin }^4\theta +{\cos }^4\theta ...\left(ii\right)$
$\alpha =1-\frac{1}{2}(\sin 2\theta {)}^2$
$\alpha \text{is minimum when}\sin 2\theta \to 1$
$\therefore {\alpha }^{\ast }={\alpha }_{min}=1-\frac{1}{2}=\frac{1}{2}$
solving $\left(i\right)$ and $\left(ii\right)$ we get,
$\beta =-[{\sin }^4\theta .{\cos }^4\theta +{\sin }^2\theta .{\cos }^2\theta +2]\beta =-(t^2+t+2),\text{where}t=\frac{{\sin }^{2}2\theta }{4}\in [0,\frac{1}{4}]\therefore {\beta }_{min}={\beta }^{\ast }=-\frac{37}{16}$
$\therefore {\alpha }^{\ast }+{\beta }^{\ast }=\frac{1}{2}-\frac{37}{16}=-\frac{29}{16}$