Using elementary row transformations, find the inverse of the following matrix:

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Introduction to Matrix

Algebra of Matrices

Inverse of a Matrix

Question

Using elementary row transformations, find the inverse of the following matrix:
$[\begin{array}{lll} 3 0 2 \\ 1 5 9 \\ 6 4 7 \end{array}]$

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let $A = [\begin{array}{lll} 3 0 2 \\ 1 5 9 \\ 6 4 7 \end{array}], A = IA \Rightarrow [\begin{array}{lll} 3 0 2 \\ 1 5 9 \\ 6 4 7 \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A$

applying $R_{3} \to R_{3} - 2 R_{1}, [\begin{array}{lll} 3 0 2 \\ 1 5 9 \\ 0 4 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A$

applying $R_{2} \to \frac{1}{5} R_{2}, [\begin{array}{lll} 3 0 2 \\ 0 3 5 \\ 0 4 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - \frac{1}{5} & \frac{3}{5} & 0 \\ - 2 & 0 & 1 \end{array}] A$

applying $R_{3} \to 3 R_{3} - 4 R_{2}, [\begin{array}{ccc} 3 & 0 & 2 \\ 0 & 3 & 5 \\ 0 & 0 & - 11 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - \frac{1}{5} & \frac{3}{5} & 0 \\ - \frac{26}{5} & - \frac{12}{5} & 3 \end{array}] A$

applying $R_{3} \to \frac{- 1}{11} R_{3}, [\begin{array}{lll} 3 & 0 & 2 \\ 0 & 3 & 5 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ \frac{- 1}{5} & \frac{3}{5} & 0 \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{array}] A$

applying $R_{2} \to R_{2} - 5 R_{3}, R_{1} \to R_{1} - 2 R_{3}, [\begin{array}{lll} 3 0 0 \\ 0 3 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} \frac{3}{55} & \frac{- 25}{55} & \frac{6}{11} \\ - \frac{141}{55} & - \frac{27}{55} & \frac{15}{11} \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{array}] A$

applying $\begin{aligned} R_{1} \to \frac{1}{3} R_{1} \\ R_{2} \to \frac{1}{3} R_{2} \end{aligned} [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} \frac{1}{55} & \frac{- 8}{55} & \frac{2}{11} \\ \frac{- 47}{55} & \frac{- 9}{55} & \frac{5}{11} \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{array}] A$

Hence $A^{- 1} = \frac{1}{55} [\begin{array}{ccc} - 1 & 8 & - 10 \\ 47 & 9 & - 25 \\ - 26 & - 12 & 15 \end{array}]$