The system of equations lambda x + y + z = 0, | vedclass

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(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.The coeff

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$0$

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(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.The coefficient matrix is given by:$A = \begin{bmatrix} \lambda & 1 & 1 \ -1 & \lambda & 1 \ -1 & -1 & \lambda \end{bmatrix}$Setting the determinant to zero:$\begin{vmatrix} \lambda & 1 & 1 \ -1 & \lambda & 1 \ -1 & -1 & \lambda \end{vmatrix} = 0$Expanding along the first row:$\lambda(\lambda^2 - (-1)) - 1(-\lambda - (-1)) + 1(1 - (-\lambda)) = 0$$\lambda(\lambda^2 + 1) - 1(-\lambda + 1) + 1(1 + \lambda) = 0$$\lambda^3 + \lambda + \lambda - 1 + 1 + \lambda = 0$$\lambda^3 + 3\lambda = 0$$\lambda(\lambda^2 + 3) = 0$This gives $\lambda = 0$ or $\lambda^2 = -3$. Since $\lambda$ must be a real value,$\lambda^2 = -3$ has no real solutions.Therefore,the only real value for $\lambda$ is $0$.

The system of equations $\lambda x + y + z = 0, -x + \lambda y + z = 0, -x - y + \lambda z = 0$ will have a non-zero solution if real values of $\lambda$ are given by

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