Question

The inverse of the $2\times 2$ matrix $\left[\begin{array}{cc}2& 3\\ 6& 8\end{array}\right]$ is

• A. $\left[\begin{array}{cc}3& -1\\ -4& \frac{3}{2}\end{array}\right]$
• B. $\left[\begin{array}{cc}-4& \frac{3}{2}\\ 3& -1\end{array}\right]$
• C. $\left[\begin{array}{cc}6& -2\\ -8& 3\end{array}\right]$
• D. $\left[\begin{array}{cc}4& -\frac{3}{2}\\ -3& 1\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}-4& \frac{3}{2}\\ 3& -1\end{array}\right]$

Inverse of $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is
${\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
$\therefore {\left[\begin{array}{cc}2& 3\\ 6& 8\end{array}\right]}^{-1}=\frac{1}{16-18}\left[\begin{array}{cc}8& -3\\ -6& 2\end{array}\right]$
$=\frac{1}{-2}\left[\begin{array}{cc}8& -3\\ -6& 2\end{array}\right]=\left[\begin{array}{cc}-4& \frac{3}{2}\\ 3& -1\end{array}\right]$