Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix}1&-1\\2&3\end{bmatrix}\)
Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix}1&-1\\2&3\end{bmatrix}\)
Solution and Explanation
Let A= \(\begin{bmatrix}1&-1\\2&3\end{bmatrix}\) We know that A = IA
\(\therefore \begin{bmatrix}1&-1\\2&3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A\)
\(\Rightarrow\begin{bmatrix}1&-1\\0&5\end{bmatrix}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}A\) (R2 \(\to\) R2-2R1)
\(\Rightarrow\begin{bmatrix}1&-1\\0&5\end{bmatrix}=\begin{bmatrix}1&0\\-\frac{2}{5}&\frac{1}{5}\end{bmatrix}A\) (R2 \(\to \frac{1}{5R_2}\))
\(\Rightarrow\begin{bmatrix}1&-1\\0&5\end{bmatrix}=\begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{2}{5}&\frac{1}{5}\end{bmatrix}A\) (R1 \(\to\) R1+R2)
so A-1= \(\begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\-\frac{2}{5}&\frac{1}{5}\end{bmatrix}\)
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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.
