MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\), then \((\mathrm{AB})^{-1}\) is
- A \(\left(\frac{1}{5}\right)\left[\begin{array}{ll}5 & -5 \\ 4 & -5\end{array}\right]\)
- B \(\left(\frac{1}{5}\right)\left[\begin{array}{cc}5 & -5 \\ -4 & 5\end{array}\right]\)
- C \(\left(\frac{1}{5}\right)\left[\begin{array}{cc}5 & -5 \\ 4 & 5\end{array}\right]\)
- D \(\left(\frac{1}{5}\right)\left[\begin{array}{cc}5 & -5 \\ -4 & -5\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{1}{5}\right)\left[\begin{array}{cc}5 & -5 \\ -4 & 5\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 2 \\ 2 & 1 \\ 0 & 1\end{array}\right]\)
\(\therefore A B\)
\(
\begin{array}{l}(A B)^{-1} \\ =\frac{1}{|A B|}\left[\begin{array}{cc}1+4+0 & 2+2+1 \\ 2+2+0 & 4+1+0\end{array}\right]=\left[\begin{array}{ll}5 & 5 \\ 4 & 5\end{array}\right]\end{array}
\)
\(\therefore A B\)
\(
\begin{array}{l}(A B)^{-1} \\ =\frac{1}{|A B|}\left[\begin{array}{cc}1+4+0 & 2+2+1 \\ 2+2+0 & 4+1+0\end{array}\right]=\left[\begin{array}{ll}5 & 5 \\ 4 & 5\end{array}\right]\end{array}
\)
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