Q74 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If $A=\left[\begin{array}{ll}3 & 4 \\ 1 & 2\end{array}\right]$, then $A^{-1}$ equal to
- A$\left[\begin{array}{cc}3 & -4 \\ -1 & 2\end{array}\right]$
- B$\left[\begin{array}{cc}\frac{3}{2} & -2 \\ \frac{-1}{2} & 1\end{array}\right]$
- C$\left[\begin{array}{cc}2 & -4 \\ -1 & 3\end{array}\right]$
- Correct answerD$\left[\begin{array}{cc}1 & -2 \\ \frac{-1}{2} & \frac{3}{2}\end{array}\right]$
Solution
We have, $A=\left[\begin{array}{ll}3 & 4 \\ 1 & 2\end{array}\right]$
$\begin{aligned}<br/>\therefore & |A|=3 \times 2-1 \times 4=6-4=2 \\<br/>\Rightarrow \operatorname{adj} A= & {\left[\begin{array}{cc}<br/>2 & -4 \\<br/>-1 & 3<br/>\end{array}\right] } \\<br/>& {\left[\text { If } A=\left[\begin{array}{ll}<br/>a & b \\<br/>c & d<br/>\end{array}\right], \text { then } \operatorname{adj} A=\left[\begin{array}{cc}<br/>d & -b \\<br/>-c & a<br/>\end{array}\right]\right] }<br/>\end{aligned}$
Hence, $A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
$A^{-1}=\frac{1}{2}\left[\begin{array}{cc}<br/>2 & -4 \\<br/>-1 & 3<br/>\end{array}\right]=\left[\begin{array}{cc}<br/>1 & -2 \\<br/>-\frac{1}{2} & \frac{3}{2}<br/>\end{array}\right]$
$\begin{aligned}<br/>\therefore & |A|=3 \times 2-1 \times 4=6-4=2 \\<br/>\Rightarrow \operatorname{adj} A= & {\left[\begin{array}{cc}<br/>2 & -4 \\<br/>-1 & 3<br/>\end{array}\right] } \\<br/>& {\left[\text { If } A=\left[\begin{array}{ll}<br/>a & b \\<br/>c & d<br/>\end{array}\right], \text { then } \operatorname{adj} A=\left[\begin{array}{cc}<br/>d & -b \\<br/>-c & a<br/>\end{array}\right]\right] }<br/>\end{aligned}$
Hence, $A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
$A^{-1}=\frac{1}{2}\left[\begin{array}{cc}<br/>2 & -4 \\<br/>-1 & 3<br/>\end{array}\right]=\left[\begin{array}{cc}<br/>1 & -2 \\<br/>-\frac{1}{2} & \frac{3}{2}<br/>\end{array}\right]$
Hence, the correct answer is option: 'D'.