Fera Academy

Paper 1 Answers

Time3 hours

Marks105

SetSet 1

PaperLevel 1 - Math I (Physics) Paper 1

Information

Section A: Differentiation and Limits
Section B: Integration
Section C: Complex Numbers
Section D: Series
Section E: Taylor Series and Matrices
Section F: Linear Systems
Section G: Eigenvalues and Matrix Structure

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Section A: Differentiation and Limits

Answer 1. Rates, limits, and first principles

[15 marks]

This question concerns differentiation rules, limits, and the derivative definition.

a) Differentiate \(f(x)=(x^2+1)\sin(3x)\), and evaluate \(f'\left(\frac{\pi}{6}\right)\).

[4 marks]

Use the product rule with \(u=x^2+1\) and \(v=\sin(3x)\). Then \(u'=2x\). For the second factor, use the chain rule with inner function \(3x\), so \(v'=3\cos(3x)\). Hence \[ f'(x)=2x\sin(3x)+3(x^2+1)\cos(3x). \] At \(x=\frac{\pi}{6}\), we have \(3x=\frac{\pi}{2}\), so \(\sin(3x)=1\) and \(\cos(3x)=0\). Therefore \[ f'\left(\frac{\pi}{6}\right)=2\left(\frac{\pi}{6}\right)(1)+3\left(\frac{\pi^2}{36}+1\right)(0)=\frac{\pi}{3}. \]

Answer 1. Rates, limits, and first principles

[15 marks]

b) Evaluate \(\displaystyle \lim_{x\to1}\frac{x^2+5x-6}{x-1}\).

[3 marks]

Both numerator and denominator tend to \(0\), so factor the numerator: \[ x^2+5x-6=(x+6)(x-1). \] For \(x\ne1\), \[ \frac{x^2+5x-6}{x-1}=x+6. \] Therefore \[ \lim_{x\to1}\frac{x^2+5x-6}{x-1}=1+6=7. \]

Answer 1. Rates, limits, and first principles

[15 marks]

c) Evaluate \(\displaystyle \lim_{t\to\infty}\frac{\sqrt{9t^2+2t}}{4t-1}\).

[3 marks]

Divide the numerator and denominator by \(t\): \[ \frac{\sqrt{9t^2+2t}}{4t-1}=\frac{\sqrt{9+\frac{2}{t}}}{4-\frac{1}{t}}, \] because \(t\to\infty\) is positive. Taking the limit gives \[ \lim_{t\to\infty}\frac{\sqrt{9+\frac{2}{t}}}{4-\frac{1}{t}}=\frac{\sqrt{9}}{4}=\frac{3}{4}. \]

Answer 1. Rates, limits, and first principles

[15 marks]

d) Using first principles, find the derivative of \(g(x)=\frac{2x-1}{x+3}\), where \(x\ne-3\).

[5 marks]

Let \(g(x)=\frac{2x-1}{x+3}\). From first principles, \[ g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}. \] Now \[ g(x+h)=\frac{2x+2h-1}{x+h+3}. \] Thus \[ \frac{g(x+h)-g(x)}{h}=\frac{1}{h}\left(\frac{2x+2h-1}{x+h+3}-\frac{2x-1}{x+3}\right). \] Using a common denominator, \[ \frac{g(x+h)-g(x)}{h}=\frac{(2x+2h-1)(x+3)-(2x-1)(x+h+3)}{h(x+h+3)(x+3)}. \] Expand only the difference in the numerator: \[ (2x+2h-1)(x+3)-(2x-1)(x+h+3) \] \[ =(2x-1)(x+3)+2h(x+3)-(2x-1)(x+3)-h(2x-1). \] So the numerator is \[ h\{2(x+3)-(2x-1)\}=h(2x+6-2x+1)=7h. \] Therefore \[ \frac{g(x+h)-g(x)}{h}=\frac{7}{(x+h+3)(x+3)}. \] Taking \(h\to0\), \[ g'(x)=\frac{7}{(x+3)^2}. \]

Section B: Integration

Answer 2. Integrals and a complex sine identity

[15 marks]

This question uses trigonometric identities, rational integration, and complex exponential definitions of trigonometric functions.

a) Evaluate \(\displaystyle \int \sin^3 x\cos^2 x\,dx\).

