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?
Best Answer
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 $ab-2b-4\neq0$.