Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006

For a matrix proportional to its transpose, applying the transpose twice gives $A^T=5A$ and therefore $A=25A$. Over the real numbers or complex numbers, $24A=0$ implies

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

A symmetric matrix that is also an orthogonal matrix satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B{)}^2=(I+B{)}^2$, hence $4B=0$. Thus there is no nonzero choice of $B$. The only symmetric case is $B=0$, giving $C=I$.

Yes, in even dimension. For example, choose the skew-symmetric matrixHere $B^2=-I$, so $(I+B{)}^{-1}=(I-B)/2$ and $C=(I-B{)}^2/2=-B$. It is an orthogonal skew-symmetric matrix. More generally the skew-symmetry condition $C=-C^{-1}$ gives $(I-B{)}^2=-(I+B{)}^2$, or $B^2=-I$; no real odd-dimensional matrix can satisfy that condition.

Let $S_m=I+D+\cdots +D^m$. The finite matrix geometric series telescopes to $(I-D)S_m=I-D^{m+1}$. By the assumed convergence, its terms $D^{m+1}$ tend to zero, so the limit $S$ satisfies $(I-D)S=I$. Consequently $S=(I-D{)}^{-1}$. This is the finite-dimensional Neumann series argument; invertibility by itself would not imply convergence.

For the stated nilpotent matrix, $D^3=0$ and only the first three terms remain:

The characteristic polynomial is $\det (\lambda I-M)=\lambda ({\lambda }^2-2)$. Solving the three eigenvector equations givesEach eigenspace consists of the scalar multiples of its displayed nonzero vector. Their pairwise dot products vanish: $v_{+}\cdot v_{-}=1-2+1=0$ and $v_{\pm }\cdot v_0=1-1=0$. This also agrees with the orthogonal diagonalization of a real symmetric matrix.

The Euclidean norms are $\Vert v_{\pm }\Vert =2$ and $\Vert v_0\Vert =\sqrt{2}$. With normalized eigenvectors as rows,The rows form an orthonormal basis, so $R{R}^T=I$ and $R$ is an orthogonal matrix. The zero eigenvalue means $M{v}_0=0$ for a nonzero vector; thus $M$ is singular and has no matrix inverse.

The upper-left block is a planar rotation matrix. Its real characteristic polynomial is $(\lambda -\cos \theta {)}^2+{\sin }^2\theta$, which is strictly positive for real $\lambda$ because the angle excludes integer multiples of $\pi$. Hence any real eigenvector has $x=y=0$. The remaining equation is $5z=\lambda z$, givingThere are infinitely many nonzero solutions, spanning one independent eigenvector direction.

Since $(1-i{)}^2=-2i$, division in the complex numbers gives

Multiplying by the complex conjugate of the denominator gives $(1+i)/(i-1)=-i$. Cubing yields

Write $-8i=8e^{-i\pi /2}$. The three distinct complex roots are $z=2e^{i(-\pi /6+2k\pi /3)}$, $k=0,1,2$. In Cartesian form,

The right-hand side has modulus $4$ and argument $\pi /3$. Its square roots have modulus $2$ and arguments $\pi /6$ and $7\pi /6$, giving

Complex exponentiation uses a branch of the complex logarithm. All logarithms of $i$ are $i(\pi /2+2\pi k)$ with $k\in \mathbb{Z}$, soSince the trigonometric factor is always $i$, all values in Cartesian form areThe principal logarithm selects $k=0$ and gives the principal value $i{e}^{\pi /2}$.

Use the complex exponential forms of the sine and cosine, and put $w=e^{2iz}$. Multiplying the equation by $2e^{iz}$ gives $(2-i)(w-1)/i+(w+1)=0$, which simplifies to $w=1-i$. Taking all branches of the complex logarithm,Thus every solution isThe exponential factor used to multiply the equation is nonzero, so no solutions were introduced or lost.

For a sufficiently smooth vector field on an oriented piecewise-smooth surface $S$ with boundary $\partial S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is the scalar surface-area element, and $d\mathbf{r}$ follows the positively oriented boundary. The boundary orientation is fixed by the right-hand rule: the fingers curl along $\partial S$ when the thumb points along $\mathbf{n}$.