[5 marks]

Use \(\sin^2 x=1-\cos^2 x\), and keep one \(\sin x\) for substitution: \[ \int \sin^3 x\cos^2 x\,dx=\int (1-\cos^2 x)\cos^2 x\sin x\,dx. \] Let \(u=\cos x\), so \(du=-\sin x\,dx\). Then \[ \int (1-\cos^2 x)\cos^2 x\sin x\,dx=-\int (1-u^2)u^2\,du. \] Therefore \[ -\int (u^2-u^4)\,du=-\left(\frac{u^3}{3}-\frac{u^5}{5}\right)+C. \] Substitute back \(u=\cos x\): \[ \int \sin^3 x\cos^2 x\,dx=\frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C. \]

Answer 2. Integrals and a complex sine identity

[15 marks]

b) Evaluate \(\displaystyle \int \frac{3x+1}{x^2+x-2}\,dx\).

[5 marks]

Factor the denominator: \[ x^2+x-2=(x-1)(x+2). \] Write \[ \frac{3x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}. \] Multiplying by \((x-1)(x+2)\) gives \[ 3x+1=A(x+2)+B(x-1). \] Equating coefficients, \[ A+B=3,\qquad 2A-B=1. \] Adding the equations gives \(3A=4\), so \(A=\frac{4}{3}\). Then \(B=3-\frac{4}{3}=\frac{5}{3}\). Hence \[ \int \frac{3x+1}{x^2+x-2}\,dx=\frac{4}{3}\int\frac{dx}{x-1}+\frac{5}{3}\int\frac{dx}{x+2}. \] Therefore \[ \int \frac{3x+1}{x^2+x-2}\,dx=\frac{4}{3}\ln|x-1|+\frac{5}{3}\ln|x+2|+C. \]

Answer 2. Integrals and a complex sine identity

[15 marks]

c) Using \(\sin z=\frac{e^{iz}-e^{-iz}}{2i}\), show that \(\sin(a+ib)=\sin a\cosh b+i\cos a\sinh b\). Hence find \(\sin(ib)\).

[5 marks]

Start from the exponential definition \[ \sin z=\frac{e^{iz}-e^{-iz}}{2i}. \] For \(z=a+ib\), \[ e^{iz}=e^{i(a+ib)}=e^{-b}e^{ia}=e^{-b}(\cos a+i\sin a), \] and \[ e^{-iz}=e^{-i(a+ib)}=e^b e^{-ia}=e^b(\cos a-i\sin a). \] Therefore \[ e^{iz}-e^{-iz}=e^{-b}(\cos a+i\sin a)-e^b(\cos a-i\sin a). \] Group real and imaginary terms: \[ e^{iz}-e^{-iz}=(e^{-b}-e^b)\cos a+i(e^{-b}+e^b)\sin a. \] Divide by \(2i\): \[ \sin(a+ib)=\frac{(e^{-b}+e^b)\sin a}{2}+\frac{e^{-b}-e^b}{2i}\cos a. \] Since \(\frac{1}{i}=-i\), this becomes \[ \sin(a+ib)=\sin a\left(\frac{e^b+e^{-b}}{2}\right)+i\cos a\left(\frac{e^b-e^{-b}}{2}\right). \] Thus \[ \sin(a+ib)=\sin a\cosh b+i\cos a\sinh b. \] At \(a=0\), \[ \sin(ib)=0\cdot\cosh b+i\cdot1\cdot\sinh b=i\sinh b. \]

Section C: Complex Numbers

Answer 3. Complex form, powers, and roots

[15 marks]

Use principal arguments in \((-\pi,\pi]\) unless a general argument is requested.

a) Find the modulus, principal argument, and rectangular form of \(z=e^{1-i\frac{2\pi}{3}}\).

[3 marks]

Write \[ z=e^{1-i\frac{2\pi}{3}}=e\,e^{-i\frac{2\pi}{3}}. \] The modulus is the positive scale factor \[ |z|=e. \] An argument is \(-\frac{2\pi}{3}\), which already lies in \((-\pi,\pi]\), so the principal argument is \[ \operatorname{Arg}(z)=-\frac{2\pi}{3}. \] In rectangular form, \[ z=e\left(\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right). \] Using \(\cos\left(-\frac{2\pi}{3}\right)=-\frac{1}{2}\) and \(\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\), \[ z=-\frac{e}{2}-i\frac{e\sqrt{3}}{2}. \]

Answer 3. Complex form, powers, and roots

[15 marks]

b) State De Moivre's theorem and use it to evaluate \(\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)^5\).

[3 marks]

De Moivre's theorem states that \[ (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta) \] for integer \(n\). Taking \(n=5\), \[ (\cos\theta+i\sin\theta)^5=\cos(5\theta)+i\sin(5\theta). \] For \(\theta=\frac{\pi}{10}\), \[ \left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)^5=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i. \]

Answer 3. Complex form, powers, and roots

[15 marks]

c) Use De Moivre's theorem to express \(\cos(4\theta)\) as a polynomial in \(\cos\theta\).

