If $\begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, then $\begin{bmatrix} x \\ y \\ z \end{bmatrix} =$

  • A
    $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + K \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}, K \in R$
  • B
    $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + K \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}, K \in R$
  • C
    $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + K \begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}, K \in R$
  • D
    $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + K \begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}, K \in R$

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Let $a, \lambda, \mu \in \mathbb{R}$. Consider the system of linear equations:
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$(D)$ If $\lambda + \mu \neq 0$,then the system has no solution for $a = -3$.

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