Let $G$ $=$ {x, y, z} and let $F(G)$ be the set of all real-valued functions defined on $G$. $f$, $g$, and $h$ are functions defined as follows:
$f(x)$ $=$ $1$ $g(x)$ $=$ $1$ $h(x)$ $=$ $1$
$f(y)$ $=$ $1$ $g(y)$ $=$ $1$ $h(y)$ $=$ $0$
$f(z)$ $=$ $1$ $g(z)$ $=$ $0$ $h(z)$ $=$ $0$
Prove that {f, g, h} is a basis of $F(G)$ and find the coordinates of $k$ relative to this basis such that $k(a)$ $=$ $2$, $k(b)$ = $-1$, $k(c)$ $=$ $2$.
So I got that the augmented matrix was:
$\begin{pmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{pmatrix}$
which simply row reduced to:
$\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$
So since the rank was $3$, the columns would be linearly independent and therefore {f, g, h} was a basis.
And for the coordinates, I got the matrix:
$\begin{pmatrix}1&1&1&2\\ 1&1&0&-1\\ 1&0&0&2\end{pmatrix}$
which row reduced to:
$\begin{pmatrix}1&0&0&2\\ 0&1&0&-3\\ 0&0&1&3\end{pmatrix}$
So I thought the coordinates would simply be:
$\begin{pmatrix}2\\ -3\\ 3\end{pmatrix}$
But I am not sure if my derivations of the augmented matrices is correct as the numbers are written as results of functions and not simply as elements in a matrix.
Any help?
Personally I would skip the matrices and do a proof directly from the definitions of $f,g,h$.
Suppose $af + bg + ch = 0$ for some $a,b,c \in \mathbb{R}$. Then by definition of $f,g,h$ we have $a+b+c=0, a+b=0, a=0$ which implies that $a=b=c=0$. So $f,g,h$ are linearly independent.
Now suppose $k \in F(G)$ and $k(x)=a, k(y)=b, k(z)=c$ for some $a,b,c \in \mathbb{R}$. Then $k = cf + (b-c)g + (a-b)h$. So $f,g,h$ spans $F(G)$.
We can also use the formula above to find the coordinates of the $k$ in your original question.