Question

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

A. $$ - \frac{{31}}{{16}}$$
B. $$ - \frac{{17}}{{16}}$$
C. $$ - \frac{{37}}{{16}}$$
D. $$ - \frac{{29}}{{16}}$$  
Answer :   $$ - \frac{{29}}{{16}}$$
Solution :
\[M = \left[ \begin{array}{l} \,\,{\sin ^4}\theta {\rm{ }} - 1\sin \theta \\ {\rm{1 + co}}{{\rm{s}}^2}\theta \,\,\,\,\,\,\,\,\,\,{\cos ^4}\theta \end{array} \right]\]
$$\eqalign{ & \left| M \right| = {\sin ^4}\theta {\cos ^4}\theta + 1 + {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta {\cos ^2}\theta \cr & = 2 + {\sin ^2}\theta {\cos ^2}\theta + {\sin ^4}\theta {\cos ^4}\theta \cr} $$
\[{M^{ - 1}} = \frac{1}{{\left| M \right|}}\left[ \begin{array}{l} \,\,\,{\cos ^4}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\sin ^2}\theta \\ - 1 - {\cos ^2}\theta \,\,\,\,\,\,\,\,\,\,\,{\sin ^4}\theta \end{array} \right]\]
$${\text{Given}}\,{\text{that}}\,M = \alpha I + \beta {M^{ - 1}}$$
\[ \Rightarrow \,\left[ \begin{array}{l} \,\,\,{\sin ^4}\theta \,\,\,\,\,\,\,\,\, - 1 - {\sin ^2}\theta \\ 1 + {\cos ^2}\theta \,\,\,\,\,\,\,\,{\cos ^4}\theta \end{array} \right] = \left[ \begin{array}{l} \alpha\,\,\,\,\,0\\ 0\,\,\,\,\,a \end{array} \right] + \frac{\beta }{{\left| M \right|}}\left[ \begin{array}{l} \,\,\,{\cos ^4}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + {\sin ^2}\theta \\ - 1 - {\cos ^2}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,{\sin ^4}\theta \end{array} \right]\]
$$\eqalign{ & \Rightarrow \,\frac{\beta }{{\left| M \right|}} = - 1\,{\text{and}}\,\alpha \, + \frac{\beta }{{\left| M \right|}}{\cos ^4}\theta = {\sin ^4}\theta \cr & \Rightarrow \,\alpha = {\sin ^4}\theta + {\cos ^4}\theta \cr & \Rightarrow \,\beta = - \left[ {2 + {{\sin }^2}\theta {{\cos }^2}\theta + {{\sin }^4}\theta {{\cos }^4}\theta } \right] \cr & {\text{Now,}}\, \alpha \, = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 2{\sin ^2}\theta {\cos ^2}\theta \cr & = 1 - 2{\sin ^2}\theta {\cos ^2}\theta = 1 - \frac{1}{2}{\sin ^2}2\theta \cr} $$
For $$\alpha $$ to be minimum $${\sin ^2}2\theta $$  is maximum i.e., 1.
$$\eqalign{ & \therefore \,{\alpha ^ * } = 1 - \frac{1}{2} = \frac{1}{2} \cr & {\text{Also}},\beta = - \left[ {2 + \frac{1}{4}{{\sin }^2}2\theta + \frac{1}{{16}}{{\sin }^4}2\theta } \right] \cr} $$
For $$\beta $$ to be minimum, $${\sin ^2}2\theta $$  is maximum i.e.,
$$\eqalign{ & \therefore \,\,{\beta ^ * } = - \left[ {2 + \frac{1}{4} + \frac{1}{{16}}} \right] = - \frac{{32 + 4 + 1}}{{16}} = \frac{{ - 37}}{{16}} \cr & \therefore \,\,{\alpha ^ * } + {\beta ^ * } = \frac{1}{2} - \frac{{37}}{{16}} = \frac{{ - 29}}{{16}} \cr} $$

Releted MCQ Question on
Algebra >> Matrices and Determinants

Releted Question 1

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$  be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$  be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

A. $$C$$ is empty
B. $$B$$  has as many elements as $$C$$
C. $$A = B \cup C$$
D. $$B$$  has twice as many elements as elements as $$C$$
Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
Releted Question 3

Let $$a, b, c$$  be the real numbers. Then following system of equations in $$x, y$$  and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$     has

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$

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