Problems

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Solution:

1. Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if \(\frac{x_1^2}9+\frac{y_1^2}{4} = 1\) we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that \((x_1, y_1)\) is on \(G_1\) is equivalent to \((x_2,y_2)\) being on the \(G_2\). Since the point \((x_2,y_2)\) is a \(45^{\circ}\) rotation of \((x_1,y_1)\) anticlockwise about the origin, this means \(G_2\) is a \(45^{\circ}\) anticlockwise rotation of \(G_1\).
2. 1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
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3. Consider the transformation \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) which is a shear, leaving the \(x\)-axis invariant. Then we must have:
   

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   Since the shear leaves lines of the form \(y = k\) invariant, the points where \(\frac{\d y}{\d x} = 0\) must also map to points where this is true, ie \((\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)\) map to points \((\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)\) where the tangent is horizontal. The line \(x = c\) map back to lines \(\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}\), ie \(y = -\frac12 x- \frac{c}{2}\). Therefore we are interested in points on the original curve where the gradient is \(-\frac12\), ie \((\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})\), these map to \((\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})\)

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Solution:

1. Suppose \((x,y)\) is on the line of invariant points, then \begin{align*} &&\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\ \Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*} \end{align*} Therefore either \(x = 0, y = 0\) or \((a-1)(d-1)-bc = 0\) \(\Rightarrow ((a-1)(d-1)-bc)xy = 0\). We also know this is true for all values \(x,y\) on the line of invariant points. If there is one where both \(x \neq 0, y \neq 0\) we are done, otherwise the line of invariant points must be one of the axes. ie but then one of \(\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) is true and we'd also be done. If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since \(\mathbf{A}\) is a linear operator) and so every point is fixed. ie \(\mathbf{A} = \mathbf{I}\).
2. Suppose \((a-1)(d-1) -bc = 0\) and \(b \neq 0\) then I claim that \(y = \frac{1-a}{b}x\) is a line of invariant points. It's clear that the first equation will be satisfied in \((*)\) so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied. If \(b = 0\) then \((a-1)(d-1) = 0\), so our matrix must look like \(\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}\) (if \(d \neq 1\))or \(\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}\). In the first case, the line \(y = \frac{c}{1-d}x\) and in the second \(x = 0\) is an invariant line.
3. Suppose the invariant line is \(y = mx+k\) then we must have that \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\ (c+dm)x + dk \end{pmatrix} \end{align*} and \((c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)\) Since this equation must be true for all values of \(x\), and \(k \neq 0\) we can say that \(mb = d-1\) and \(-c+(a-d)m+m^2b = 0\), ie \(-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0\) if \(m \neq 0\) then \((a-1)\frac{(d-1)}{b} - c = 0\) ie our desired relation is true. If \(m = 0\) then we must have that \(y = k\) is an invariant line, ie \(d-1=0\) and \(c=0\) which also satisfies our relation.

