The value of the determinant left|begin{array | vedclass

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Question

(D) Let $\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}ight|$

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

Vedclass · Answer

(D) Let $\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}ight|$.Apply the column operations $C_{1} ightarrow C_{1} - bC_{3}$ and $C_{2} ightarrow C_{2} + aC_{3}$.$\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} + 2b^{2} & 2 a b - 2ab & -2 b \ 2 a b - 2ab & 1-a^{2}+b^{2} + 2a^{2} & 2 a \ 2 b - b(1-a^{2}-b^{2}) & -2 a + a(1-a^{2}-b^{2}) & 1-a^{2}-b^{2}\end{array}ight|$$\Delta = \left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -2 b \ 0 & 1+a^{2}+b^{2} & 2 a \ b(1+a^{2}+b^{2}) & -a(1+a^{2}+b^{2}) & 1-a^{2}-b^{2}\end{array}ight|$Taking common factors $(1+a^{2}+b^{2})$ from $C_{1}$ and $C_{2}$:$\Delta = (1+a^{2}+b^{2})^{2} \left|\begin{array}{ccc}1 & 0 & -2 b \ 0 & 1 & 2 a \ b & -a & 1-a^{2}-b^{2}\end{array}ight|$Expanding along $R_{1}$:$\Delta = (1+a^{2}+b^{2})^{2} [1(1-a^{2}-b^{2} + 2a^{2}) - 0 + (-2b)(0 - b)]$$\Delta = (1+a^{2}+b^{2})^{2} [1+a^{2}-b^{2} + 2b^{2}]$$\Delta = (1+a^{2}+b^{2})^{2} (1+a^{2}+b^{2}) = (1+a^{2}+b^{2})^{3}$.

The value of the determinant $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$ is

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