Using elementary transformations, find the inverse of the matrix,

Engineering

Mathematics

Introduction to Matrix

Algebra of Matrices

Inverse of a Matrix

Question

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

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

A = AI

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

C₂ → C₂ + C₃

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

applying C₁ → C₃ – C₁

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

applying C₃ → C₃ – 3C₁

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

applying C₂ → C₂/5 ; C₃ → C₃/5

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

applying C₁ → C₁ – C₂

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

C₁ → C₁ + C₃

$[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = A [\begin{array}{ccc} - 2 / 5 & 0 & 3 / 5 \\ - 115 & 1 / 5 & 0 \\ 215 & 1 / 5 & - 2 / 5 \end{array}]$

It is in the form of I = AA^–1

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