[5 marks]

By De Moivre's theorem, \[ (\cos\theta+i\sin\theta)^4=\cos(4\theta)+i\sin(4\theta). \] Expand the left-hand side using the binomial theorem: \[ (\cos\theta+i\sin\theta)^4=\cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta+4i^3\cos\theta\sin^3\theta+i^4\sin^4\theta. \] Since \(i^2=-1\), \(i^3=-i\), and \(i^4=1\), the real part is \[ \cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta. \] Equating real parts gives \[ \cos(4\theta)=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta. \] Using \(\sin^2\theta=1-\cos^2\theta\), \[ \cos(4\theta)=\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-\cos^2\theta)^2. \] Expand and collect powers of \(\cos\theta\): \[ \cos(4\theta)=\cos^4\theta-6\cos^2\theta+6\cos^4\theta+1-2\cos^2\theta+\cos^4\theta. \] Thus \[ \cos(4\theta)=8\cos^4\theta-8\cos^2\theta+1. \]

Answer 3. Complex form, powers, and roots

[15 marks]

d) Solve \(z^4+4z^2+16=0\), giving all roots in exponential form and rectangular form.

[4 marks]

Let \(y=z^2\). Then the quartic equation becomes \[ y^2+4y+16=0. \] Use the quadratic formula: \[ y=\frac{-4\pm\sqrt{4^2-4(1)(16)}}{2}=\frac{-4\pm\sqrt{-48}}{2}. \] Since \(\sqrt{-48}=4i\sqrt{3}\), \[ y=-2\pm2i\sqrt{3}. \] Thus the original equation splits into \[ z^2=-2+2i\sqrt{3} \] and \[ z^2=-2-2i\sqrt{3}. \] For \(-2+2i\sqrt{3}\), the modulus is \(4\) and an argument is \(\frac{2\pi}{3}\), so \[ -2+2i\sqrt{3}=4e^{i2\pi/3}. \] Its square roots have modulus \(2\) and arguments \[ \frac{1}{2}\left(\frac{2\pi}{3}+2\pi k\right)=\frac{\pi}{3}+k\pi,\qquad k=0,1. \] These give \(2e^{i\pi/3}\) and \(2e^{i4\pi/3}\). For \(-2-2i\sqrt{3}\), the modulus is \(4\) and an argument is \(-\frac{2\pi}{3}\), so its square roots are \(2e^{-i\pi/3}\) and \(2e^{i2\pi/3}\). Therefore the four roots are \[ z=2e^{i\pi/3},\quad 2e^{i4\pi/3},\quad 2e^{-i\pi/3},\quad 2e^{i2\pi/3}. \] In rectangular form, \[ z=1+i\sqrt{3},\quad -1-i\sqrt{3},\quad 1-i\sqrt{3},\quad -1+i\sqrt{3}. \]

Section D: Series

Answer 4. Convergence tests

[15 marks]

This question concerns convergence tests for numerical and power series.

a) Use the ratio test to determine whether \(\displaystyle \sum_{n=1}^{\infty}\frac{3^n}{n!}\) converges.

[4 marks]

Let \[ a_n=\frac{3^n}{n!}. \] Use the ratio test: \[ \frac{a_{n+1}}{a_n}=\frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n}=\frac{3}{n+1}. \] Hence \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{3}{n+1}=0. \] Since \(0<1\), the ratio test shows that \[ \sum_{n=1}^{\infty}\frac{3^n}{n!} \] converges absolutely, and therefore converges.

Answer 4. Convergence tests

[15 marks]

b) Use the root test to determine whether \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{2n+1}{5n-4}\right)^n\) converges.

[4 marks]

Let \[ b_n=\left(\frac{2n+1}{5n-4}\right)^n. \] Use the root test: \[ \sqrt[n]{|b_n|}=\frac{2n+1}{5n-4}. \] Therefore \[ \lim_{n\to\infty}\sqrt[n]{|b_n|}=\lim_{n\to\infty}\frac{2n+1}{5n-4}=\frac{2}{5}. \] Since \(\frac{2}{5}<1\), the root test shows that the series \[ \sum_{n=1}^{\infty}\left(\frac{2n+1}{5n-4}\right)^n \] converges.

Answer 4. Convergence tests

[15 marks]

c) Use the alternating series test, and then test absolute convergence, for \(\displaystyle \sum_{n=2}^{\infty}(-1)^n\frac{1}{\sqrt n}\).

