Problems

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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: \begin{array}{c|ccccc} a & 0 & 1 & 2 & 3 & 4 \\ a^2 & 0 & 1 & 4 & 4 & 1 \end{array} Therefore \(a^2 \in \{0,1,4\}\) and so we can have \begin{array} $2a^2+b^2 & 0 & 1 & 4 \\ \hline 0 & 0 & 1 & 4 \\ 1 & 2 & 3 & 1 \\ 4 & 3 & 4 & 2 \end{array} Therefore the only solution must have \(5 \mid a,b\), but then we can write them has \(5a'\) and \(5b'\) so the equation becomes \(2\cdot25 a'^2 + 25b'^2 = 5c^2\) ie \(5 \mid c^2 \Rightarrow 5 \mid c\). But that means we can always divide \((a,b,c)\) by \(5\), which is clearly a contradiction if we consider the lowest power of \(5\) dividing \(a,b,c\) for any solution.

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Solution: \begin{align*} && 0 &\leqslant (a_1 - a_2)^2 \\ &&&= a_1^2 -2a_1a_2 + a_2^2 \\ &&&= (a_1+a_2)^2 -4a_1a_2 \\ \Leftrightarrow && a_1a_2 &\leqslant \left ( \frac{a_1+a_2}2 \right)^2 \end{align*} Claim: \((*)\) is true Proof: (By induction) We have already proven the base case. Suppose it is true for some \(m\), then consider \(m+1\) \begin{align*} && a_1 \cdots a_{2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right)^{2^m} \tag{by (*)} \\ && a_{2^m+1} \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right)^{2^m} \tag{by (*)} \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right) \\ && (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right) \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} \cdot (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{ (a_1 \cdots a_{2^m})^{1/2^m} +(a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ \frac{a_1 + \cdots + a_{2^m}}{2^m}+\frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ a_1 + \cdots + a_{2^m}+a_{2^m+1} + \cdots + a_{2^{m+1}} }{2^{m+1}} \right )^2 \\ \Rightarrow && a_1 \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^{m+1}}}{2^{m+1}} \right)^{2^{m+1}} \end{align*} Which is precisely \((*)\) for \(m+1\). Therefore our statement is true by induction. Suppose \(n < 2^m\) and \(b_1 = a_1, b_2 = a_2, \cdots b_n = a_n\) and \(b_{n+1} = \cdots = b_{2^m} = A\) where \(A = \frac{a_1 + \cdots + a_n}{n}\) then \begin{align*} && b_1 \cdots b_n \cdot b_{n+1} \cdots b_{2^m} &\leq \left ( \frac{b_1 + \cdots + b_n + b_{n+1} + \cdots + b_{2^m}}{2^{m}} \right)^{2^m} \\ \Leftrightarrow && a_1 \cdots a_n \cdot A^{2^m-n} &\leq \left ( \frac{a_1 + \cdots + a_n + (2^m-n)A}{2^m} \right)^{2^m} \\ &&&= \left ( \frac{nA + (2^m - n)A}{2^m} \right)^{2^m} \\ &&&= A^{2^m} \\ \Rightarrow && a_1 \cdots a_n &\leq A^n \\ \Rightarrow && (a_1 \cdots a_n)^{1/n} &\leq A = \frac{a_1 + \cdots + a_n}{n} \end{align*}

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Solution: \begin{align*} \mathbf{PAP} &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix}\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\begin{pmatrix}15 & -10\\ 30 & 5 \end{pmatrix} \\ &= \begin{pmatrix}75 & 0\\ 0 & -25 \end{pmatrix} \end{align*} Which is diagonal as required. Letting \(\mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a}\) \begin{align*} && \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11&=0 \\ \Leftrightarrow && (\mathbf{P}\mathbf{x}_{1}+\mathbf{a})^T\mathbf{A}(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 + \mathbf{x}_{1}^T\mathbf{PAa} + \mathbf{a}^T\mathbf{AP}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T(\mathbf{P}\mathbf{x}_{1}+\mathbf{a}) - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \end{align*} It would be nice if we picked \(\mathbf{a}\) such that \(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T = 0\), if \(\mathbf{a} = \begin{pmatrix} a_1 \\a_2 \end{pmatrix}\) then this equation becomes: \begin{align*} && 2\begin{pmatrix}-a_1 + 8a_2 & 8a_1+11a_2 \end{pmatrix} + \begin{pmatrix}18 & 6 \end{pmatrix} &= 0 \\ \Rightarrow && a_1 = 1, a_2 = -1 \end{align*} So our equation is now \begin{align*} && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1 +(2\mathbf{a}^T\mathbf{A}+\mathbf{b}^T)\mathbf{P}\mathbf{x}_{1}+\mathbf{a}^T\mathbf{Aa} + \mathbf{b}^T\mathbf{a} - 11 &= 0 \\ \Leftrightarrow && \mathbf{x}_{1}^T\mathbf{PAP}\mathbf{x}_1-6 +12 - 11 &= 0 \\ \Leftrightarrow && 25(3x_1^2 - y_1^2) &= 5 \\ \Leftrightarrow && 3x_1^2 - y_1^2 &= \frac{1}{5} \end{align*}

