Using elementary row transformations, find the inverse of the following matrix:

Engineering

Mathematics

Introduction to Matrix

Multiplication of Matrices

Inverse of a Matrix

Question

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

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

we know A = IA

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

R₁ → R₁ – R₂

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

R₁ ↔ R₂

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

$R_{2} \to \frac{R}{- 5}, R_{1} \to R_{1} + R_{3}$

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

$R_{1} \to \frac{R_{1}}{5}$

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

R₃ → R₃ + 2R₂ – 3R₁

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

R₁ → R₁ + R₃ , R₃ → R₃ + (– 1)

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

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