Answer
$$[x]_B=\left[ \sqrt{5} \quad 2\sqrt{5}\right]^T.$$
Work Step by Step
Let $x=(-3,4)$ and $B=\{ (\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5}),(-\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5}) \}$
To find the coordinates of $x$ relative to $B$, we have to find the following;
$$\langle (-3,4), (\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5})\rangle =-\frac{3\sqrt{5}}{5}+\frac{8\sqrt{5}}{5}= \sqrt{5} $$
$$\langle (-3,4),(-\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5})\rangle =\frac{6\sqrt{5}}{5}+\frac{4\sqrt{5}}{5} =2\sqrt{5}.$$
Then, the coordinate matrix of $x$ relative to $B$ is given by
$$[x]_B=\left[ \sqrt{5} \quad 2\sqrt{5}\right]^T.$$
