Given $x= \left ( 6, 2, -3 \right ),$ how to find the coordinate $x$ up to the basis $V$ In the vector space $\mathbb{R}^{3},$ given two systems of vectors
$$U= \left \{ u_{1}= \left ( 4, 2, 5 \right ), u_{2}= \left ( 2, 1, 3 \right ), u_{3}= \left ( 3, 1, 3 \right ) \right \}$$
$$V= \left \{ v_{1}= \left ( 5, 2, 1 \right ), v_{2}= \left ( 6, 2, 1 \right ), v_{3}= \left ( -1, 7, 4 \right ) \right \}$$
Proved that $U$ and $V$ are two bases of $\mathbb{R}^{3}.$
Source: StackMath/@haidangel_ in.edit
In the edited part, I gave a bonus question:

Given $x= \left ( 6, 2, -3 \right ).$ How to find the coordinate $x$ up to the basis $V.$

Now I have two approaches but I don't know which one is true ? I need to the help.
First approach. Consider the linear combination
$$\alpha_{1}v_{1}+ \alpha_{2}v_{2}+ \alpha_{3}v_{3}= x$$
This is equivalent to the matrix equation
$$\begin{bmatrix} 5 & 6 & -1\\ 2 & 2 & 7\\ 1 & 1 & 4 \end{bmatrix}\begin{bmatrix} \alpha_{1}\\ \alpha_{2}\\ \alpha_{3} \end{bmatrix}= \begin{bmatrix} 8\\ 2\\ -3 \end{bmatrix}$$
To find the solution, consider the augmented matrix.
Applying elementary row operations, we obtain
$$\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 2 & 2 & 7 & 2\\ 1 & 1 & 4 & -3 \end{array} \right ]\xrightarrow{R_{3}\leftrightarrow R_{2}}\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 2 & 2 & 7 & 2 \end{array} \right ]\xrightarrow{2R_{2}- R_{3}}\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]$$
$$\left [ \begin{array}{rrr|r} 5 & 6 & -1 & 8\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]\xrightarrow{6R_{2}- 25R_{3}- R_{1}}\left [ \begin{array}{rrr|r} 1 & 0 & 0 & 174\\ 1 & 1 & 4 & -3\\ 0 & 0 & 1 & -8 \end{array} \right ]$$
It follows that the solution is $\alpha_{1}= 174, \alpha_{3}= -8, \alpha_{2}= -3- \alpha_{1}- 4\alpha_{3}= -145.$ We obtain
$$\left [ x \right ]_{V}= \begin{bmatrix} 174\\ -145\\ -8 \end{bmatrix}$$
Second approach. By the coordinate transformation equation
$$\left [ x \right ]_{V}= P_{V\rightarrow E}\cdot\left [ x \right ]_{E}= \left ( P_{E\rightarrow V} \right )^{-1}\cdot\left [ x \right ]_{E}= \begin{bmatrix} 5 & 6 & -1\\ 2 & 2 & 7\\ 1 & 1 & 4 \end{bmatrix}^{-1}\begin{bmatrix} 6\\ 2\\ -3 \end{bmatrix}= \begin{bmatrix} 176\\ -147\\ -8 \end{bmatrix}$$

In the first approach, you wrote $8$ in place of $6$. Otherwise, both approaches are correct.
