Question
If $A =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then value of $A ^{-1}$ :

Answer

(B)
$A_{11}=4, A_{12}=-3, A_{21}=-2, A_{22}=1$$
\operatorname{adj} A=\left[\begin{array}{cc}
4 & -3 \\
-2 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
$
$
\begin{aligned}
|A| & =4-6=-2 \neq 0 \\
A^{-1}=\frac{\operatorname{adj} A}{|A|} & =\frac{-1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]=\left[\begin{array}{cc}
-2 & 1 \\
\frac{3}{2} & -1 / 2
\end{array}\right]
\end{aligned}
$
Hence correct option is (B).

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