Let $v_1 = (1, 0, 2), v_2 = (1, 1, a)$, and $v_3 = (a, 1, −1)$. Find the value(s) of $a$ for which $v_1, v_2$, and $v_3$ are linearly dependent

Question

I'm struggling to solve this question without the use of the determinant (I'm not allowed to use it). I've tried setting up a matrix with the vectors and putting that matrix into reduced row echelon form and solving that, but no matter what I do I always end up with a value of $a$ that, when substituted back into the vectors, results in a matrix that is linearly independent.

I've also tried considering other properties of linearly dependent vectors, such as the fact $2$ of those vectors are linearly dependent if they are scalar multiples of each other or that one of them is a linear combination of the others, but I am still unable to produce a valid value for $a$. Any help would be appreciated, thank you in advance.

My attempt: matrix in rref $$ \begin{pmatrix} 1 & 0 & a-1 \\ 0 & 1 & 1 \\ 0 & 0 & -3a-1 \\ \end{pmatrix} $$ I believe -3a -1 = 0 should yield the correct solution, but I seem to be wrong and I'm not quite sure why.

Answer 1

Suppose there was a relation so that $xv_1+yv_2=v_3$ (this is the definition of linear dependence). From the second entries in the vectors, you know $y$ must be $1$. So for the first entry you have $x+1=a$ and for the third entry you have $2x+a=-1 \implies 2x+1=-a$. Adding the two results, $3x+2=0 \implies x=\frac{-2}{3}$. Plugging this back in we have $\frac{-2}{3}+1=a \implies a=\frac{1}{3}$.

Answer 2

You can choose any two pairs from $v_1, v_2, v_3$ (in this case, I choose $v_1$ and $v_2$) and set up an equation as below: $$c \cdot v_1 + c' \cdot v_2 = v_3$$

Bringing the scalars into the matrix representation of the vector,

$$\left[ \begin{matrix}c + c' \\ c' \\ 2c + ac'\end{matrix}\right] = \left[\begin{matrix}a \\ 1 \\ -1\end{matrix}\right]$$.

Then we have that $c' = 1$, $c = a - 1$ and the final vector component $2c + ac'$ gives us the equation $3a - 1 = 0$. We solve, and $a = \frac{1}{3}$.