Question

Let $A=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$ and matrix $B$ is defined such that $B=A+3A^2+3A^3+A^4$ . If $\left|B\right|=8$ , then the number of values of $\alpha$ in $\left[0,10\pi \right]$ is

• A. $10$
• B. $12$
• C. $5$
• D. $3$

Answer 1

Answer: (A)

$B=A\left(I+3A+3A^2+A^3\right)=A{\left(I+A\right)}^3$
$\left|B\right|=\left|A\right|{\left(\left|I+A\right|\right)}^3=8\dots \left(i\right)$
$A=\left|\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right|=co{s}^2\alpha +si{n}^2\alpha =1$
$\left|I+A\right|=\left|\begin{array}{cc}1+cos\alpha & sin\alpha \\ -sin\alpha & 1+cos\alpha \end{array}\right|={\left(1+cos\alpha \right)}^2+{\left(sin\right)}^2\alpha =2+2cos\alpha$
From $\left(i\right)$
$1{\left(2+2cos\alpha \right)}^3=8\Rightarrow 2+2cos\alpha =2$
$\Rightarrow cos\alpha =0$
$\alpha =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2},\frac{9\pi }{2},\frac{11\pi }{2},\frac{13\pi }{2},\frac{15\pi }{2},\frac{17\pi }{2},\frac{19\pi }{2}$
Total number of values $=10$