The Cartesian curl is obtained by cross-differentiating the components:

No scalar potential exists on the full three-dimensional domain. A sufficiently smooth gradient field has zero curl, since mixed partial derivatives commute. The calculated curl is not identically zero. The fact that it vanishes at some individual points does not make the whole field a gradient field.

Orient the triangle by the path $A\to B\to C\to A$. The cross product $(B-A)\times (C-A)=(0,1,0)$ gives the positive normal vector. In the plane $y=1$, the triangular region is $x,z\ge 0$, $x+z\le 1$. The surface integral in Stokes theorem is thereforeFor the boundary line integral, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\le t\le 1$. Its contribution isThus both oriented integrals equal $1/15$. Reversing the orientation would reverse both signs.

The coordinate intercepts suggest the normal vectorIndeed $\mathbf{n}\cdot \mathbf{a}=\mathbf{n}\cdot \mathbf{b}=\mathbf{n}\cdot \mathbf{c}=1$, so the two independent edge vectors in the plane are both orthogonal to $\mathbf{n}$.

The plane equation in normal form isAn equivalent vector parametrization is $\mathbf{r}=\mathbf{a}+s(\mathbf{b}-\mathbf{a})+t(\mathbf{c}-\mathbf{a})$, with $s,t\in \mathbb{R}$.

The distance from a point to a plane, applied at the origin, is the plane constant divided by the Euclidean norm of its normal vector. ThereforeThe closest point on a plane lies on the normal through the origin; this is an orthogonal projection onto an intercept plane.

Taking the dot product with $\mathbf{n}$ gives $\mathbf{n}\cdot \mathbf{r}=\lambda +\mu +\nu$. The necessary and sufficient condition is thereforeThese coefficients describe an affine combination of the three vertices, or their barycentric coordinate system. Nonnegativity would additionally restrict the point to the triangle; it is not required for the whole plane.

Put $S=a^{-2}+b^{-2}+c^{-2}$. The orthogonal projection onto an intercept plane is $\mathbf{d}=\mathbf{n}/S$. Comparing its coordinates with $(\lambda a,\mu b,\nu c)$ givesTheir sum is one, as the plane equation requires.

The vector $\mathbf{a}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$. Also $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}=1/S$, hence $\mathbf{d}\cdot (\mathbf{b}-\mathbf{c})=0$. Subtraction gives $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Both vectors are nonzero, soThis is the altitude property of the orthogonal projection onto an intercept plane: the projection point is the triangle's orthocenter.

The scalar triple product is a determinant:A cyclic permutation of its three columns is an even permutation, so it leaves the determinant unchanged. Thus $\det (\mathbf{a},\mathbf{b},\mathbf{c})=\det (\mathbf{b},\mathbf{c},\mathbf{a})=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, proving the identity.

Use the vector triple product identity twice:Subtracting cancels the common term, yieldingIt is generally nonzero: the cross product does not satisfy the associative property.

With $s=\sin {\theta }_0$ and $c=\cos {\theta }_0$, conversion from spherical coordinates gives $\mathbf{a}=r_0(s,0,c)$, $\mathbf{b}=r_0(-s/2,\sqrt{3}s/2,c)$ and $\mathbf{c}=r_0(-s/2,-\sqrt{3}s/2,c)$. The volume of the parallelepiped is the absolute value of their scalar triple product:The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

Let $G$ denote carrying the gene and $+$ a positive result. The law of total probability givesThe last value is to three significant figures. The much larger non-carrier population contributes most of the positive tests in this mathematical model.

By Bayes' theorem, the required conditional probability isThis is about $1.94\%$; it is a different conditioning direction from the test's $99\%$ positive probability among carriers.

The negative-test probability is $P(-)=1-0.05094=0.94906$. Applying Bayes' theorem again,

For $B$ to make a first throw, $A$ must first pass, with probability $1/2$. For $B$ then to win on that throw, $B$ must roll one of two winning faces, with probability $1/3$. The throws are independent, givingA first throw of $3$ by $A$ would make $B$ win without throwing, and is excluded by the requested event.

