Find the inverse of each of the matrices,if it exists.\(\begin{bmatrix} 1 & 3 & -2\\ -3 & 0 & -5\\ 2&5&0 \end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix} 1 & 3 & -2\\ -3 & 0 & -5\\ 2&5&0 \end{bmatrix}\)
We know that A = IA
\(\begin{bmatrix} 1 & 3 & -2\\ -3 & 0 & -5\\ 2&5&0 \end{bmatrix}\)=\(\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0&0&1 \end{bmatrix}\)A
Applying \(R_2 → R_2 + 3R_1\) and \(R_3 → R_3 − 2R_1\), we have:
\(\begin{bmatrix} 1 & 3 & -2\\ 0 & 9 & -11\\ 0&-1&4 \end{bmatrix}\)=\(\begin{bmatrix} 1 & 0 & 0\\ 3 & 1 & 0\\ -2&0&1 \end{bmatrix}\)A .
Applying \(R_1 → R_1 + 3R_3\) and \(R_2 → R_2 +8R_3\), we have:
\(\begin{bmatrix} 1 & 0 & 10\\ 0 & 1 & 21\\ 0&-1&4 \end{bmatrix}\)=\(\begin{bmatrix} -5 & 0 & 3\\ -13 & 1 & 8\\ -2&0&1 \end{bmatrix}\)A
Applying \(R_3 → R_3 + R_2\) , we have
\(\begin{bmatrix} 1 & 0 & 10\\ 0 & 1 & 21\\ 0&0&25 \end{bmatrix}\)=\(\begin{bmatrix} -5 & 0 & 3\\ -13 & 1 & 8\\ -15&1&9 \end{bmatrix}\)A
Applying \(R_3 → \frac{1}{25}R_3\) , we have
\(\begin{bmatrix} 1 & 0 & 10\\ 0 & 1 & 21\\ 0&0&1 \end{bmatrix}\)=\(\begin{bmatrix} -5 & 0 & 3\\ -13 & 1 & 8\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}\)
Applying \(R_1 → R_1 -10R_3\) and \(R_2 → R_2 -21R_3\), we have:
\(\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0&0&1 \end{bmatrix}\)=\(\begin{bmatrix} 1 & -\frac25 & -\frac35\\ -\frac25 & \frac{4}{25} & \frac{11}{25}\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}\)A
therefore A-1=\(\begin{bmatrix} 1 & -\frac25 & -\frac35\\ -\frac25 & \frac{4}{25} & \frac{11}{25}\\ -\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}\)
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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.
