- Det- 1- a2 -b2 2ab -2b- 2ab 1- a2-b2 2a- 2b -2a 1- a2-b2 -

The correct option is B (1+a^2+b^2)^3 Performing R_1→ R_1+bR_3 and R_2→ R_2 aR_3 [ 1+a^2+b^2 amp; 0 amp; b(1+a^2+b^2 ...

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Summation by Sigma Method

. Det[ 1+...

Question

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

(a^{2} + b^{2})^{3}

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(1 + a^{2} + b^{2})^{3}

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(2 - a^{2} - b^{2})

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(2 - a^{2} - b^{2})^{2}

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Solution

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By using properties of determinants, show that:

∣ ∣ ∣ ∣ ∣ \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}

∣ ∣ ∣ ∣ ∣ \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}

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