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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$, then $(A B)^{-1}=$
MathematicsMatricesMHT CETMHT CET 2020 (12 Oct Shift 2)
Options:
  • A $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
  • B $\left[\begin{array}{cc}2 & -3 \\ 7 & 11\end{array}\right]$
  • 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:
2058 Upvotes Verified Answer
The correct answer is: $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
$\begin{aligned} A B &=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]=\left[\begin{array}{cc}11 & 3 \\ 7 & 2\end{array}\right] \Rightarrow|A B|=\left|\begin{array}{cc}11 & 3 \\ 7 & 2\end{array}\right|=22-21=1 \\ \operatorname{adj}(A B) &=\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right] \Rightarrow(A B)^{-1}=\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right] \end{aligned}$

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