Answer
$$[x]_B=\left[ \frac{58}{13} \quad -1 \quad -\frac{4}{13} \quad 3\right]^T.$$
Work Step by Step
Let $x=(2,-1,4,3)$ and $B=\{ (\frac{5}{13},0,\frac{12}{13},0),(0,1,0,0),(-\frac{12}{13},0,\frac{5}{13},0),(0,0,0,1) \}$
To find the coordinates of $x$ relative to $B$, we have to find the following;
$$\langle (2,-1,4,3),(\frac{5}{13},0,\frac{12}{13},0)\rangle =\frac{10}{13}+\frac{48}{13}=\frac{58}{13}$$
$$\langle (2,-1,4,3),(0,1,0,0)\rangle = -1$$
$$\langle (2,-1,4,3),(-\frac{12}{13},0,\frac{5}{13},0)\rangle = -\frac{24}{13}+\frac{20}{13}=-\frac{4}{13}$$
$$\langle (2,-1,4,3),(0,0,0,1)\rangle = 3.$$
Then, the coordinate matrix of $x$ relative to $B$ is given by
$$[x]_B=\left[ \frac{58}{13} \quad -1 \quad -\frac{4}{13} \quad 3\right]^T.$$
