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(A) Given system: $\begin{bmatrix} 1 \ 2 \ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmat

Vedclass · Accepted Answer

$\forall \lambda \in(-\infty, \infty)$,the given system has non-trivial solution

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(A) Given system: $\begin{bmatrix} 1 \ 2 \ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = 0$ Multiplying the matrices,we get: $\begin{bmatrix} 1 & 2 & \lambda \ 2 & 4 & 2\lambda \ \lambda & 2\lambda & \lambda^2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = 0$ Let $A = \begin{bmatrix} 1 & 2 & \lambda \ 2 & 4 & 2\lambda \ \lambda & 2\lambda & \lambda^2 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = 1(4\lambda^2 - 4\lambda^2) - 2(2\lambda^2 - 2\lambda^2) + \lambda(4\lambda - 4\lambda) = 0 - 0 + 0 = 0$. Since the determinant of the coefficient matrix is $0$,the system $AX = 0$ always has infinitely many solutions (non-trivial solutions) for any value of $\lambda \in (-\infty, \infty)$.

Consider the following system of equations in matrix form $\begin{bmatrix} 1 \\ 2 \\ \lambda \end{bmatrix} \begin{bmatrix} 1 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$. Then which one of the following statements is true?

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