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Adjoint of a Matrix

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Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is C $(1 + a^{2} + b^{2})^{3}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} + b R_{3}; R_{2} \to R_{2} - a R_{3}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + b^{2} & 0 & - b (1 + a^{2} + b^{2}) 0 & 1 + a^{2} + b^{2} & a (1 + a^{2} + b^{2}) 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - b 0 & 1 & a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ (1 + a^{2} + b^{2})^{2} R_{3} \to R_{3} - 2 b R - 1 + 2 a R - 2$
$= (1 + a^{2} + b^{2})^{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - b 0 & 1 & a 0 & 0 & 1 + a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}$

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Similar questions

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∣ ∣ ∣ ∣ ∣ \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} ∣ ∣ ∣ ∣ ∣

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

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