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(A) The system is inconsistent if the determinant $D = 0$ and at least one of $D_{1}, D_{2}, D_{3} 
eq 0$.First,calculate the determinant $D$:$D = \b

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exactly one negative value of $\lambda$.

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(A) The system is inconsistent if the determinant $D = 0$ and at least one of $D_{1}, D_{2}, D_{3} 
eq 0$.First,calculate the determinant $D$:$D = \begin{vmatrix} 2 & -4 & \lambda \ 1 & -6 & 1 \ \lambda & -10 & 4 \end{vmatrix} = 2(-24 + 10) + 4(4 - \lambda) + \lambda(-10 + 6\lambda)$$D = 2(-14) + 16 - 4\lambda - 10\lambda + 6\lambda^{2} = 6\lambda^{2} - 14\lambda - 12 = 2(3\lambda^{2} - 7\lambda - 6) = 2(3\lambda + 2)(\lambda - 3)$.Setting $D = 0$,we get $\lambda = 3$ or $\lambda = -\frac{2}{3}$.Now calculate $D_{1}, D_{2}, D_{3}$:$D_{1} = \begin{vmatrix} 1 & -4 & \lambda \ 2 & -6 & 1 \ 3 & -10 & 4 \end{vmatrix} = 1(-24 + 10) + 4(8 - 3) + \lambda(-20 + 18) = -14 + 20 - 2\lambda = 6 - 2\lambda = -2(\lambda - 3)$.For $\lambda = 3$,$D = 0$ and $D_{1} = 0$. Checking $D_{2}$ and $D_{3}$ for $\lambda = 3$ also yields $0$,implying infinitely many solutions.For $\lambda = -\frac{2}{3}$,$D = 0$ but $D_{1} = -2(-\frac{2}{3} - 3) = -2(-\frac{11}{3}) = \frac{22}{3} 
eq 0$.Since $D = 0$ and $D_{1} 
eq 0$ at $\lambda = -\frac{2}{3}$,the system is inconsistent for exactly one negative value of $\lambda$.

Let $\lambda \in R$. The system of linear equations
$2x_{1} - 4x_{2} + \lambda x_{3} = 1$
$x_{1} - 6x_{2} + x_{3} = 2$
$\lambda x_{1} - 10x_{2} + 4x_{3} = 3$
is inconsistent for:

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Let $\lambda \in R$. The system of linear equations$2x_{1} - 4x_{2} + \lambda x_{3} = 1$$x_{1} - 6x_{2} + x_{3} = 2$$\lambda x_{1} - 10x_{2} + 4x_{3} = 3$is inconsistent for: