Question

The inverse of the matrix $A=\left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right]$ is

• A. $\left[\begin{array}{cc}4& -1\\ -7& 2\end{array}\right]$
• B. $\left[\begin{array}{cc}4& 1\\ 7& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}2& 7\\ 1& 4\end{array}\right]$
• D. $\left[\begin{array}{cc}7& 2\\ 1& 4\end{array}\right]$

Answer 1

Answer: (A)

Let $A=\left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right]$. We know that, $A=IA$
$\therefore \left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A\Rightarrow \left[\begin{array}{cc}1& 1/2\\ 7& 4\end{array}\right]=\left[\begin{array}{cc}1/2& 0\\ 0& 1\end{array}\right]A$ $($ using $R_1\to \frac{1}{2}{R}_1)$
$\Rightarrow \left[\begin{array}{cc}1& 1/2\\ 0& 1/2\end{array}\right]=\left[\begin{array}{cc}1/2& 0\\ -7/2& 1\end{array}\right]A$ (using $R_2\to R_2-7R_1)$
$\Rightarrow \left[\begin{array}{cc}1& 1/2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1/2& 0\\ -7& 2\end{array}\right]A$ (using $R_2\to 2R_2$ )
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}4& -1\\ -7& 2\end{array}\right]A($ using $R_1\to R_1-\frac{1}{2}{R}_2)$
$\therefore A^{-1}=\left[\begin{array}{cc}4& -1\\ -7& 2\end{array}\right]$