Find the inverse of the following matrices by using elementary row transformations :

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Find the inverse of the following matrices by using elementary row transformations :
$[\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}]$

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$[\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}]$

Let A = AI

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

Applying R₁ → R₁ – R₂

$[\begin{array}{ccc} 1 & - 3 & - 1 \\ 1 & 2 & 4 \\ 3 & 1 & 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 1 \end{array}]$

Applying R₂ → R₂ – R₁ and R₃ → R₃ – 2R₂

$[\begin{array}{ccc} 1 & - 3 & - 1 \\ 0 & 5 & 5 \\ 0 & 10 & 4 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & 0 \\ - 3 & 3 & 1 \end{array}]$ and $[\begin{array}{ccc} 1 & - 3 & - 1 \\ 0 & 5 & 5 \\ 0 & 0 & - 6 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & 0 \\ - 1 & - 1 & 1 \end{array}]$

Applying R₂ → 1/5 R₂ and R₃ → 1/6 R₃

$[\begin{array}{ccc} 1 & - 3 & - 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{cccc} 1 & - 1 & 0 \\ - 1 / 5 & 2 / 5 & 0 \\ 1 / 6 & 1 / 6 & - 1 / 6 \end{array}]$

Applying R₁ → R₁ + 3R₂ and R₂ → R₂ – R₃

$[\begin{array}{lll} 1 0 2 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} 8 / 5 & 1 / 5 & 0 \\ - 11 / 30 & - 7 / 30 & 1 / 6 \\ 11 / 6 & 1 / 6 & - 1 / 6 \end{array}]$

Applying R₁ → R₁ – 2R₃

$[\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} 115 & - 2 / 15 & 1 / 3 \\ - 11 / 30 & 7 / 30 & 1 / 6 \\ 1 / 6 & 1 / 6 & - 1 / 6 \end{array}]$

So $A^{- 1} = - \frac{1}{30} [\begin{array}{ccc} - 2 & 4 & - 10 \\ 11 & - 7 & - 5 \\ - 5 & - 5 & 5 \end{array}]$

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