Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\)
We know that \(A = AI \)
\(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\)=A\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}\)=A \(\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\) \((C_2\rightarrow C_2+3C_1) \)
⇒ \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}\) A \(\begin{bmatrix} -2 & 3 \\ -1 & 1 \end{bmatrix}\) \((C_1\rightarrow C_1-C_2)\)
⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) =A \(\begin{bmatrix} -1 & 3 \\ -\frac12 & 1 \end{bmatrix}\) \((C_1\rightarrow \frac{1}{2}C_1)\)
\(\therefore\)A-1=\(\begin{bmatrix} -1 & 3 \\ -\frac12 & 1 \end{bmatrix}\)
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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.
