Find the inverse of matrices by elementary row transformation.

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Find the inverse of matrices by elementary row transformation.
$A = ∣ \begin{array}{ccc} 1 & 2 & 1 \\ - 1 & 0 & 2 \\ 2 & 1 & - 3 \end{array} ∣$

$B = ∣ \begin{array}{ccc} cosθ & - sinθ & 0 \\ sinθ & cosθ & 0 \\ 0 & 0 & 1 \end{array} ∣$

$A^{- 1} = (\begin{array}{ccc} cosθ & sinθ & 0 \\ - sinθ & cosθ & 0 \\ 0 & 0 & 1 \end{array})$

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$A [\begin{array}{ccc} 1 & 0 & 0 \\ - \tan & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} cosθ & - sinθ & 0 \\ 0 & \frac{1}{cosθ} & 0 \\ 0 & 0 & 1 \end{array}]$

R₂ → R₂cosθ
$A ∣ \begin{array}{ccc} 1 & 0 & 0 \\ - sinθ & cosθ & 0 \\ 0 & 0 & 1 \end{array} ∣ = ∣ \begin{array}{ccc} cosθ & - sinθ & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$
R₁ → R₁ + sinθR₂

$A ∣ \begin{array}{ccc} 1 - \sin & sinθ & 0 \\ - sinθ & cosθ & 0 \\ 0 & 0 & 1 \end{array} ∣ = ∣ \begin{array}{ccc} cosθ & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

R₁ → R₁ / cosθ

$A ∣ \begin{array}{ccc} cosθ & sinθ & 0 \\ - sinθ & cosθ & 0 \\ 0 & 0 & 1 \end{array} ∣ = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

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