MHT CET · Maths · Matrices
If \(\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\), then \(A=\)
- A \(\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)
- B \(\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\)
- C \(\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)
- D \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & {\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]} \\ & \Rightarrow A=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]^{-1}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right]^{-1} \\ & \Rightarrow A=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & \Rightarrow A=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & \Rightarrow A=\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\end{aligned}\)
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