Question
Find the inverse of the following matrices $\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$
✓
Answer
$\begin{array}{l}\text { Let } A=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right] \\ \therefore|A|=\left|\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right|=-1-4=-5 \neq 0\end{array}$
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{rr}1 & 2 \\2 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -5\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right]$
By $\left(-\frac{1}{5}\right) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\2 / 5 & -1 / 5\end{array}\right]$
By $R_1-2 R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right]} \\ \therefore A^{-1}=\frac{1}{5}\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right] .\end{array}$
The answer can be checked by finding the product $\mathrm{AA}^{-1}$
$\mathrm{AA}^{-1}=\left(\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right] $
$=\left[\begin{array}{ll}1\left(\frac{1}{5}\right)+2\left(\frac{2}{5}\right) & 1\left(\frac{2}{5}\right)+2\left(-\frac{1}{5}\right) \\ 2\left(\frac{1}{5}\right)-1\left(\frac{2}{5}\right) & 2\left(\frac{2}{5}\right)-1\left(-\frac{1}{5}\right)\end{array}\right] $
$ =\left[\begin{array}{lll}\frac{1}{5}+\frac{4}{5} & \frac{2}{5}-\frac{2}{5} \\ \frac{2}{5}-\frac{2}{5} & \frac{4}{5}+\frac{1}{5}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I} $
Hence, $\mathrm{A}^{-1}$ is the required answer.
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{rr}1 & 2 \\2 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -5\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right]$
By $\left(-\frac{1}{5}\right) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\2 / 5 & -1 / 5\end{array}\right]$
By $R_1-2 R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right]} \\ \therefore A^{-1}=\frac{1}{5}\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right] .\end{array}$
The answer can be checked by finding the product $\mathrm{AA}^{-1}$
$\mathrm{AA}^{-1}=\left(\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right] $
$=\left[\begin{array}{ll}1\left(\frac{1}{5}\right)+2\left(\frac{2}{5}\right) & 1\left(\frac{2}{5}\right)+2\left(-\frac{1}{5}\right) \\ 2\left(\frac{1}{5}\right)-1\left(\frac{2}{5}\right) & 2\left(\frac{2}{5}\right)-1\left(-\frac{1}{5}\right)\end{array}\right] $
$ =\left[\begin{array}{lll}\frac{1}{5}+\frac{4}{5} & \frac{2}{5}-\frac{2}{5} \\ \frac{2}{5}-\frac{2}{5} & \frac{4}{5}+\frac{1}{5}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I} $
Hence, $\mathrm{A}^{-1}$ is the required answer.
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