KCET · Maths · Matrices
If \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\), then \(\mathrm{A}^{-1}\) is equal to
- 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 & Solution
Correct Answer
(A) \(-\frac{1}{2}\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\)
Now, \(|A|=4-6=-2\)
\[
\begin{aligned}
\operatorname{adj}(A) &=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right] \\
\therefore \quad A^{-1} &=\frac{\operatorname{adj}(A)}{|A|}=-\frac{1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
\end{aligned}
\]
Now, \(|A|=4-6=-2\)
\[
\begin{aligned}
\operatorname{adj}(A) &=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right] \\
\therefore \quad A^{-1} &=\frac{\operatorname{adj}(A)}{|A|}=-\frac{1}{2}\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]
\end{aligned}
\]
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