Q96 / 180 · COMEDK 2019 · Mathematics
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then $A^{-1}=$
- Correct answerA$\frac{-1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
- 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$\left[\begin{array}{ll}2 & 4 \\ 1 & 3\end{array}\right]$
Solution
If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then
$$
A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$$
So, if $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then
$$
\begin{array}{r}
A^{-1}=\frac{1}{4-6}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right] \\
=-\frac{1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
\end{array}
$$
Hence, the correct answer is option: 'A'.