Find the inverse of each of the matrices, if it exists\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
We know that A = IA
\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)A
Applying \(R_1\rightarrow R_1-\frac{1}{2}R_2\), we have:
\(\begin{bmatrix} 0 & 0 \\ 4 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 1 & -\frac12 \\ 0 & 1 \end{bmatrix}\)A
Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S. Therefore, A−1 does not exist.
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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.
