Question

Using elementary transformations, find the inverse of the followng matrix.

$\left[\begin{array}{cc}4& 5\\ 3& 4\end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{cc}4& 5\\ 3& 4\end{array}\right]$, We know that A=IA
$\therefore \left[\begin{array}{cc}4& 5\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A\Rightarrow \left[\begin{array}{cc}1& \frac{5}{4}\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}\frac{1}{4}& 0\\ 0& 1\end{array}\right]A$ (Using $R_1\to \frac{1}{4}{R}_1)$
$\Rightarrow \left[\begin{array}{cc}1& \frac{5}{4}\\ 0& \frac{1}{4}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{4}& 0\\ -\frac{3}{4}& 1\end{array}\right]A$ (Using $R_2\to R_2-3R_1)$
$\Rightarrow \left[\begin{array}{cc}1& \frac{4}{5}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{4}& 0\\ -3& 4\end{array}\right]A$ (Using $R_2\to 4R_2)$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}4& -5\\ -3& 4\end{array}\right]A$ (Using $R_1\to R_1-\frac{5}{4}{R}_2)$
Hence, $A^{-1}=\left[\begin{array}{cc}4& -5\\ -3& 4\end{array}\right](\because A{A}^{-1}=I)$