If ${x^2} + {y^2} + {z^2} \ne 0,\;x = cy + bz,\;y = az + cx$ and $z = bx + ay,$ then ${a^2} + {b^2} + {c^2} + 2abc = $

JEE MATHEMATICS MCQ: MATRICES

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{x^2} + {y^2} + {z^2} \ne 0,\;x = cy + bz,\;y = az + cx

and

z = bx + ay,

then

{a^2} + {b^2} + {c^2} + 2abc =

View answer and explanation

Correct option: C

Option C (1) is correct because the determinant of the coefficient matrix must be zero for non-trivial solutions, leading to the given expression equaling 1.

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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

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\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|

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\omega

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

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

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