Using properties of determinants, prove the following .

Engineering

Mathematics

Introduction to Determinants

Question

Using properties of determinants, prove the following $∣ \begin{array}{ccc} 1 + a^{2} - b^{2} & 2 ab & - 2 b \\ 2 ab & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{array} ∣ = (1 + a^{2} + b^{2})^{3}$ .

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LHS :

∣ \begin{array}{ccc} 1 + a^{2} - b^{2} & 2 ab & - 2 b \\ 2 ab & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{array} ∣

Applying C₁ → C₁ – bC₃, C₂ → C₂ + aC₃, we get,

= ∣ \begin{array}{ccc} 1 + a^{2} + b^{2} & 0 & - 2 b \\ 0 & 1 + a^{2} + b^{2} & 2 a \\ b (1 + a^{2} + b^{2}) & - a (1 + a^{2} + b^{2}) & 1 - a^{2} - b^{2} \end{array} ∣

Taking out (1 + a² + b²) common from C₁ and C₂, we get,

= (1 + a^{2} + b^{2})^{2} ∣ \begin{array}{ccc} 1 & 0 & - 2 b \\ 0 & 1 & 2 a \\ b & - a & 1 - a^{2} - b^{2} \end{array} ∣

Applying R₃ → R₃ – bR₁ + aR₂, we get,

= (1 + a^{2} + b^{2})^{2} ∣ \begin{array}{ccc} 1 & 0 & - 2 b \\ 0 & 1 & 2 a \\ 0 & 0 & 1 + a^{2} + b^{2} \end{array} ∣

Expanding along C₁, we get,

= (1 + a² + b²)² (1 + a² + b²) = (1 + a² + b²)³ = RHS

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