How do i get a/b?

Question

Consider the vectors $\mathbf{v} = \begin{pmatrix} 1\\3 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 3\\2 \end{pmatrix}$.

If the vectors $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, answer with $0$. If they aren't, find coefficients $a$ and $b$, not both $0$, such that $a \begin{pmatrix} 1 \\ 3 \end{pmatrix} + b \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$and answer with $\dfrac{a}{b}$.

I know that vectors v and w are not linearly independent but im not sure how to get $\dfrac{a}{b}$.

Answer 1

A sequence of vectors $\{\vec{v},\vec{w}\}$ are linearly independent if the equation

$$a\vec{v}+b\vec{w}=\vec{0}$$ is only satisfied by $a=b=0$. Therefore, we represent the vectors by the homogenuous linear system

$$\begin{bmatrix}1 & 3\\3&2\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}0 \\ 0\end{bmatrix}$$

solving for $a$ and $b$ we find $$\begin{cases} a+3b=0\\ 3a+2b=0 \end{cases}\implies \begin{cases} a=-3b\\ a=-\frac{2}{3}b \end{cases} \implies \begin{cases} a=0\\ b=0 \end{cases}$$

therefore the sequence of vectors $\{\vec{v},\vec{w}\}$ are linearly independent.

Answer 2

Hint

$$a+3b=0=3a+2b\implies b=2a$$

Answer 3

$a \begin{pmatrix} 1 \\ 3 \end{pmatrix} + b \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}.$This becomes the system of equations \begin{align*} a + 3b &= 0, \\ 3a + 2b &= 0. \end{align*}The first equation gives $a = -3b$, and substituting that into the second equation gives $3(-3b) + 2b = 0,$which becomes $-7b = 0$ and therefore, $b= 0$. Substituting back into the expression, we see that $a = 0$.