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:
$[\begin{array}{lll} 0 1 2 \\ 1 2 3 \\ 3 1 1 \end{array}]$

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

we can write A = IA ⇒ A^–1A = A^–1IA.

⇒ I = A^–1A. [we use this idea]

$∵ [\begin{array}{lll} 0 1 2 \\ 1 2 3 \\ 3 1 1 \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A .$

applying R₁ → R₁ + R₂, we get.

$[\begin{array}{lll} 1 3 5 \\ 1 2 3 \\ 3 1 1 \end{array}] = [\begin{array}{lll} 1 1 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A .$

applying R₂ → R₂ – R₁and R₃ → R₃ – 3R₁ we get.

$[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & - 1 & - 2 \\ 0 & - 8 & - 14 \end{array}] = [\begin{array}{ccc} 1 & 1 & 0 \\ - 1 & 0 & 0 \\ - 3 & - 3 & 1 \end{array}] A$

applying R₂ → – R₂ , we get.

$[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & 2 \\ 0 & - 8 & - 14 \end{array}] = [\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 0 \\ - 3 & - 3 & 1 \end{array}] A$

applying R₃ → R₃ + 8R₂ , we get.

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

applying R₁ → R₁ – 3R₂ , we get.

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

applying R₁ → R₁ + R₃ and R₂ → R₂ – 2R₃ we get.

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

Therefore , $A^{- 1} = [\begin{array}{ccc} 1 / 2 & - 1 / 2 & 1 / 2 \\ - 4 & 3 & - 1 \\ 5 / 2 & - 3 / 2 & 1 / 2 \end{array}] = \frac{1}{2} [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 6 & - 2 \\ 5 & - 3 & 1 \end{array}]$

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