Math GRE - #7

If A is a 3 x 3 matrix such that $A\left(\begin{array}{c}
0\\
1\\
2\end{array}\right)=\left(\begin{array}{c}
1\\
0\\
0\end{array}\right)$ and $A\left(\begin{array}{c}
3\\
4\\
5\end{array}\right)=\left(\begin{array}{c}
0\\
1\\
0\end{array}\right)$, then $A\left(\begin{array}{c}
6\\
7\\
8\end{array}\right)=$

• $\left(\begin{array}{c}
  0\\
  0\\
  1\end{array}\right)$
• $\left(\begin{array}{c}
  -1\\
  2\\
  0\end{array}\right)$
• $\left(\begin{array}{c}
  1\\
  -1\\
  0\end{array}\right)$
• $\left(\begin{array}{c}
  9\\
  10\\
  11\end{array}\right)$

Solution :

This is a basic application of linearity. Note that: \[
A\left(\begin{array}{c}
6\\
7\\
8\end{array}\right)=2A\left(\begin{array}{c}
3\\
4\\
5\end{array}\right)-A\left(\begin{array}{c}
0\\
1\\
2\end{array}\right)=2\left(\begin{array}{c}
0\\
1\\
0\end{array}\right)-\left(\begin{array}{c}
1\\
0\\
0\end{array}\right)=\left(\begin{array}{c}
-1\\
2\\
0\end{array}\right)\]