Find the coordinate vector of $x = \begin{bmatrix}-5\\-2\\0\end{bmatrix}$ with respect to the basis $B = \{ \begin{bmatrix}1\\5\\2\end{bmatrix}, \begin{bmatrix}0\\1\\-4\end{bmatrix}, \begin{bmatrix}0\\0\\1\end{bmatrix} \}$ for $\mathbb{R}^3 $
$[x]_B = ?$
So I think I have an idea, but i'm not quite sure.. Should I just put this in matrix form and then put it in RREF form?
Hint:
You are searching three coordinates $[x]_B=[x,y,z]^T$ such that: $$ x [1,5,2]^T+y[0,1,-4]^T+z[0,0,1]^T=[-5,-2,0]^T $$ this means: $$ \begin{cases} x=-5\\ 5x+y=-2\\ 2x-4y+z=0 \end{cases} $$
The aim of the exercise is to find real numbers $\lambda_1,\lambda_2,\lambda_3$ such that $$\begin{bmatrix}-5\\-2\\0\end{bmatrix}=\lambda_1 \begin{bmatrix}1\\5\\2\end{bmatrix}+\lambda_2 \begin{bmatrix}0\\1\\-4\end{bmatrix}+\lambda_3 \begin{bmatrix}0\\0\\1\end{bmatrix} $$ Now the task is quite easy, isn't it? Simply compare the coefficients of both sides of the equality, starting with $\lambda_1$, then $\lambda_2$ and lastly $\lambda_3$.
You vector in the new basis will then be given by $$ \begin{bmatrix}\lambda_1\\\lambda_2\\ \lambda_3\end{bmatrix}.$$
As this is equivalent to saying that $$\begin{bmatrix}-5\\-2\\0\end{bmatrix}=\underbrace{\begin{bmatrix}1& 0 & 0\\5 & 1 & 0\\2& -4 & 1\end{bmatrix}}_{:=T}\begin{bmatrix}\lambda_1\\\lambda_2\\ \lambda_3\end{bmatrix}. $$ You could as well find the inverse of the matrix $T$ and then you obtain $$\begin{bmatrix}\lambda_1\\\lambda_2\\ \lambda_3\end{bmatrix}=T^{-1}\begin{bmatrix}-5\\-2\\0\end{bmatrix}.$$
Basically I found that you need to just Put the Basis B matrices together with the vector x at the end and you get $[x]_B = \begin{bmatrix}-5\\23\\102\end{bmatrix}$
