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Question: Given \(a_{i}^{2}\) + \(b_{i}^{2}\) + \(c_{i}^{2}\) = 1 (i = 1, 2, 3) and \(a_{i}a_{j} + b_{i}b_{j} ...

Given ai2a_{i}^{2} + bi2b_{i}^{2} + ci2c_{i}^{2} = 1 (i = 1, 2, 3) and aiaj+bibj+cicja_{i}a_{j} + b_{i}b_{j} + c_{i}c_{j} = 0 (i ¹ j; i, j = 1, 2, 3) then the value of ∣a1a2a3b1b2b3c1c2c3∣2\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right|^{2} is –

A

0

B

12\frac{1}{2}

C

1

D

2

Answer

1

Explanation

Solution

∣a1a2a3b1b2b3c1c2c3∣\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right|× ∣a1a2a3b1b2b3c1c2c3∣\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right|

= ∣a1b1c1a2b2c2a3b3c3∣\left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|× ∣a1a2a3b1b2b3c1c2c3∣\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right| =∣100010001∣\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| = 1