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Using elementary row transformations, find the inverse of each of the following matrices: \left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]
CBSE | Class 12 | Exercise 5E | Maths | Matrices | RS Aggarwal
Using elementary row transformations, find the inverse of each of the following matrices: \left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]

Solution:

We have A=\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)
To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:
(A \mid I) \rightarrow\left(I \mid A^{-1}\right)
Let us proceed

\left(\begin{array}{lll|lll}0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 2 & 3 & 0 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1\end{array}\right) \stackrel{R_{2 \leftrightarrow R_{1}}}{\longrightarrow}\left(\begin{array}{lll|lll}1 & 2 & 3 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0\end{array}\right)

\stackrel{R_{3} \rightarrow R_{3}-3 R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1\end{array}\right)

\stackrel{R_{3} \rightarrow R_{3}+5 R_{2}}{\longrightarrow}\left(\begin{array}{lll|ccc}1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 5 & -3 & 1\end{array}\right) \stackrel{R_{3} \rightarrow \frac{1}{2} R_{3}}{\longrightarrow}\left(\begin{array}{lll|ccc}1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right)

\stackrel{R_{1} \rightarrow R_{1}-2 R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 0 & -1 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right)

\stackrel{R_{1} \rightarrow R_{1}+R_{3} ; R_{2} \rightarrow R_{2}-2 R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & -4 & 3 & -1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right)

Thus we have the inverse of the matrix as
A^{-1}=\left(\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right)

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