Before $A$'s second throw, $A$ can win either immediately, with probability $1/3$, or by passing to $B$ and having $B$ throw $3$, with probability $(1/2)(1/6)$. These are disjoint events, so

This is a two-turn renewal game. A full pair of throws has probability $(1/2{)}^2=1/4$ of passing twice and returning to the original state. The probability that $A$ wins before that restart is $5/12$. Thus first-step analysis gives $p=5/12+p/4$, orEquivalently sum the geometric series $(5/12){\sum }_{j\ge 0}(1/4{)}^j$. The probability of surviving infinitely many pairs is zero, so the eventual winner is well defined with probability one.

For the Lagrange multiplier equations, use $\mathcal{L}=x^{2}y+2y-\lambda (x^2+y^2-1)$. They give $2x(y-\lambda )=0$, $x^2+2=2\lambda y$, and $x^2+y^2=1$. If $x\ne 0$, then $\lambda =y$, implying $x^2+2=2y^2$ and therefore $y^2=1$, contradicting $x\ne 0$. Thus only $x=0$, $y=\pm 1$ remain.

The circle is compact and the function is continuous, so global extrema exist. On it, eliminating $x^2$ gives $f=3y-y^3$, which is increasing for $-1<y<1$ since its derivative is $3(1-y^2)>0$. ThereforeThese constrained extrema are flat to quadratic order along the circle, so a vanishing second derivative alone would not classify them.

Minimize $\frac{1}{2}\Vert \mathbf{r}{\Vert }^2$ subject to ${\mathbf{n}}_1\cdot \mathbf{r}=1$, ${\mathbf{n}}_2\cdot \mathbf{r}=2$, where ${\mathbf{n}}_1=(2,1,-1)$ and ${\mathbf{n}}_2=(1,-1,1)$. The Lagrange multiplier equation gives $\mathbf{r}=\lambda {\mathbf{n}}_1+\mu {\mathbf{n}}_2$. Since ${\mathbf{n}}_1\cdot {\mathbf{n}}_2=0$, $\Vert {\mathbf{n}}_1{\Vert }^2=6$, and $\Vert {\mathbf{n}}_2{\Vert }^2=3$, the constraints imply $\lambda =1/6$, $\mu =2/3$. HenceNo explicit parametrization of the line was needed. For any other feasible point, $\mathbf{h}=\mathbf{r}-{\mathbf{r}}_{\ast }$ is orthogonal to both normals and therefore to ${\mathbf{r}}_{\ast }$. The Pythagorean theorem gives $\Vert \mathbf{r}{\Vert }^2=\Vert {\mathbf{r}}_{\ast }{\Vert }^2+\Vert \mathbf{h}{\Vert }^2$, proving the unique global minimum. This is the geometric principle behind the minimum-norm least-squares solution.

Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.

Put $h=x-1$. The Taylor series of the natural logarithm gives the first four nonzero terms:The corresponding infinite series converges for $|h|<1$.

For the arctangent, put $h=x-1$. Its derivative is $(d/dh)\arctan (1+h)=1/(2+2h+h^2)=\frac{1}{2}-\frac{h}{2}+\frac{h^2}{4}+O(h^4)$. Integrating and using $\arctan 1=\pi /4$ givesThe constant counts as the first nonzero term. The quartic coefficient vanishes; the next nonzero term is $-h^5/40$.

The removable value at zero is one. Expand the denominator as $e^x-1=x(1+x/2+x^2/6+\cdots )$ and determine the reciprocal coefficients by multiplication. The first four nonzero terms areThis is the exponential generating function of the Bernoulli numbers, with convention $B_1=-1/2$. The coefficients of $x^3$ and $x^5$ are zero.

The homogeneous solution is $A\cos 2x+B\sin 2x$. A trial $C\cos x$ gives the particular solution $\cos x/3$. The two initial conditions set $A=-1/3$, $B=0$, giving

Use ${\cos }^{2}x=(1+\cos 2x)/2$. The constant forcing has particular solution $1/8$, while the $\cos 2x$ term satisfies the resonance condition and needs the trial $Cx\sin 2x$. Since $(d^2/d{x}^2+4)(x\sin 2x)=4\cos 2x$, the forced term is $x\sin 2x/8$. Adding the homogeneous solution and imposing the initial conditions gives