<s>[INST] Linear algebra-Coordinate vector Given Base $A=(a_1,a_2,a_3) $
$$ B=B=\left(b_{1}=\left(\begin{array}{c}
1\\
-1\\
0
\end{array}\right),b_{2}=\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right),b_{3}=\left(\begin{array}{c}
0\\
-1\\
1
\end{array}\right)\right) $$
Given transformation matrix : $ \left[I\right]_{A}^{B}=\left[\begin{array}{ccc}
c & b & a\\
b & a & c\\
a & c & b
\end{array}\right] $ and also $ \left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{A}+\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\vec{0} $
I have to find a base $A $ that fulfills these conditions.
In the solution they did: $$ \left(\begin{array}{c}
0\\
0\\
1
\end{array}\right)=b_{2}+b_{3}\Rightarrow\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\left(\begin{array}{c}
0\\
1\\
1
\end{array}\right)\Rightarrow\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{A}=-\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\left(\begin{array}{c}
0\\
-1\\
-1
\end{array}\right) $$
It doesn't add up to me, as I thought that $ \left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B} $ means taking the linear combination of $$\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}= 0b_1+0b_2+1b_3 = b_3 =\left(\begin{array}{c}
0\\
-1\\
1
\end{array}\right) $$
Would appreciate any insight
 [/INST] I think $\begin{bmatrix}0\\ 0\\ 1\\\end{bmatrix}_{B} $ means the coordinate vector of  $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} $ in the basis $B$. And since $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} = 0b_1 + 1b_2 + 1b_3$, it does seem right.
</s>