Using elementary transformations, find the inverse of the matrix

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

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

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

for finding inverse using elementary row operation

∴ A = IA

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

R₂ → R₂ + 3R₁ & R₃ → R₃ – 2R₁, we get

$\Rightarrow [\begin{array}{ccc} 1 & 3 & - 2 \\ 0 & 9 & 7 \\ 0 & - 5 & 4 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A$

$\Rightarrow [\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 9 & - 7 \\ 0 & - 5 & 4 \end{array}] = [\begin{array}{ccc} 0 & - 1 / 3 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A [\begin{aligned} ∵ R_{1} \to R_{1} \\ - \frac{1}{3} R_{2} \end{aligned}]$

$R_{2} \to \frac{1}{9} R_{2}$

$\Rightarrow [\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 1 & - 7 / 9 \\ 0 & - 5 & 4 \end{array}] = [\begin{array}{ccc} 0 & - 1 / 3 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ - 2 & 0 & 1 \end{array}] A$

R₃ → R₃ + 5R₂

$\Rightarrow [\begin{array}{ccc} 1 & 0 & 1 / 3 \\ 0 & 1 & - 7 / 9 \\ 0 & 0 & 1 / 9 \end{array}] = [\begin{array}{ccc} 0 & - 1 / 3 & 0 \\ 1 / 3 & 1 / 9 & 0 \\ - 2 & 0 & 1 \end{array}] A$

R₁ → R₁ – 3R₃ , R₂ → R₂ + 7R₃

$\Rightarrow [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 / 9 \end{array}] = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 1 / 3 & + 5 / 9 & 1 \end{array}] A$

R₃ → 9R₃

$\Rightarrow \underset{I}{\underset{⏟}{[\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}]}} = \underset{A^{- 1}}{\underset{⏟}{[\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{array}]}} \underset{A}{\underset{⏟}{A}}$

Hence, $A^{- 1} = [\begin{array}{ccc} 1 & - 2 & - 3 \\ - 2 & 4 & 7 \\ - 3 & 5 & 9 \end{array}]$

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