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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2 \\ 1 & 2\end{array}\right]$, then $(A B)^{-1}=$
MathematicsMatricesMHT CETMHT CET 2022 (06 Aug Shift 2)
Options:
  • A $\left[\begin{array}{cc}\frac{-17}{5} & \frac{9}{5} \\ 2 & -1\end{array}\right]$
  • B $\left[\begin{array}{cc}\frac{17}{5} & \frac{9}{5} \\ 2 & 1\end{array}\right]$
  • C $\left[\begin{array}{cc}\frac{-17}{5} & 2 \\ \frac{-9}{5} & -1\end{array}\right]$
  • D $\left[\begin{array}{cc}\frac{-17}{5} & 2 \\ \frac{9}{5} & 1\end{array}\right]$
Solution:
2237 Upvotes Verified Answer
The correct answer is: $\left[\begin{array}{cc}\frac{-17}{5} & \frac{9}{5} \\ 2 & -1\end{array}\right]$
$\begin{aligned} & A B=\left[\begin{array}{ll}2+1+1 & 3+4+2 \\ 6+1+3 & 9+2+6\end{array}\right] \\ & =\left[\begin{array}{cc}5 & 9 \\ 10 & 17\end{array}\right] \\ & (A B)-1=\frac{1}{-5}\left[\begin{array}{cc}17 & -9 \\ -10 & 5\end{array}\right]=\left[\begin{array}{cc}-\frac{17}{5} & \frac{9}{5} \\ 2 & -1\end{array}\right]\end{aligned}$

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