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Solution: While descending, the body experiences the force \(-\frac15mg - 2mv^2\). \begin{align*} \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\ \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T \end{align*} We care about when \(v_1 = 0\), ie \(\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4\) seconds. If the body enters at speed \(1\ \mathrm{ms}^{-1}.\) then for the first part of it's journey it will experience forces \(-\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\ \Rightarrow && \frac14 \ln (1 + v^2) &= -x \end{align*} Therefore the depth is \(\frac14 \ln 2\) metres. When the body is rising, it experiences forces of: \(\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\ \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\ \Rightarrow && 1-v^2 &= \frac12 \\ \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1} \end{align*} This will take \begin{align*} \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\ \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\ \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T \end{align*} Since \(v = \frac{1}{\sqrt{2}}\) \begin{align*} T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\ &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*} and therefore the total time will be: \begin{align*} & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\ =& \frac{\pi}{8} + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*}

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Solution:

Since the sphere is on the point of slipping at both \(Q_1\) and \(Q_2\), \(F_{r1} = \mu_1 R_1\) and \(F_{r2} = \mu_2 R_2\) \begin{align*} \text{N2}(\uparrow): && -mg-Mg-\mu_1 R_1 + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\ \text{N2}(\rightarrow): && -R_1 + R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= 0 \\ \\ \Rightarrow && R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= R_1 \\ % && -mg-Mg+\mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\ % \\ \overset{\curvearrowleft}{O}: && mg - \mu_1 R_1 - \mu_2R_2 &= 0 \\ \Rightarrow && \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= -mg \\ && \mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \\ && \quad \quad \mu_2 R_2 \cos \alpha - \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= Mg \\ \Rightarrow && \frac{\mu_2+\mu_1 \l \cos \alpha - \mu_2 \sin \alpha \r }{\mu_1 ( \cos \alpha - \mu_2 \sin \alpha) + \sin \alpha + \mu_2 \cos \alpha - \mu_1 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 } &= \frac{m}{M} \\ && \frac{\mu_2+\mu_1 \cos \alpha - \mu_1\mu_2 \sin \alpha }{\cos \alpha (-2\mu_1+\mu_2) + \sin \alpha (1 +2\mu_1\mu_2) -\mu_2} &= \frac{m}{M} \end{align*} If instead the sphere is about to roll about \(Q_2\), then the forces at \(Q_1\) will be \(0\), we can then take moments about \(Q_2\). Looking at perpendicular distances from \(Q_2\) to \(O\) and \(P\) we have \(r \cos \alpha\) and \(r(1-\cos \alpha)\) \begin{align*} \overset{\curvearrowleft}{Q_2}: && mg (1 - \cos \alpha) - Mg \cos \alpha &= 0 \\ \Rightarrow && \frac{1}{\sec \alpha-1} &= \frac{m}{M} \end{align*}

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Solution: \begin{align*} && \text{energy when projected} &= \frac12 m(2kgh) \\ &&&= kghm \\ && \text{energy when hitting ceiling the first time} &= mgh + \frac12 m v^2 \\ \text{COE}: && kghm &= mgh + \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh(k-1) \end{align*} It will rebound with speed \(\sqrt{2gh(k-1)}a\). \begin{align*} && \text{energy when rebounding from ceiling} &=gh(k-1)a^2 + mgh \\ && \text{energy before hitting the floor} &= \frac12 mv^2 \\ \text{COE}: && gh(k-1)a^2 + mgh &= \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh((k-1)a^2+1) \end{align*} The ball will rebound with kinetic energy \(m gh((k-1)a^2+1)a^2 = mgh((k-1)a^4+a^2)\) And will reach the ceiling with kinetic energy \(mgh((k-1)a^4+a^2-1)\). When \(n = 1\), the kinetic energy (before hitting the ceiling for the first time) is \(mgh(k-1)\). Suppose \(s_n\) is the expression for the kinetic energy divided by \(mgh\), ie \(s_1 = k-1\), then: Clearly \(s_1 = k-1 = a^{4\cdot1-4}(k-1) + \frac{a^{4\cdot-4}-1}{a^2+1}\), so our hypothesis holds for \(n=1\). Suppose it is true for \(n\), then the \(n+1\)th time it will be: \begin{align*} s_{n+1} &= s_n a^4+a^2-1 \\ &= \left ( a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \right) a^4 + a^2 - 1 \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4}{a^2+1} + \frac{a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4+a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-1}{a^2+1} \end{align*} Which is our desired expression, therefore it is true by induction. We wont reach the ceiling if this energy is not positive, ie: \begin{align*} && 0 &\leq a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \\ \Rightarrow && \frac{1}{a^2+1}&\geq a^{4n-4}\left (k - 1 + \frac{1}{a^2+1} \right) \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{a^2+1} \cdot \frac{1}{k - 1 + \frac{1}{a^2+1}} \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{(k-1)(a^2+1)+1} \\ \Rightarrow && 4(n-1) \ln a &\geq - \ln[(k-1)(a^2+1)+1] \\ \underbrace{\Rightarrow}_{\ln a < 0} && (n-1) &\leq \frac{ - \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ \Rightarrow && n & \leq 1 -\frac{ \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ &&&= 1 -\frac{ \ln[(k-1)a^2+k]}{4\ln a} \end{align*}