If $a_i^2 + b_i^2 + c_i^2 = 1\left( {i = 1,2,3} \right)$ and ${a_i}{a_j} + {b_i}{b_j} + {c_i}{c_j} = 0\left( {i \ne j\,\,{\rm{and}}\,\,\,i,j = 1,2,3} \right),$ then the value of $\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$ is

JEE MATHEMATICS MCQ: MATRICES

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Application +4 | -1

a_i^2 + b_i^2 + c_i^2 = 1\left( {i = 1,2,3} \right)

and

{a_i}{a_j} + {b_i}{b_j} + {c_i}{c_j} = 0\left( {i \ne j\,\,\mathrm{and}\,\,\,i,j = 1,2,3} \right),

then the value of

\left| {\begin{array}{cccccccccccccccccccc} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|

View answer and explanation

Correct option: C

Option C (1) is correct because the determinant's square equals 1, derived from the given orthogonality conditions and normalization of vectors.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|