Université de Montréal MAT1600 tutoring

Test linear independence

Adding coordinates gives $$u+v=(1,3,1)=w$$. Hence $u+v-w=(0,0,0)$ is a linear combination equal to the zero vector, with coefficients that are not all zero. This is exactly why the three vectors are dependent.

Reduce to two independent vectors and compare coefficients

The target is $(2,5,1)=2u+v$, and replacing $w$ by $u+v$ gives $$\begin{aligned}&(a+c)u+(b+c)v\\&=2u+v\end{aligned}$$.

To check that coefficients can be compared, suppose $\alpha u+\beta v=(0,0,0)$, where $\alpha$ and $\beta$ are real numbers. The first coordinate forces $\alpha=0$, and the third forces $\beta=0$. Thus $u$ and $v$ are independent, so $a+c=2$ and $b+c=1$.

Describe every solution and the free parameter

Neither equation fixes $c$, so let $t$ denote its freely chosen real value. Then $c=t$, $a=2-t$, and $b=1-t$.

The complete solution is $$\begin{aligned}&(a,b,c)\\&=(2-t,1-t,t)\end{aligned}$$ for $t\in\mathbb{R}$. Here the parameter labels infinitely many coefficient triples; it is not an extra equation or an unknown to solve for.

Check the coordinates and compare an impossible target

$$\begin{aligned}&(a+c,2a+b+3c,b+c)\\&=(2,5,1)\end{aligned}$$ after substitution, for every $t$.

For comparison, any combination of these vectors has second coordinate equal to twice its first plus its third. The target $(2,5,2)$ fails that condition because $5\ne 2(2)+2$, so that different target would have no solution.