Question:

Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)

Updated On: Aug 16, 2024
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Solution and Explanation

Let A=\(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)

We know that\(A = IA\)  

\(\therefore\) \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)\(\begin{bmatrix} 1 & 0 \\ 0 & 1  \end{bmatrix}\) A 

⇒ \(\begin{bmatrix} 1 & \frac12 \\ 7 & 4  \end{bmatrix}\)=\(\begin{bmatrix} \frac12 & 0 \\ 0 & 1  \end{bmatrix}\)A            (R1\(→\) \(\frac{1}{2}R_1\)

\(\begin{bmatrix} 1 & \frac12 \\ 0 & \frac12  \end{bmatrix}\)=\(\begin{bmatrix} \frac12 &0\\  -\frac72  & 1  \end{bmatrix}\)A           (R2→R2-7R1

⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & \frac12  \end{bmatrix}\)=\(\begin{bmatrix} 4 & -1 \\ -\frac72 & 1  \end{bmatrix}\)A        (R1->R1-R2

⇒ \(\begin{bmatrix} 1 & 0 \\ 0 & 2  \end{bmatrix}\) = \(\begin{bmatrix} 4 & -1 \\ -7 & 2  \end{bmatrix}\)A  (R2->2R2

\(\therefore\) A-1=\(\begin{bmatrix} 4 & -1 \\ -7 & 2  \end{bmatrix}\)
 

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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.