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(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the augmented mat

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$\lambda^2 - \lambda - 6 = 0$

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(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the augmented matrix must satisfy the condition for consistency.The coefficient matrix is:$D = \begin{vmatrix} 1 & 1 & 1 \ 4 & \lambda & -\lambda \ 3 & 2 & -4 \end{vmatrix}$Expanding along the first row:$D = 1(-4\lambda - (-2\lambda)) - 1(-16 - (-3\lambda)) + 1(8 - 3\lambda) = 0$$D = (-4\lambda + 2\lambda) - (-16 + 3\lambda) + (8 - 3\lambda) = 0$$-2\lambda + 16 - 3\lambda + 8 - 3\lambda = 0$$-8\lambda + 24 = 0 \Rightarrow \lambda = 3$Now,check the options to see which quadratic equation has $\lambda = 3$ as a root:For option $A$: $\lambda^2 - \lambda - 6 = 0 \Rightarrow (\lambda - 3)(\lambda + 2) = 0$. Here,$\lambda = 3$ is a root.Thus,the correct option is $A$.

Let $\lambda$ be a real number for which the system of linear equations $x + y + z = 6$,$4x + \lambda y - \lambda z = \lambda - 2$,and $3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:

Similar Questions

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