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Question: Answered & Verified by Expert
If $A^{-1}=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$ and $B^{-1}=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$, then $(A B)^{-1}=$
MathematicsMatricesMHT CETMHT CET 2021 (20 Sep Shift 2)
Options:
  • A $\left[\begin{array}{cc}2 & 7 \\ 3 & -1\end{array}\right]$
  • B $\left[\begin{array}{cc}2 & -7 \\ -3 & 11\end{array}\right]$
  • C $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
  • D $\left[\begin{array}{cc}2 & 3 \\ 7 & -11\end{array}\right]$
Solution:
2382 Upvotes Verified Answer
The correct answer is: $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
$$
\begin{aligned}
& (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1} \\
& =\left[\begin{array}{cc}
1 & 0 \\
-3 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+0 & -3+0 \\
-6-1 & 9+2
\end{array}\right]=\left[\begin{array}{cc}
2 & -3 \\
-7 & 11
\end{array}\right]
\end{aligned}
$$

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