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 to

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

Answer 1

Answer: (C)

$4$

$A=\left[\begin{array}{cc}2& 0\\ -a& 2\end{array}\right]\Rightarrow adjA=\left[\begin{array}{cc}2& 0\\ a& 2\end{array}\right]\Rightarrow A^{-1}=\frac{1}{4}\left[\begin{array}{cc}2& 0\\ a& 2\end{array}\right]$

$\therefore A^{-2}=A^{-1}.A^{-1}=\frac{1}{16}\left[\begin{array}{cc}2& 0\\ a& 2\end{array}\right]\left[\begin{array}{cc}2& 0\\ a& 2\end{array}\right]=\frac{1}{16}\left[\begin{array}{cc}4& 0\\ 4a& 4\end{array}\right]=\left[\begin{array}{cc}1/4& 0\\ a/4& 1/4\end{array}\right]\Rightarrow x=\frac{a}{4}$

$\therefore \frac{a}{x}=4$