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Using elementary row transformations, find the inverse of each of the following matrices: \left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \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}{ll} 6 & 7 \\ 8 & 9 \end{array}\right]

Solution:

We have A=\left(\begin{array}{ll}6 & 7 \\ 8 & 9\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’s proceed
\begin{array}{l} \left(\begin{array}{lll} 6 & 7 \\ 8 & 9 \end{array} \mid \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \stackrel{R_{2} \rightarrow R_{2}-\frac{8}{6} R_{1}}{\longrightarrow}\left(\begin{array}{cc} 6 & 7 \\ 0 & -\frac{1}{3} \end{array} \mid \begin{array}{cc} 1 & 0 \\ -\frac{4}{3} & 1 \end{array}\right)^{R_{2} \rightarrow-3 R_{2} ; R_{1} \rightarrow \frac{1}{6} R_{1}} \\ \left(\begin{array}{cc|cc} 1 & \frac{7}{6} & \frac{1}{6} & 0 \\ 0 & 1 & -4 & -3 \end{array}\right) \stackrel{R_{1} \rightarrow R_{1}-\frac{7}{6} R_{2}}{\longrightarrow}\left(\begin{array}{cc|cc} 1 & 0 & -9 & \frac{7}{2} \\ 0 & 1 & 4 & -3 \end{array}\right) \end{array}
Therefore we got the inverse of the matrix as
A^{-1}=\left(\begin{array}{cc} \frac{-9}{2} & \frac{7}{2} \\ 4 & -3 \end{array}\right)

i

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