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Solution: \begin{questionparts} \item \begin{align*} R_{-\alpha}AR_{\alpha} &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} a & b \\ b & c \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \\ &= \begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} a\cos \alpha + b \sin \alpha & -a\sin\alpha + b \cos\alpha \\ b\cos\alpha + c \sin\alpha & c\cos\alpha-b\sin\alpha \end{pmatrix} \\ &= \begin{pmatrix} a\cos^2\alpha+2b\sin\alpha\cos\alpha+c\sin^2\alpha & -a\sin\alpha\cos \alpha+b\cos^2\alpha +c\sin\alpha\cos\alpha-b\sin^2 \alpha\\ (c-a)\sin\alpha\cos \alpha +b(\cos^2\alpha-\sin^2 \alpha) & a\sin^2 \alpha -2b\sin\alpha\cos\alpha+c\cos^2\alpha \end{pmatrix} \\ &= \begin{pmatrix} * & \frac{c-a}{2}\sin2\alpha+b \cos 2\alpha\\\frac{c-a}{2}\sin2\alpha+b \cos 2\alpha & * \end{pmatrix} \end{align*} Therefore this will be diagonal if \(\tan 2\alpha = \frac{2b}{a-c} \Rightarrow \alpha = \frac12 \tan^{-1} \l \frac{2b}{a-c} \r\) \item \begin{align*} x^2+(y+2x\cot2\theta)^2 &= x^2(1 + 4\cot^22\theta) + 4\cot2\theta xy + y^2 \\ &= \begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix} 1 + 4\cot^22\theta & 2\cot 2\theta \\ 2\cot 2\theta & 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} \end{align*} Plugging this \(\mathbf{A}\) in our result from before we discover \begin{align*} \frac12 \tan^{-1} \l \frac{2b}{a-c} \r &= \frac12 \tan^{-1} \l \frac{4\cot 2\theta}{1 + 4\cot^22\theta-1} \r \\ &= \frac12 \tan^{-1} \l \tan 2 \theta \r \\ &= \theta \end{align*} Therefore, the matrix will be: \begin{align*} & \textrm{diag}\begin{pmatrix} (1+4\cot^2 2\theta)\cos^2 \theta + 4\cot2\theta \sin\theta\cos\theta + \sin^2 \theta \\ (1+4\cot^2 2\theta)\sin^2 \theta - 4\cot2\theta \sin\theta\cos\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cos^2\theta + \frac{\cos^2 2\theta}{\sin^2 \theta} + 2\cos 2\theta + \sin^2 \theta \\ \sin^2\theta + \frac{\cos^2 2\theta}{\cos^2 \theta} - 2\cos 2\theta + \cos^2 \theta \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos2\theta}{\sin^2 \theta} + 2\r \\ 1 + \cos 2\theta \l \frac{\cos2\theta}{\cos^2 \theta} - 2\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + \cos 2\theta \l \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta}\r \\ 1 -\cos 2\theta \l \frac{-\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\r \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} 1 + (\cos^2\theta - \sin^2 \theta) \cosec^2 \theta \\ 1 - (\cos^2\theta - \sin^2 \theta) \sec^2 \theta \\ \end{pmatrix} \\ =& \textrm{diag}\begin{pmatrix} \cot^2 \theta \\ \tan^2 \theta \\ \end{pmatrix} \\ \end{align*} Therefore this is a rotation of an ellipse with equation: \((\cot \theta x)^2 + (\tan \theta y)^2 = 1\), ie the shortest side and longest side are \(\cot \theta\) and \(\tan \theta\) respectively, but we know since \(0 < \theta < \tfrac{1}{4}\pi\) the shortest will be \(\tan \theta\) and the longest \(\cot \theta\).

