Finding a $t$ for which three vectors are linearly dependent I am trying to solve the following problem:

Find a $t \in \mathbb{R}$ so that $(3,1,4)$, $(2,-3,5)$ and $(5,9,t)$ are linearly dependent.

I know how to solve for $t$, but the approach didn't seem very systematic to me. Instead of finding $a_1, a_2, a_3$, not all $0$, such that $a_1(3,1,4) + a_2(2,-3,5) + a_3 (5,9,t) = 0$, I set up
$$
a(3,1,4) + b(2,-3,5) = (5,9,t)
$$
for $a,b \in \mathbb{R}$. Then
$$
(3a + 2b, a - 3b, 4a + 5b) = (5,9,t),
$$
so we extract the system of equations
\begin{align*}
3a + 2b & = 5 \\ 
a - 3b & = 9 \\ 
4a + 5b & = t. 
\end{align*}
Solving the first two equations, we get $a = 3$, $b = -2$. Plugging into the third equation, we get $t = 2$.

My only explanation for why I did this is "because it worked" and allowed me to avoid computing or row-reducing with $t$ directly. I think this guarantees that these three equations are consistent because $a$ and $b$ are determined by the first two equations, and I can select $t$ to force them to satisfy the third.
Could someone help me understand the sequencing of steps? The first thing I would have done if given the problem without having tried anything is to select $a_1, a_2, a_3$ as above and try to row-reduce, but that's quite messy (and not covered at this point in Axler's book). I'd like to have a better intuitive understanding than this than "because it works."
 A: So, what you seem to be exploiting here is Axler's Linear Dependence Lemma:

A list of vectors $(v_1, \ldots, v_n)$ is linearly dependent if and only if some $v_i \in \operatorname{span}(v_1, \ldots, v_{i-1})$.

It's not difficult to see that $(3, 1, 4)$ and $(2, -3, 5)$ are linearly independent, as neither is a scalar multiple of the other. So, if we add a third vector $(5, 9, t)$ to the list, it will form a linearly dependent list if and only if it is a linear combination of the other two.
This is what you're checking here. By making the system consistent, you are guaranteeing a linear combination of $(3, 1, 4)$ and $(2, -3, 5)$ that produces $(5, 9, t)$. If you can form such a linear combination (i.e. if the system is consistent), then you have linear dependence. If not, then you have linear independence.
A: Theo's answer perfectly justifies why your method works. However since you mentioned that you want to reduce computation and row reducing, you might be interested in Cramer's Rule.
So it follows that the system of equation
$a_1(3,1,4) + a_2(2,-3,5) + a_3 (5,9,t) = 0$ will have a non-zero solution if the determinant is $0$. If the determinant is $0$ then it will have a unique solution , i.e $a_{1}=a_{2}=a_{3}=0$ which would imply linear independence.
So it suffices to just find $t$ such that :-
$$\begin{vmatrix} 3&1&4\\2&-3&5\\ 5&9&t\end{vmatrix}=0$$
A: For a more intuitive understanding of linear (in)dependence, please see my answer here.
It’s also useful to think about the geometry of the problem. As $t$ varies, the points $(5,9,t)$ trace out a line. In fact it’s a line parallel to the $z$-axis. The vectors that are linear combinations of $(3,1,4)$ and $(2,-3,5)$ form a plane. Essentially what you’re doing is finding the value of $t$ where the line intersects the plane.
The logic you used makes perfect sense, even though you didn’t use the standard (rather unintuitive) definition of linear independence.
