MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]\), then the value of determinant of \(A^{-1}\) is
- A \(-6\)
- B \(\frac{-1}{6}\)
- C \(\frac{1}{36}\)
- D 36
Answer & Solution
Correct Answer
(B) \(\frac{-1}{6}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { We have } A=\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right] \\
& \therefore|A|=(2-6)+(0-2)=-6
\end{aligned}
\)
We know that \(\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}\)
\(
\therefore\left|\mathrm{A}^{-1}\right|=\frac{1}{-6}
\)
\begin{aligned}
& \text { We have } A=\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right] \\
& \therefore|A|=(2-6)+(0-2)=-6
\end{aligned}
\)
We know that \(\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}\)
\(
\therefore\left|\mathrm{A}^{-1}\right|=\frac{1}{-6}
\)
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