Let the system of linear equations 4x + lambd | vedclass

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(A) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to $0$.$\begin{vmatrix} 4

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$\mu = 6, \lambda \in R$

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(A) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to $0$.$\begin{vmatrix} 4 & \lambda & 2 \ 2 & -1 & 1 \ \mu & 2 & 3 \end{vmatrix} = 0$Expanding along the first row:$4((-1)(3) - (1)(2)) - \lambda((2)(3) - (1)(\mu)) + 2((2)(2) - (-1)(\mu)) = 0$$4(-3 - 2) - \lambda(6 - \mu) + 2(4 + \mu) = 0$$4(-5) - 6\lambda + \lambda\mu + 8 + 2\mu = 0$$-20 - 6\lambda + \lambda\mu + 8 + 2\mu = 0$$\lambda\mu - 6\lambda + 2\mu - 12 = 0$$\lambda(\mu - 6) + 2(\mu - 6) = 0$$(\lambda + 2)(\mu - 6) = 0$For this equation to hold for any $\lambda \in R$,we must have $\mu - 6 = 0$,which implies $\mu = 6$.Thus,the condition is $\mu = 6$ for any $\lambda \in R$.

Let the system of linear equations $4x + \lambda y + 2z = 0$,$2x - y + z = 0$,and $\mu x + 2y + 3z = 0$ (where $\lambda, \mu \in R$) have a non-trivial solution. Then which of the following is true?

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