What values of a is the set of vectors linearly dependent? The question is is "determine conditions on the scalars so that the set of vectors is linearly dependent".
$$ v_1 = \begin{bmatrix} 1 \\  2 \\  1\\  \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}
$$
When I reduce the matrix I get
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & a-2 & 0 \\ 0 & 0 & b - \frac{4}{(a-2)} \end{bmatrix}$$
If the matrix is linearly independent then shouldn't $a-2 = 0$ and $b - \frac{4}{(a-2)} = 0$? So, I said the solution is when $a-2 \neq 0 $ and $b - \frac{4}{(a-2)} \neq 0$. The textbooks says the answer is when $ b(a-2) = 4 $. I understand how they got to $ b(a-2) = 4 $ but why is it equals instead of not equals?
 A: Note that 
$$
\det\begin{bmatrix}
1 & 1 & 0 \\
2 & a & 2 \\
1 & 3 & b
\end{bmatrix}=ab-2b-4
$$
The vectors $v_1$, $v_2$, and $v_3$ are linearly independent if and only if $ab-2b-4\neq0$.
A: The Determinant Test is appropriate here, since you have three vectors from $\mathbb{R}^{3}$. The set of vectors is linearly dependent if and only if $det(M) = 0$, where $M = [v_{1} v_{2} v_{3}]$. 
A: let be $\{v_1,v_2,v_3\} \subseteq \Bbb{R}^3$, with
$$ v_1 = \begin{bmatrix} 1 \\  2 \\  1\\  \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}
$$
$\{v_1,v_2,v_3\}$  is linearly dependent iff $\{v_1,v_2,v_3\}$ is not independent, therefore if $$\forall \alpha_1,\alpha_2,\alpha_3 \in \Bbb{R}(\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3 \cdot v_3=0_{\Bbb{R}^3} \to \alpha_1=\alpha_2=\alpha_3=0)$$ is false. I consider $$\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3 \cdot v_3=\alpha_1 \cdot \begin{bmatrix} 1 \\  2 \\  1\\  \end{bmatrix} + \alpha_2 \cdot \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix} + \alpha_3 \cdot \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}=$$$$=\begin{bmatrix} \alpha_1 \\  2\alpha_1 \\  \alpha_1\\  \end{bmatrix} + \begin{bmatrix} \alpha_2 \\ a\alpha_2 \\ 3\alpha_2 \\ \end{bmatrix} +  \begin{bmatrix} 0 \\ 2\alpha_3 \\ b\alpha_3 \\ \end{bmatrix}=\begin{bmatrix} \alpha_1+\alpha_2 \\  2\alpha_1+a\alpha_2+2\alpha_3 \\  \alpha_1+3\alpha_2+b\alpha_3 \\  \end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$$ I consider the linear system $\Sigma:=\left\{\begin{matrix}
\alpha_1+\alpha_2=0\\ 
2\alpha_1+a\alpha_2+2\alpha_3=0\\ 
\alpha_1+3\alpha_2+b\alpha_3=0
\end{matrix}\right.$ it is homogeneous system and $$\mathbf{rnk}(\Sigma)\neq3 \leftrightarrow Sol(\Sigma)\neq\{(0,0,0)\}$$
A: You said the vectors are independent 

...when $a-2 \neq 0 $ and $b - \frac{4}{(a-2)} \neq 0$. 

However, the question was about "conditions on the scalars so that the set of vectors is linearly dependent"...
