If $(B A)^{-1}=C$ where $B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$…

If $(B A)^{-1}=C$ where $B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$ and $C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ is given by
  1. $\left[\begin{array}{ccc}-3 & -3 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
  2. $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
  3. $\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6\end{array}\right]$
  4. $\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

Solution

$(B A)^{-1}=C$ $\begin{aligned} & \Rightarrow A^{-1} B^{-1}=C \\ & \Rightarrow A^{-1} B^{-1} B=C B \\ & \Rightarrow A^{-1}=C B \\ & \Rightarrow A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]=\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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