By using properties of determinants- show that- 1-a2-b2 2 a b -2 b- 2 a b 1-a2-b2 2 a- 2 b -2 a 1-a2-b2 -1-a2-b23

Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have: Expanding along R1, we have: ...

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Evaluation of a Determinant

By using prop...

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By using properties of determinants, show that:

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D e t ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 1 + & a^{2} & - b^{2} & 2 a b & - 2 b 2 a b & 1 - & a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - & a^{2} - b^{2} \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & - 2 b 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(1 + a^{2} + b^{2})}^{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(1 + a^{2} + b^{2})}^{3}

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