Recipe

Find the coefficient that produces a target vector

Use when: you're given vectors and a target vector, and asked to find a scalar coefficient c so that a linear combination of the given vectors equals the target exactly

Goal: the coefficient value that makes the combination hit the target — or a clear demonstration that no such value exists

Taught in Linear combinations, first look

1. 1

   Expand the linear combination entry by entry, substituting the known vectors' numbers but leaving the coefficient as a letter.

   Example

   With \mathbf a=(1,3), \mathbf b=(2,-1), and target (7,-3): \mathbf a+c\mathbf b = (1+2c,\ 3-c).

   Why: this turns one vector equation into an ordinary algebra problem in the unknown coefficient, which you already know how to solve entry by entry.

2. 2

   Every entry of the combination must equal the corresponding entry of the target vector — that's one scalar equation per coordinate, not one equation total.

   Example

   1+2c=7 \qquad\text{and}\qquad 3-c=-3

   Why: a vector equation is really several scalar equations bundled together, one per entry — solving only one of them ignores what the other entry demands.

3. 3

   Solve each scalar equation for the coefficient on its own, without assuming the two will agree.

   Example

   1+2c=7 \Rightarrow c=3; \quad 3-c=-3 \Rightarrow c=6.

   • every equation gives the same value of the coefficient — that shared value is the answer; the target is reachable.
   • the equations give different values for the same coefficient — no single coefficient satisfies every entry at once, so the target is not reachable by this combination.

   Why: one free coefficient is fixing every entry of the result with a single number — if the entries demand different values of that number, no value can satisfy all of them together.

4. 4

   Report the coefficient that works, or state plainly that none does — either way, substitute back into the original combination and check every entry, not just the one you solved from.

   Example

   c=3 gives (7, 0), which matches the target's first entry but not its second; c=6 gives (13,-3), matching the second entry but not the first — confirming no single c works.

   Why: substituting back is the step that actually confirms the conclusion — a correctly solved equation can still hide an arithmetic slip that only a full check, against every entry, catches.