(A) Given the system of equations:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$Since $x^2+y^2+z^2 \neq 0$, the system has a non-trivial soluti

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(A) Given the system of equations:$x - cy - bz = 0$$-cx + y - az = 0$$-bx - ay + z = 0$Since $x^2+y^2+z^2 \neq 0$, the system has a non-trivial solution. Therefore, the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(1 - a^2) - (-c)(-c - ab) + (-b)(ca + 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$

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $x^2+y^2+z^2 \neq 0, \quad x=cy+bz, \quad y=az+cx$ and $z=bx+ay$, then $a^2+b^2+c^2+2abc$ is equal to

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3 and 4 .Determinants and Matrices

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $x^2+y^2+z^2 \neq 0, \quad x=cy+bz, \quad y=az+cx$ and $z=bx+ay$, then $a^2+b^2+c^2+2abc$ is equal to

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If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
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$x + y = 1$
is consistent,then the value of $k$ is