Problem

6. If the set $\left\{\left[\begin{array}{c}4 \\ 3 \\ 11\end{array}\right],\left[\begin{array}{c}-2 \\ 0 \\ -4\end{array}\right],\left[\begin{array}{l}3 \\ a \\ 5\end{array}\right]\right\}$ in $\mathbb{R}^{3}$ is linearly independent, what value can $a$ not be?

Answer

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Answer

Set the determinant equal to zero and solve for 'a': $-6a - 6 = 0 \Rightarrow a = \boxed{-1}$

Steps

Step 1 ：Form the matrix and calculate its determinant: $\begin{vmatrix} 4 & -2 & 3 \\ 3 & 0 & a \\ 11 & -4 & 5 \end{vmatrix} = -6a - 6$

Step 2 ：Set the determinant equal to zero and solve for 'a': $-6a - 6 = 0 \Rightarrow a = \boxed{-1}$

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