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

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

we know that

A = IA

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

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

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

R₂ → R₂ – 5R₁

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

R₃ → R₃ – R₂

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

R₃ → 2R₃

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

$R_{1} \to R_{1} + \frac{1}{2} R_{3}$

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

$R_{2} \to R_{2} - \frac{5}{2} R_{3}$

$[\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 30 / 2 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}] A$

I = A^–1A

$I = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}] A$

$∴ A^{- 1} = [\begin{array}{ccc} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}]$

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