[4 marks]

The series is \[ \sum_{n=2}^{\infty}(-1)^n\frac{1}{\sqrt n}. \] The magnitudes are \[ c_n=\frac{1}{\sqrt n}. \] They satisfy \(c_n>0\), they decrease as \(n\) increases, and \[ \lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{1}{\sqrt n}=0. \] Therefore the alternating series test shows that the given series converges. To test absolute convergence, consider \[ \sum_{n=2}^{\infty}\left|(-1)^n\frac{1}{\sqrt n}\right|=\sum_{n=2}^{\infty}\frac{1}{\sqrt n}. \] This is a \(p\)-series with \(p=\frac{1}{2}\le1\), so it diverges. Hence the original series is conditionally convergent.

Answer 4. Convergence tests

[15 marks]

d) Find the interval of convergence of \(\displaystyle \sum_{n=1}^{\infty}\frac{(x-2)^n}{n4^n}\).

[3 marks]

Let \[ a_n(x)=\frac{(x-2)^n}{n4^n}. \] For \(x\ne2\), use the ratio test on absolute values: \[ \left|\frac{a_{n+1}(x)}{a_n(x)}\right|=\left|\frac{(x-2)^{n+1}}{(n+1)4^{n+1}}\cdot\frac{n4^n}{(x-2)^n}\right|=\frac{n}{n+1}\cdot\frac{|x-2|}{4}. \] Taking the limit gives \[ L=\frac{|x-2|}{4}. \] The series converges absolutely when \(L<1\), so \[ |x-2|<4,\qquad -2<x<6. \] Test the endpoints in the original series. At \(x=6\), \[ \sum_{n=1}^{\infty}\frac{4^n}{n4^n}=\sum_{n=1}^{\infty}\frac{1}{n}, \] which diverges. At \(x=-2\), \[ \sum_{n=1}^{\infty}\frac{(-4)^n}{n4^n}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}, \] which converges by the alternating series test. Therefore the interval of convergence is \[ [-2,6). \]

Section E: Taylor Series and Matrices

Answer 5. Approximation and matrix conditions

[15 marks]

This question combines Taylor approximation with elementary vector and determinant reasoning.

a) Find the cubic Taylor polynomial for \(f(x)=\ln(1+x)\) about \(0\), and use it to approximate \(\ln(1.2)\).

[5 marks]

For \(f(x)=\ln(1+x)\), compute derivatives: \[ f'(x)=\frac{1}{1+x},\qquad f''(x)=-\frac{1}{(1+x)^2},\qquad f'''(x)=\frac{2}{(1+x)^3}. \] At \(x=0\), \[ f(0)=0, \quad f'(0)=1, \quad f''(0)=-1, \quad f'''(0)=2. \] The cubic Taylor polynomial about \(0\) is \[ P_3(x)=0+1x+\frac{-1}{2!}x^2+\frac{2}{3!}x^3. \] Thus \[ P_3(x)=x-\frac{x^2}{2}+\frac{x^3}{3}. \] At \(x=0.2\), \[ P_3(0.2)=0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}=0.2-0.02+\frac{0.008}{3}. \] Therefore \[ P_3(0.2)=0.182666\ldots. \]

Answer 5. Approximation and matrix conditions

[15 marks]

b) Use the Lagrange remainder to bound the error in the approximation from part (a).

[4 marks]

The Lagrange remainder after the cubic term is \[ R_3(x)=\frac{f^{(4)}(\xi)}{4!}x^4 \] for some \(\xi\) between \(0\) and \(x\). Since \[ f^{(4)}(x)=-\frac{6}{(1+x)^4}, \] for \(0\le x\le0.2\), \[ |f^{(4)}(x)|=\frac{6}{(1+x)^4}\le6. \] So we may take \(M=6\). Therefore \[ |R_3(0.2)|\le\frac{6}{4!}(0.2)^4=\frac{6}{24}(0.0016). \] Hence \[ |R_3(0.2)|\le0.0004. \] This means the approximation from part (a) differs from \(\ln(1.2)\) by at most \(0.0004\).

Answer 5. Approximation and matrix conditions

[15 marks]

c) Find the value of \(k\) for which \((1,-2,3)\) and \((2,k,6)\) are linearly dependent.

[3 marks]

The vectors are linearly dependent exactly when one is a scalar multiple of the other. Suppose \[ (2,k,6)=\lambda(1,-2,3). \] From the first component, \(2=\lambda\), so \(\lambda=2\). The second component then requires \[ k=2(-2)=-4. \] The third component gives \[ 6=2(3)=6, \] which is consistent. Therefore the vectors are linearly dependent exactly when \[ k=-4. \]

Answer 5. Approximation and matrix conditions

[15 marks]

d) For \(A=\begin{pmatrix}1&a&0\\2&1&1\\0&3&a\end{pmatrix}\), find the real values of \(a\) for which \(A\) is singular.

