Find the inverse of each of the matrices, if it exists.
\(\begin{bmatrix} 3 & 10\\ 2 & 7\end{bmatrix}\)
\(\begin{bmatrix} 3 & 10\\ 2 & 7\end{bmatrix}\)
Solution and Explanation
Let A=\(\begin{bmatrix} 3 & 10\\ 2 & 7\end{bmatrix}\)
We know that \(A = IA \)
\(\begin{bmatrix} 3 & 10\\ 2 & 7\end{bmatrix}\)= \(\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\)A
⇒ \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)= \(\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}\)A \((R_1\rightarrow R_1-R_2)\)
⇒ \(\begin{bmatrix} 1 & 3\\ 0 & 1\end{bmatrix}\)= \(\begin{bmatrix} 1 & -1\\ -2 & 3\end{bmatrix}\) A \((R_2\rightarrow R_2-2R_2)\)
⇒ \(\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\)= \(\begin{bmatrix} 7 & -10\\ -2 & 3\end{bmatrix}\)\(A\)et \((R_1\rightarrow R_1-3R_2)\)
\(\therefore A^{-1}=\) \(\begin{bmatrix} 7 & -10\\ -2 & 3\end{bmatrix}\)
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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.
