Question

If the square of the matrix $\left(\genfrac{}{}{0ex}{}{ab}{a-a}\right)$ is the unit matrix, then $b$ is equal to

• A. $\frac{a}{1+a^2}$
• B. $\frac{1-a^2}{a}$
• C. $\frac{1+a^2}{a}$
• D. $\frac{a}{1-a^2}$
• E. $1+a^2$

Answer 1

Answer: (B)

Given, ${\left[\begin{array}{cc}a& b\\ a& -a\end{array}\right]}^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}a& b\\ a& -a\end{array}\right]\left[\begin{array}{cc}a& b\\ a& -a\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}{a}^2+ab& ab-ab\\ a^2-a^2& ab+a^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}{}^2+ab& 0\\ 0& ab+a^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
On equating the elements, we get
$a^2+ab=1$
$\Rightarrow ab=1-a^2\Rightarrow b=\frac{1-a^2}{a}$