9. Let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be a linear transformation that maps $u=\left[\begin{array}{c}-23 \\ 1\end{array}\right]$ into $\left[\begin{array}{l}1 \\ 2\end{array}\right]$ and maps $v=\left[\begin{array}{l}0 \\ 1\end{array}\right]$ into $\left[\begin{array}{l}0 \\ 1\end{array}\right]$. Use the fact that $T$ is linear to find $T(3 u)$ and $T(3 u+2 v)$.
Solution
Step 1 ：Given that the transformation of the vectors u and v under T are Tu = [1, 2] and Tv = [0, 1] respectively.
Step 2 ：Using the property of linear transformations T(cu) = cT(u) for any scalar c and vector u, we can find T(3u) by multiplying the transformation of u under T by 3. This gives us T_3u = [3, 6].
Step 3 ：Similarly, using the property of linear transformations T(u+v) = T(u) + T(v) for any vectors u and v, we can find T(3u+2v) by adding 3 times the transformation of u under T to 2 times the transformation of v under T. This gives us T_3u_2v = [3, 8].
Step 4 ：Hence, the transformation of the vector 3u under T is \(\boxed{\left[\begin{array}{l}3 \\ 6\end{array}\right]}\) and the transformation of the vector 3u+2v under T is \(\boxed{\left[\begin{array}{l}3 \\ 8\end{array}\right]}\).
