Which among the following is inverse of the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\) ?(a) \(\begin{bmatrix}\frac{1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)(b) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)(c) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\1&\frac{-2}{13}\end{bmatrix}\)(d) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&-2\end{bmatrix}\)I had been asked this question in examination.I need to ask this question from Invertible Matrices topic in section Matrices of Mathematics – Class 12

Which among the following is inverse of the matrix A=\(\begin{bmatrix}

1.	Which among the following is inverse of the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\) ?(a) \(\begin{bmatrix}\frac{1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)(b) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)(c) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\1&\frac{-2}{13}\end{bmatrix}\)(d) \(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&-2\end{bmatrix}\)I had been asked this question in examination.I need to ask this question from Invertible Matrices topic in section Matrices of Mathematics – Class 12
Answer» The CORRECT option is (b) \(\begin{BMATRIX}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\) The explanation: Consider the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\) Using ELEMENTARY row operation, we write A=IA. \(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A \(\begin{bmatrix}-13&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&-3\\0&1\end{bmatrix}\)A(Applying R1→R1-3R2) \(\begin{bmatrix}1&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\0&1\end{bmatrix}\)A(Applying \(R_1 \rightarrow -\frac{R_1}{13}\)) \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)A(Applying R2→R2-5R1) \(A^{-1}=\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\).

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