Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)
We know that \(A = IA\)
\(\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)\(=A\)\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}\)\(= A\)\(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\) \((C_1\rightarrow C_1+2C_2)\)
⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\)\(=A\)\(\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}\) \((C_2\rightarrow C_2+C_1)\)
⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)\(=A\) \(\begin{bmatrix} 1 & \frac12 \\ 2 & \frac32 \end{bmatrix}\) \((C_2\rightarrow \frac{1}{2}C_2) \)
\(\therefore A^{-1}=\) \(\begin{bmatrix} 1 & \frac12 \\ 2 & \frac32 \end{bmatrix}\)
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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.
