Question

If $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, then $A^{-1}=$

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

Answer 1

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

For $2\times 2$ matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, we swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant $(ad-bc).$

So for $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ inverse of matrice is

$ad-bc=1\times 4-2\times 3=-2$

$A^{-1}=$ $\frac{1}{-2}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]$