Enlarge the triangle

Let $\vec{a}$ and $\vec{b}$ be the vectors joining $(0,0)$ to $(0,2)$ and $(0,0)$ to $(3,3)$ respectively. $\vec{a} = \begin{pmatrix} 0\\2 \end{pmatrix} \quad \vec{b} = \begin{pmatrix} 3\\3 \end{pmatrix}$ $\vec{a}$ points in the direction $\begin{pmatrix} 0\\1\end{pmatrix}$ while $\vec{b}$ points in the direction $\begin{pmatrix} 1\\1 \end{pmatrix}$ . The angle between these vectors is clearly $\frac{\pi}{4}$ . This can be shown mathematically: $\vec{a} \cdot \vec{b} = ||\vec{a}||||\vec{b}||\cos\theta$ $6 = 2\cdot 3\sqrt{2}\cdot \cos\theta$ $\cos\theta = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$ $\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$ The area of the triangle prior to application of $T$ is $A = \frac{1}{2}||\vec{a}||||\vec{b}||\sin\theta$ . This can be simplified as $A = \frac{1}{2}||\vec{a}||||\vec{b}||\cos\theta\cdot \frac{sin\theta}{\cos\theta} = \frac{1}{2}\vec{a}\cdot \vec{b} \cdot \tan\theta$ $A = \frac{1}{2}\cdot 6 \cdot \tan\left(\frac{\pi}{4}\right) = 3$

Every time we apply $T$ to the $xy$ plane, all areas are scaled by $\det(T) = 1\cdot3 - 2\cdot 5 = -7$ . The negative sign just means that the orientation of space is flipped; the absolute value of all areas is scaled by $7$ . To find the number of applications required: $3 \times 7^n = 1000$ $7^n = \frac{1000}{3}$ $n = \log_7\left(\frac{1000}{3}\right) \approx 2.99$ Therefore, we need $\boxed{3}$ applications to exceed $A = 1000$ .