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I know vectors are linearly dependent if one can be written as a linear combination of it's predecessors; so I have the following three equations:

(1)ax = 1

(2)2a + 3b = 2

(3)bx = -1

By adding (1) + (3): x(a + b) = 0, so either x = 0 or a + b = 0, but x cannot be 0 since ax = 1.

By solving a + b = 0 and 2a + 3b = 2 simultaneously gives a = -2 and b = 2, hence x = -0.5.

Is x = -0.5 the only answer to this, or is there something I'm missing? Are there any more solutions?

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    You assume that if they are dependent then the last one can be written as a combination of the others, but that's not the only way they can fail to be independent. It's clear that if $x=0$, then the second vector can be written as a multiple of the first and thus they are also not independent. – 2017-02-01

2 Answers 2

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Consider the matrix $A=\begin{pmatrix} x&2&0\\0&3&x\\1&2&-1 \end{pmatrix}$. Then $\det(A)\neq 0$ if and only if the rows are linearly independent. Hence it suffices to calculate $\det(A)$. A short calculation yields $\det(A)=-x(2x+1)$. Hence the vectors are linearly dependent if and only if $x=0$ or $x=-\frac{1}{2}$.

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Yes your answer is correct.

When x = 0. Then a = b = 0. That not satisfies equation (2).

So x = -0.5 is the only solution.