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

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

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

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

Let A = IA

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

Applying R₂ → R₂ + R₁ , R₃ → R₃ + 3R₁

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

Applying R₂ → R – R₃

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

Applying R₁ → R₁ – R₂ and R₃ → R₃ + 4R₂

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

R₂ → R₂ (– 1) and R₂ → R₂ + 2R₃

$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & 4 & - 3 \end{array}] A$

Applying R₃ → (– 1)R₃

$[\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{array}] A$

which is the FOM I = BA

So, $A^{- 1} = [\begin{array}{ccc} 1 & - 1 & 1 \\ - 8 & 7 & - 5 \\ 5 & - 4 & 3 \end{array}]$

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