1. **Problem statement:** Given the vector $$x=\begin{bmatrix}\frac{15}{7} \\ \frac{16}{7}\end{bmatrix}$$, find the transformation matrix for expanding it on the x-axis by 4 units and then compressing it on the x-axis by 1 unit. Also, graph the resulting vectors. 2. **Formula and rules:** - To expand or compress a vector on the x-axis, we multiply the x-component by the scale factor and leave the y-component unchanged. - The transformation matrix for scaling on the x-axis by a factor $k$ is: $$\begin{bmatrix}k & 0 \\ 0 & 1\end{bmatrix}$$ 3. **Expansion by 4 units on x-axis:** - Scale factor $k=4$ - Transformation matrix: $$A=\begin{bmatrix}4 & 0 \\ 0 & 1\end{bmatrix}$$ - Apply to vector $x$: $$Ax=\begin{bmatrix}4 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}\frac{15}{7} \\ \frac{16}{7}\end{bmatrix}=\begin{bmatrix}4 \times \frac{15}{7} \\ 1 \times \frac{16}{7}\end{bmatrix}=\begin{bmatrix}\frac{60}{7} \\ \frac{16}{7}\end{bmatrix}$$ 4. **Compression by 1 unit on x-axis:** - Scale factor $k=1$ (no change) - Transformation matrix: $$B=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ - Apply to vector $x$: $$Bx=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}\frac{15}{7} \\ \frac{16}{7}\end{bmatrix}=\begin{bmatrix}\frac{15}{7} \\ \frac{16}{7}\end{bmatrix}$$ 5. **Explanation:** - Expanding by 4 stretches the vector horizontally by 4 times, changing the x-component from $\frac{15}{7}$ to $\frac{60}{7}$. - Compressing by 1 means no change, so the vector remains the same. 6. **Final answers:** - Expansion matrix: $$\begin{bmatrix}4 & 0 \\ 0 & 1\end{bmatrix}$$ - Expanded vector: $$\begin{bmatrix}\frac{60}{7} \\ \frac{16}{7}\end{bmatrix}$$ - Compression matrix: $$\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ - Compressed vector: $$\begin{bmatrix}\frac{15}{7} \\ \frac{16}{7}\end{bmatrix}$$