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Properties of Determinants

The value of ...

Question

The value of the determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 + a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

A

0

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B

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

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C

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

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D

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

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Solution

The correct option is C $(1 + a^{2} + b^{2})^{3}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} - b^{2} & 2 a 2 b & - 2 a & 1 + a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Let $a = 1, b = 2$
$∣ ∣ ∣ ∣ \begin{matrix} - 2 & 4 & - 4 4 & 4 & 2 4 & - 2 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 216$
Option D $= {(1 + 1 + 4)}^{3} = 216$

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Similar questions

∣ ∣ ∣ ∣ ∣ \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}

Q. Using properties of determinant, prove the following:

∣ ∣ ∣ ∣ ∣ \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}

Q. Using properties of determinants, prove 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}

Q. The value of determinant

∣ ∣ ∣ ∣ ∣ \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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