1. **Problem statement:** We have two linear transformations represented by matrices: $$T=\begin{pmatrix}4 & 1 \\ a & b\end{pmatrix}$$ and $$S=\begin{pmatrix}1 & 2 \\ -4 & -1\end{pmatrix}$$ For transformation $T$, the line $y=-2x$ is twinned with $y=3x$. We need to find $a$ and $b$. For transformation $S$, we need to find the line twinned with $y=3x$. --- 2. **Understanding twinned lines under linear transformations:** A line $y=mx$ can be represented by the vector $\mathbf{v} = \begin{pmatrix}1 \\ m\end{pmatrix}$. Under transformation $M$, the image of $\mathbf{v}$ is $M\mathbf{v}$. The twinned line means the image of one line under $M$ is another line, so: $$M \begin{pmatrix}1 \\ m_1\end{pmatrix} = k \begin{pmatrix}1 \\ m_2\end{pmatrix}$$ for some scalar $k \neq 0$. This means the transformed vector is a scalar multiple of the vector representing the twinned line. --- 3. **Part (a): Find $a$ and $b$ for $T$ given lines $y=-2x$ and $y=3x$ are twinned.** Let $m_1 = -2$ and $m_2 = 3$. Write the equation: $$T \begin{pmatrix}1 \\ -2\end{pmatrix} = k \begin{pmatrix}1 \\ 3\end{pmatrix}$$ Calculate left side: $$\begin{pmatrix}4 & 1 \\ a & b\end{pmatrix} \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}4 \times 1 + 1 \times (-2) \\ a \times 1 + b \times (-2)\end{pmatrix} = \begin{pmatrix}4 - 2 \\ a - 2b\end{pmatrix} = \begin{pmatrix}2 \\ a - 2b\end{pmatrix}$$ Set equal to $k \begin{pmatrix}1 \\ 3\end{pmatrix} = \begin{pmatrix}k \\ 3k\end{pmatrix}$: From first component: $$2 = k \implies k = 2$$ From second component: $$a - 2b = 3k = 3 \times 2 = 6$$ So: $$a - 2b = 6 \quad (1)$$ --- 4. **Use the fact that $T$ is a linear transformation matrix, and the twinning is mutual.** Since $y=3x$ is twinned with $y=-2x$ under $T$, the inverse transformation $T^{-1}$ should map $y=3x$ back to $y=-2x$. Alternatively, the vector $\begin{pmatrix}1 \\ 3\end{pmatrix}$ under $T$ should be a scalar multiple of $\begin{pmatrix}1 \\ -2\end{pmatrix}$: $$T \begin{pmatrix}1 \\ 3\end{pmatrix} = l \begin{pmatrix}1 \\ -2\end{pmatrix}$$ Calculate left side: $$\begin{pmatrix}4 & 1 \\ a & b\end{pmatrix} \begin{pmatrix}1 \\ 3\end{pmatrix} = \begin{pmatrix}4 + 3 \\ a + 3b\end{pmatrix} = \begin{pmatrix}7 \\ a + 3b\end{pmatrix}$$ Set equal to: $$l \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}l \\ -2l\end{pmatrix}$$ From first component: $$7 = l \implies l = 7$$ From second component: $$a + 3b = -2l = -2 \times 7 = -14$$ So: $$a + 3b = -14 \quad (2)$$ --- 5. **Solve the system of equations (1) and (2):** $$\begin{cases} a - 2b = 6 \\ a + 3b = -14 \end{cases}$$ Subtract (1) from (2): $$(a + 3b) - (a - 2b) = -14 - 6$$ $$a + 3b - a + 2b = -20$$ $$5b = -20 \implies b = -4$$ Substitute $b = -4$ into (1): $$a - 2(-4) = 6$$ $$a + 8 = 6$$ $$a = 6 - 8 = -2$$ --- 6. **Answer for part (a):** $$a = -2, \quad b = -4$$ --- 7. **Part (b): Find the line twinned with $y=3x$ under transformation $S$.** Given: $$S = \begin{pmatrix}1 & 2 \\ -4 & -1\end{pmatrix}$$ Let the twinned line be $y = mx$. We want: $$S \begin{pmatrix}1 \\ 3\end{pmatrix} = k \begin{pmatrix}1 \\ m\end{pmatrix}$$ Calculate left side: $$\begin{pmatrix}1 & 2 \\ -4 & -1\end{pmatrix} \begin{pmatrix}1 \\ 3\end{pmatrix} = \begin{pmatrix}1 + 6 \\ -4 - 3\end{pmatrix} = \begin{pmatrix}7 \\ -7\end{pmatrix}$$ Set equal to: $$k \begin{pmatrix}1 \\ m\end{pmatrix} = \begin{pmatrix}k \\ km\end{pmatrix}$$ From first component: $$7 = k \implies k = 7$$ From second component: $$-7 = 7m \implies m = -1$$ --- 8. **Answer for part (b):** The line twinned with $y=3x$ under $S$ is: $$y = -x$$