Precalc help

Let $\langle \cdot, \cdot \rangle$ denote the inner product. Then 

$\dfrac{\langle \mathbf u, \mathbf v \rangle}{\|\mathbf u\|\|\mathbf v\|} = \cos 60^\circ = \dfrac12\\ \langle \mathbf u ,\mathbf v \rangle = \dfrac{3\cdot 2}2 = 3$

$A = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix}\\ A \mathbf u = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix} \mathbf u = \begin{pmatrix}\langle \mathbf u,\mathbf u \rangle\\\langle \mathbf v, \mathbf u\rangle\end{pmatrix} = \begin{pmatrix}\|\mathbf u\|^2\\\langle \mathbf u, \mathbf v\rangle\end{pmatrix} = \begin{pmatrix}9\\3\end{pmatrix}\\$

Similarly we have

$A \mathbf v = \begin{pmatrix}\langle \mathbf u, \mathbf v\rangle \\\|\mathbf v\|^2\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}$

If this is not the notation you are used to, $\langle \mathbf u,\mathbf v \rangle$ is just $\mathbf u \cdot \mathbf v$.