Question

The inverse of the 2 $\times$ 2 matrix $\left[\begin{array}{cc}1& 2\\ 5& 7\end{array}\right]$ is

• A. $\frac{1}{3}\left[\begin{array}{cc}-7& 2\\ 5& -1\end{array}\right]$
• B. $\frac{1}{3}\left[\begin{array}{cc}7& 2\\ 5& 1\end{array}\right]$
• C. $\frac{1}{3}\left[\begin{array}{cc}7& -2\\ -5& 1\end{array}\right]$
• D. $\frac{1}{3}\left[\begin{array}{cc}-7& -2\\ -5& -1\end{array}\right]$

Answer 1

The correct option is A $\frac{1}{3}\left[\begin{array}{cc}-7& 2\\ 5& -1\end{array}\right]$

Formula, A= $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$A^{-1}=\frac{1}{(ad-bc)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

For $A=\left[\begin{array}{cc}1& 2\\ 5& 7\end{array}\right]$

$A^{-1}=\frac{1}{(1\times 7-5\times 2)}\left[\begin{array}{cc}7& -2\\ -5& 1\end{array}\right]$

= $\frac{1}{3}\left[\begin{array}{cc}-7& 2\\ 5& -1\end{array}\right]$