Find $x$ in $\mathbb{R}^2$ whose cordinate vector relative to the basis $B$

Question

Consider the basis $B$ of $\mathbb{R}^2$ consisting of vectors $\begin{bmatrix}5\\-1\end{bmatrix}$ and $\begin{bmatrix}-7\\-4\end{bmatrix}$. Find $x$ in $\mathbb{R}^2$ whose coordinate vector relative to the basis $B$ is $[x]_B = \begin{bmatrix}2\\5\end{bmatrix}$

$x = ?$

I put the matrices together obtaining a $3 \times 3$ matrix that I row reduced to get $\begin{bmatrix}1&0&-1\\0&1&-7/11\end{bmatrix}$ but then when I tried $x = \begin{bmatrix}-1\\-7/11\end{bmatrix}$ it said it was incorrect. I'm confused what i'm doing wrong

Answer 1

Hint:

Simply, your vector is: $$ 2 \begin{bmatrix} 5\\-1 \end{bmatrix} +5\begin{bmatrix} -7\\-4 \end{bmatrix} $$

Answer 2

Look at this question (How to find the coordinate vector x with respect to the basis B for R^3?). You are trying to to the reverse operation, which is even easier: $$[x]_B=\begin{bmatrix}2 \\ 5\end{bmatrix}_B=2\begin{bmatrix}5\\-1\end{bmatrix}_{\mathbb{R^2}}+5\begin{bmatrix}-7\\-4\end{bmatrix}_{\mathbb{R^2}}=\dots$$

Answer 3

Generally, when the coordinate of a vector, say $v$, in terms of basis of $B=\{v_1,v_2,\cdots, v_n\}$ is $C=(a_1,a_2,\cdots,a_n)$, then you have $$v=a_1v_1+a_2v_2+\cdots+a_nv_n.$$

Therefore, as the other friends said, for your case the answer is $$2 \begin{bmatrix} 5\\-1 \end{bmatrix} +5\begin{bmatrix} -7\\-4 \end{bmatrix}=\begin{bmatrix} -25\\-22 \end{bmatrix}$$