Find the inverse of each of the matrices,if it exists.\(\begin{bmatrix}4&5\\3&4\end{bmatrix}\)
Solution and Explanation
Let \(A=\begin{bmatrix}4&5\\3&4\end{bmatrix}\) We know that \(A = IA\)
so \(\begin{bmatrix}4&5\\3&4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
\(\implies \begin{bmatrix}1&1\\3&4\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A (R_1\rightarrow R_1-R_2)\)
\(\implies\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\-3&4\end{bmatrix} A (R_2\rightarrow R_2-3R_1)\)
\(\implies\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}4&-5\\3&4\end{bmatrix} A (R_1\rightarrow R_1-R_2)\)
therefore \(A^{-1}=\begin{bmatrix}4&-5\\-3&4\end{bmatrix}\)
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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.
