Find the inverse of each of the matrices, if it exists.\(\begin{bmatrix}2&5\\1&3\end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix}2&5\\1&3\end{bmatrix}\) We know that \(A = IA\)
so \([\begin{bmatrix}2&5\\1&3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A\)
\(\implies\begin{bmatrix}1& \frac{5}{2}\\ 1&3\end{bmatrix}=\begin{bmatrix}\frac{1}{2}& 0\\ 0&1\end{bmatrix}A (R_1\rightarrow\frac{1}{2R_1})\)
\(\implies \begin{bmatrix}1& \frac{5}{2}\\ 0& \frac{1}{2}\end{bmatrix}]=\begin{bmatrix}\frac{1}{2}& 0\\ \frac{-1}{2}& 1\end{bmatrix}A (R_2\rightarrow R_2-R_1)\)
\(\implies\begin{bmatrix}1&0\\0&\frac{1}{2}\end{bmatrix}=\begin{bmatrix}3&-5\\ \frac{-1}{2}& 1\end{bmatrix}A (R_1\rightarrow R_2-5R_2)\)
\(\implies\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}A (R_2\rightarrow2R_2)\)
therefore \(A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}\)
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Concepts Used:
Invertible matrices
A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In is an identity matrix of order n × n.
For example,
It can be observed that the determinant of the following matrices is non-zero.
