Question

Let a matrix $A=\left[\begin{array}{cc}2{\sin }^{2}x& {\cos }^{2}x\\ -2& 1\end{array}\right]$ is such that sum of elements in $\text{Adj}\left(A\right)$ is equal to $\frac{11}{4}.$ Then the value of $x$ can be

• A. $n\pi +\frac{\pi }{6},n\in Z$
• B. $n\pi +\frac{\pi }{4},,n\in Z$
• C. $n\pi -\frac{\pi }{3},n\in Z$
• D. $n\pi -\frac{\pi }{2},n\in Z$

Answer 1

The correct option is A $n\pi +\frac{\pi }{6},n\in Z$

Given : $A=\left[\begin{array}{cc}2{\sin }^{2}x& {\cos }^{2}x\\ -2& 1\end{array}\right]$
$\Rightarrow \text{Adj}\left(A\right)=\left[\begin{array}{cc}1& -{\cos }^{2}x\\ 2& 2{\sin }^{2}x\end{array}\right]$
$\Rightarrow$Sum of elements in $\text{Adj}\left(A\right)$ is $3+2{\sin }^{2}x-{\cos }^{2}x=\frac{11}{4}\Rightarrow 3{\sin }^{2}x=\frac{3}{4}\Rightarrow \sin x=\pm \frac{1}{2}$
$\Rightarrow x=n\pi \pm \frac{\pi }{6},n\in Z$