Problems

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Solution:

1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.

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Solution: \begin{align*} \mathbf{M}^2 &= \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}^2 \\ &= \begin{pmatrix} 0 \cdot 0 + (-1) \cdot 1 & 0 \cdot (-1) + (-1) \cdot 0 \\ 1 \cdot 0 + 0 \cdot 1 & 1 \cdot (-1) + 0 \cdot 0 \end{pmatrix} \\ &= \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} \\ &= - \mathbf{I} \end{align*} \begin{align*} \exp(\theta \mathbf{M}) &= \sum_{r=0}^\infty \frac1{r!} (\theta \mathbf{M})^r \\ &= \sum_{r=0}^\infty \frac{1}{r!} \theta^r \mathbf{M}^r \\ &= \cos \theta \mathbf{I} + \sin \theta \mathbf{M} \\ &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{align*} This is a rotation of \(\theta\) degrees about the origin. \begin{align*} && \mathbf{N}^2 &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}^2 \\ && &= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \\ \Rightarrow && \exp(s\mathbf{N}) &= \sum_{r=0}^\infty \frac{1}{r!} (s\mathbf{N})^r \\ &&&= \mathbf{I} + s \mathbf{N} \\ &&&= \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \end{align*} This is a shear, leaving the \(y\)-axis invariant, sending \((1,1)\) to \((1+s, 1)\). Suppose those matrices commute, for all \(s\), ie \begin{align*} && \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} \cos \theta - s \sin \theta & -\sin \theta + s \cos \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & s \cos \theta - \sin \theta \\ \sin \theta & s \sin \theta + \cos \theta \end{pmatrix} \\ \Rightarrow && \sin \theta &= 0 \\ \Rightarrow && \theta &=n \pi, n \in \mathbb{Z} \end{align*} Clearly it doesn't matter when we do nothing. If we are rotating by \(\pi\) then it also doesn't matter which order we do it in as the stretch happens in both directions equally.

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Solution:

1. This is false, for example let \(\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), then \begin{align*} \mathbf{AB} &= \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ &= \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \\ \mathbf{BA} &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ \end{align*}
2. This is also false, using the same matrices from part (i), we find: \begin{align*} (\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\ &= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ &\neq \mathbf{A}^2-\mathbf{B}^2 \end{align*}
3. This is true. Claim: The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\) Proof: \((\Rightarrow)\) Suppose \(\det\mathbf{A} \neq 0\) then \(\mathbf{A}\) has an inverse, and so we must have \(\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}\). \((\Leftarrow)\) Suppose \(\det \mathbf{A} = 0\) then \(ad-bc=0\), so consider the matrix \(\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}\) (or if this is zero, \(\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}\))
4. This is false. Consider \(\mathbf{A} = \mathbf{B} = \mathbf{0}\), then \(\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}\) has the property that \(\mathbf{X}^2 = \mathbf{0}\) for all \(x\), so at least more than 2 values

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Solution: Claim \(G\) is a group under matrix multiplication

• (Closure) Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are matrices of that form, then \(\begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{pmatrix}\), this is clearly of the required form since if \(a_1, a_2, c_1, c_2 \neq 0\) then \(a_1a_2, c_1c_2 \neq 0\)
• (Associative) By inheritance from matrix multiplication
• (Identity) Consider \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) also clearly of the required form.
• (Inverse) Consider \((ac)^{-1}\begin{pmatrix} c & -b \\ 0 & a \end{pmatrix}\), since \(ac \neq 0\) we can assume it has an inverse mod \(5\). therefore we have another matrix of the required form.

There are \(4\) possible values for \(a\) and \(c\) and \(5\) possible values for \(b\), so \(4 \times 4 \times 5 = 80\) elements, so the group is order \(80\). \(G\) is not commutative, consider \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}\) \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix}\) The elements of order \(1\) or \(2\) satisfy \(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^{-1} & -ba^{-1}c^{-1} \\ 0 & c^{-1} \end{pmatrix}\) Therefore \(a^2 = 1, c^2 = 1 \Rightarrow a, c = 1, 4\) and \(b = -ba^{-1}c^{-1} \Rightarrow b = 0\) or , \(ac = -1\), so we have \((a,b,c) = (1,0,1), (4,0,4), (1, *, 4), (4, *, 1)\) So there are \(12\) elements of order \(1\) or \(2\). But this can't be a subgroup since \(12 \not \mid 80\)

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Solution: \begin{questionparts} \item \((ab)^2 = abab = e\) (since \(ab\) has order \(2\)), but \(a^2 = e, b^2 = e \Rightarrow a^{-1} = a, b^{-1} = b\) (since \(a\) and \(b\) have order 2) so \(ba = ab\) by multiplication on the left by \(a\) and right by \(b\). \item Suppose \((cd)^n = e \Leftrightarrow d(cd)^nc = dc \Leftrightarrow (dc)^n(dc) = e \Leftrightarrow (dc)^n = e\) Therefore any number for which \((cd)^n = e\) has the property that \((dc)^n = e\) and vice-versa, in particular the smallest number for either \(cd\) or \(dc\) will also be the smallest number for the other. \item Given \(c^{-1}bc=b^r\), then \(b^{rs} = (b^r)^s = (c^{-1}bc)^s =\underbrace{(c^{-1}bc)(c^{-1}bc) \cdots (c^{-1}bc)}_{s \text{ times}} = c^{-1}\underbrace{bb\cdots b}_{s \text{ times}}c = c^{-1}b^sc\) We proceed by induction on \(n\). When \(n = 0\), we have \(b^s = b^{sr^0}\) so the base case is true. Suppose it is true for some \(n = k\), ie \(c^{-k}b^sc^k = b^{sr^k}\). Now consider \(c^{-{k+1}}b^sc^{k+1} = c^{-1}c^{-k}b^sc^kc = c^{-1}b^{sr^k}c = (b^{sr^k \cdot r}) = b^{sr^{k+1}}\) (where the second to last equality was by the previous part). Therefore if our statement is true for \(n=k\) it is true for \(n = k+1\). Therefore, since it is also true for \(n=0\), by the principle of mathematical induction it is true for all non-negative integers \(n\).

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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