\(\text{This post is a mess. I'm assuming you mean}\\ v= \begin{pmatrix}1\\3\end{pmatrix}, ~u = \begin{pmatrix}3\\2\end{pmatrix}\\ \text{write }u = \begin{pmatrix}1\\2\end{pmatrix} = a v + b u \text{ if possible}\)
\(\text{well we have to establish that }u \text{ and }v \text{ are linearly inedpendent}\\ \text{we put them into a matrix and check that the determinant is non-zero}\\ \left|\begin{pmatrix}1&3\\3&2\end{pmatrix}\right| = 2-9=-7\neq 0\\ \text{so the two vectors are linearly independent}\)
\(\text{we have the matrix equation }\\ \begin{pmatrix}1 &3 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}\)
\(\text{using Gauss elimination}\\ \begin{pmatrix}1&3&|&1\\3&2&|&2\end{pmatrix}\\ \begin{pmatrix}1&3&|&1\\0&-7&|&-1\end{pmatrix}\\ \begin{pmatrix}1&3&|&1\\0&1&|&\dfrac 1 7\end{pmatrix}\\ \begin{pmatrix}1&0&|&\dfrac 4 7\\0&1&|&\dfrac 1 7\end{pmatrix}\\\)
\(a = \dfrac 4 7,~b = \dfrac 1 7\\ \dfrac a b = 4\)
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