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

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Using elementary row transformations, find the inverse of the following matrix:
$[\begin{array}{ccc} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}]$

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We have $[\begin{array}{ccc} 3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \cdot A$

$\Rightarrow [\begin{array}{ccc} 1 & - 3 & 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & 0 \\ 0 & 0 & 1 \end{array}] \cdot A$

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

applying $R_{2} \to R_{2} - 2 R_{1}; [\begin{array}{ccc} 1 & - 3 & - 1 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 2 & 3 & - 2 \\ 0 & 1 & 1 \end{array}] \cdot A$

applying $\begin{aligned} R_{1} \to R_{1} + 3 R_{2}, \\ R_{3} \to R_{3} - 4 R_{2} \end{aligned} [\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} - 5 & 8 & - 6 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}] \cdot A$

applying $R_{1} \to R_{1} + R_{3}; [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}] \cdot A$

Hence $A^{- 1} = [\begin{array}{ccc} 3 & - 4 & 3 \\ - 2 & 3 & - 2 \\ 8 & - 12 & 9 \end{array}]$

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