If x = cy + bz,y = az + cx,z = bx + ay (where | vedclass

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(C) The given system of homogeneous linear equations is:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$For this system to have a non-trivial sol

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$a^2 + b^2 + c^2 + 2abc = 1$

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(C) The given system of homogeneous linear equations is:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$For this system to have a non-trivial solution (where $x, y, z$ are not all zero),the determinant of the coefficient matrix must be zero:$\Delta = \begin{vmatrix} 1 & -c & -b \ -c & 1 & -a \ -b & -a & 1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(1 - a^2) - (-c)(-c - ab) + (-b)(ac + b) = 0$$1 - a^2 - c^2 - abc - abc - b^2 = 0$$1 - a^2 - b^2 - c^2 - 2abc = 0$Rearranging the terms,we get:$a^2 + b^2 + c^2 + 2abc = 1$

If $x = cy + bz$,$y = az + cx$,$z = bx + ay$ (where $x, y, z$ are not all zero) have a solution other than $x = 0, y = 0, z = 0$,then $a, b$,and $c$ are connected by the relation:

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