Question

If $\left[\begin{array}{cc}\frac{1}{4}& 0\\ x& \frac{1}{4}\end{array}\right]={\left[\begin{array}{cc}2& 0\\ -a& 2\end{array}\right]}^{-2}$, then the value of $\frac{a}{x}$ is equal of

• A. $\frac{1}{2}$
• B. $2$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $4$

$\left[\begin{array}{cc}\frac{1}{4}& 0\\ x& \frac{1}{4}\end{array}\right]={\left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]}^{-2}$

Multiplying by matrix $\to \left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]=A$

$\left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]\left[\begin{array}{cc}\frac{1}{4}& 0\\ x& \frac{1}{4}\end{array}\right]={\left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]}^{-1}\Rightarrow \left[\begin{array}{cc}\frac{1}{2}& 0\\ (\frac{-\alpha }{4}-2x)& \frac{1}{2}\end{array}\right]={\left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]}^{-1}$

Again multiplying by $-A$

$\Rightarrow \left[\begin{array}{cc}2& 0\\ -\alpha & 2\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& 0\\ (\frac{-\alpha }{4}-2x)& \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}1& 0\\ \frac{-\alpha }{2}+\frac{-\alpha }{4}-4x& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

On comparing

$-\alpha -4x=0-\alpha =-4x\frac{\alpha }{x}=4$