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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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