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Solution: \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}1\\ 2 \end{pmatrix} \\ &= \frac25\begin{pmatrix}9 - 4\\ -2+12 \end{pmatrix} \\ &= \begin{pmatrix}2\\ 4 \end{pmatrix} \\ &= 2 \begin{pmatrix}1\\ 2 \end{pmatrix} \end{align*} \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}2\\ -1 \end{pmatrix} \\ &= \frac25\begin{pmatrix}18+2\\ -4-6 \end{pmatrix} \\ &= \begin{pmatrix}8\\ -4 \end{pmatrix} \\ &= 4 \begin{pmatrix}2\\ -1 \end{pmatrix} \end{align*} Consider $T^{-1} = \frac{5}{2} \frac{1}{50}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\(, so \)T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} = \begin{pmatrix}x\\ y \end{pmatrix}$ and so: \begin{align*} x^2 + y^2 & = \begin{pmatrix}x& y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} \\ &= \begin{pmatrix}X& Y \end{pmatrix} (T^{-1})^T T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}6X+2Y\\ 2X+9Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}6(6X+2Y)+2(2X+9Y)\\ 2(6X+2Y)+9(2X+9Y) \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}40X+30Y\\ 30X +85Y \end{pmatrix} \\ &= \frac{1}{80}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}8X+6Y\\ 6X +17Y \end{pmatrix} \\ &= \frac{1}{80} \l 8X^2 + 12XY + 17Y^2\r \end{align*} Therefore \(8X^2 + 12XY + 17Y^2 = 80\). The area will be \(\det T \cdot \pi = \frac{4}{25} \cdot 50 \cdot \pi = 8 \pi\). Differentiating we obtain \(2 \cdot 8 \cdot X \cdot \frac{dX}{dY} + 2 \cdot 6 \cdot X + 2 \cdot 6 \cdot Y \cdot \frac{dX}{dY} + 2 \cdot 17 Y \Rightarrow \frac{dX}{dY} = -\frac{6X + 17Y}{8X+6Y}\), at a maximum (or minimum, \(6X = -17Y\)). Therefore \begin{align*} \Rightarrow && 8X^2 + 12 \cdot \frac{6}{17}X^2 + 17 ( -\frac{6}{17} X)^2 &= 80 \\ \Rightarrow && \frac{100}{17}X^2 &= 80 \\ \Rightarrow &&X^2 &= \frac{17 \cdot 4}{5} \\ \Rightarrow && |X| = 2 \sqrt {\frac{17}{5}} \end{align*} The point \(\frac15 (3,4)\) maps to \begin{align*} \frac{2}{5}\frac{1}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}3\\ 4 \end{pmatrix} &= \frac{2}{25} \begin{pmatrix}19\\ 18 \end{pmatrix} \end{align*} So the point is \((\frac{38}{25}, \frac{36}{25})\), with gradient \(\frac{dY}{dX} = -\frac{8X+6Y}{6X + 17Y}\) which is \(-\frac{8 \cdot 19+6 \cdot 18}{6\cdot 19 + 17 \cdot 18} = -\frac{13}{21}\) therefore the equation is \(21Y+13X = 50\)

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Solution:

We can see that \(\mathbf{M}\) sends \(\begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) to itself, and \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix}\) to \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix} + k \begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) Therefore, we have: \begin{align*} && \mathbf{M} \begin{pmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ && \sec \alpha \mathbf{M} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ \Rightarrow && \mathbf{M} &= \cos \alpha\begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \\ &&&= \cos\alpha \begin{pmatrix}\cos \alpha -k\sin\alpha + \frac{\sin^2 \alpha}{\cos \alpha} & \sin \alpha + k \cos \alpha - \sin \alpha \\ \sin \alpha - \sin \alpha - k\frac{\sin^2 \alpha}{\cos \alpha} & \frac{\sin^2 \alpha}{\cos \alpha} + \cos\alpha + k \sin \alpha \end{pmatrix} \\ &&&= \begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\ -k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha \end{pmatrix} \end{align*} Suppose $\begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} = \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix}$ then, \begin{align*} && \begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} &= \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix} \\ \Leftrightarrow && \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) & pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha)\\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) & rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha)\end{pmatrix} &= \\ && \qquad \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha & q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ -pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) & -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{pmatrix} \\ \Leftrightarrow && \begin{cases} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) &= p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha \\ pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha) &=q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) &=-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) \\ rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha) &= -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ p\cos^{2}\alpha + q\sin\alpha\cos\alpha &=-q\sin\alpha\cos\alpha + t\cos^{2}\alpha \\ -r\sin\alpha\cos\alpha + -t\sin^{2}\alpha &=-p\sin^{2}\alpha + r\sin\alpha\cos\alpha \\ r &= -q\tan^{2}\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\sin\alpha\cos\alpha &=(t-p)\cos^{2}\alpha \\ (p-t)\sin^{2}\alpha &=2r\sin\alpha\cos\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\tan \alpha &=(t-p) \end{cases} \\ \end{align*} as required

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Solution: In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.