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Find the Inverse by Using Elementary Row Transformations: ⎡ ⎢ ⎣ 1 3 − 2 − 3 0 − 1 2 1 0 ⎤ ⎥ ⎦ - Mathematics

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Question

Find the inverse by using elementary row transformations:

\[\begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]

Sum
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Solution

\[A = \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix}\]
We know
\[A = IA \]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ - 3 & 0 & - 1 \\ 2 & 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} A\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + 3 R_1\text{ and }R_3 \to R_3 - 2 R_1 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 3 & - 2 \\ 0 & 1 & - \frac{7}{9} \\ 0 & - 5 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 2 & 0 & 1\end{bmatrix} A \left[\text{ Applying }R_2 \to \frac{1}{9} R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & \frac{1}{9}\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - \frac{1}{3} & \frac{5}{9} & 1\end{bmatrix} A \left[\text{ Applying }R_1 \to R_1 - 3 R_2\text{ and }R_3 \to R_3 + 5 R_2 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & \frac{1}{3} \\ 0 & 1 & - \frac{7}{9} \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}0 & - \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_3 \to 9 R_3 \right]\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} A \left[\text{ Applying }R_2 \to R_2 + \frac{7}{9} R_3\text{ and }R_1 \to R_1 - \frac{1}{3} R_3 \right]\]
\[ \Rightarrow A^{- 1} = \begin{bmatrix}1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9\end{bmatrix} \]

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Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.2 [Page 34]

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RD Sharma Mathematics [English] Class 12
Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.2 | Q 15 | Page 34

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