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Monday 8 August 2011

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: \[
A\left(\begin{array}{c}
6\\
7\\
8\end{array}\right)=2A\left(\begin{array}{c}
3\\
4\\
5\end{array}\right)-A\left(\begin{array}{c}
0\\
1\\
2\end{array}\right)=2\left(\begin{array}{c}
0\\
1\\
0\end{array}\right)-\left(\begin{array}{c}
1\\
0\\
0\end{array}\right)=\left(\begin{array}{c}
-1\\
2\\
0\end{array}\right)\]

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