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:
$[\begin{array}{ccc} 3 & - 1 & - 2 \\ 2 & 0 & - 1 \\ 3 & - 5 & 0 \end{array}]$

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We have $[\begin{array}{ccc} 3 & - 1 & - 2 \\ 2 & 0 & - 1 \\ 3 & - 5 & 0 \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \cdot A$

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

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

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

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

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

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

$A^{- 1} = \frac{1}{8} [\begin{array}{llc} - 5 & 5 (2) & 1 \\ - 3 & 3 (2) & - 1 \\ - 5 (2) & 3 (4) & 1 (2) \end{array}]$

$A^{- 1} = \frac{- 1}{8} [\begin{array}{ccc} 5 & - 10 & - 1 \\ 3 & - 6 & 1 \\ 10 & - 12 & - 2 \end{array}]$

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