The number of real values of lambda for which | vedclass

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(B) For a homogeneous system of linear equations $AX = 0$ to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e

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$1$

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(B) For a homogeneous system of linear equations $AX = 0$ to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.The coefficient matrix is:$A = \begin{bmatrix} 2 & 4 & -\lambda \ 4 & \lambda & 2 \ \lambda & 2 & 2 \end{bmatrix}$Setting the determinant to zero:$|A| = 2(\lambda \cdot 2 - 2 \cdot 2) - 4(4 \cdot 2 - 2 \cdot \lambda) + (-\lambda)(4 \cdot 2 - \lambda \cdot \lambda) = 0$$|A| = 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) = 0$$|A| = 4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3 = 0$$|A| = \lambda^3 + 4\lambda - 40 = 0$Let $f(\lambda) = \lambda^3 + 4\lambda - 40$. Since $f'(\lambda) = 3\lambda^2 + 4 > 0$ for all real $\lambda$,the function $f(\lambda)$ is strictly increasing.Therefore,it can have only one real root.Thus,the number of real values of $\lambda$ is $1$.

Market	$x, y, z$
$I$	$10,000, 2,000, 18,000$
$II$	$6,000, 20,000, 8,000$

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is:

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