[3 marks]

Compute the determinant by expanding along the first row: \[ \det A=\det\begin{pmatrix}1&a&0\\2&1&1\\0&3&a\end{pmatrix}. \] Thus \[ \det A=1\det\begin{pmatrix}1&1\\3&a\end{pmatrix}-a\det\begin{pmatrix}2&1\\0&a\end{pmatrix}+0. \] The two minors are \[ \det\begin{pmatrix}1&1\\3&a\end{pmatrix}=a-3,\qquad \det\begin{pmatrix}2&1\\0&a\end{pmatrix}=2a. \] Therefore \[ \det A=(a-3)-a(2a)=a-3-2a^2. \] The matrix is singular when \(\det A=0\): \[ a-3-2a^2=0. \] Equivalently, \[ 2a^2-a+3=0. \] The discriminant is \[ (-1)^2-4(2)(3)=1-24=-23. \] There are no real roots, so for every real \(a\), \(\det A\ne0\). Therefore the matrix is nonsingular for all \(a\in\mathbb R\).

Section F: Linear Systems

Answer 6. A parameter-dependent system

[15 marks]

Consider the parameter-dependent linear system \[ \begin{aligned} x+y+z&=2,\\ 2x+3y+5z&=7,\\ x+2y+az&=b. \end{aligned} \]

a) Find the values of \(a\) for which the system has a unique solution.

[4 marks]

The coefficient matrix is \[ A=\begin{pmatrix}1&1&1\\2&3&5\\1&2&a\end{pmatrix}. \] Compute its determinant: \[ \det A=1\det\begin{pmatrix}3&5\\2&a\end{pmatrix}-1\det\begin{pmatrix}2&5\\1&a\end{pmatrix}+1\det\begin{pmatrix}2&3\\1&2\end{pmatrix}. \] The minors are \[ 3a-10, \quad 2a-5, \quad 4-3=1. \] Therefore \[ \det A=(3a-10)-(2a-5)+1=a-4. \] The system has a unique solution exactly when the coefficient matrix is invertible, so \[ a\ne4. \]

Answer 6. A parameter-dependent system

[15 marks]

b) Classify the non-unique cases as no solution or infinitely many solutions.

[5 marks]

Use the augmented matrix \[ \left(\begin{array}{ccc|c}1&1&1&2\\2&3&5&7\\1&2&a&b\end{array}\right). \] Apply \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-R_1\): \[ \left(\begin{array}{ccc|c}1&1&1&2\\0&1&3&3\\0&1&a-1&b-2\end{array}\right). \] Then apply \(R_3\leftarrow R_3-R_2\): \[ \left(\begin{array}{ccc|c}1&1&1&2\\0&1&3&3\\0&0&a-4&b-5\end{array}\right). \] When \(a=4\), the final row becomes \[ \left(\begin{array}{ccc|c}0&0&0&b-5\end{array}\right). \] If \(b\ne5\), this row means \(0=b-5\), a contradiction. Hence there is no solution when \[ a=4, \quad b\ne5. \] If \(b=5\), the final row is \(0=0\), so one variable is free and there are infinitely many solutions. Hence infinitely many solutions occur when \[ a=4, \quad b=5. \]

Answer 6. A parameter-dependent system

[15 marks]

c) For the unique case \(a=5\), find the inverse of the coefficient matrix.

[4 marks]

Take \(a=5\). Then \[ A=\begin{pmatrix}1&1&1\\2&3&5\\1&2&5\end{pmatrix}. \] Use Gauss-Jordan elimination on \((A|I)\): \[ \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\2&3&5&0&1&0\\1&2&5&0&0&1\end{array}\right). \] Apply \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-R_1\): \[ \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&3&-2&1&0\\0&1&4&-1&0&1\end{array}\right). \] Apply \(R_3\leftarrow R_3-R_2\): \[ \left(\begin{array}{ccc|ccc}1&1&1&1&0&0\\0&1&3&-2&1&0\\0&0&1&1&-1&1\end{array}\right). \] Eliminate above the pivot in column 3: \(R_2\leftarrow R_2-3R_3\), \(R_1\leftarrow R_1-R_3\): \[ \left(\begin{array}{ccc|ccc}1&1&0&0&1&-1\\0&1&0&-5&4&-3\\0&0&1&1&-1&1\end{array}\right). \] Now eliminate above the pivot in column 2: \(R_1\leftarrow R_1-R_2\): \[ \left(\begin{array}{ccc|ccc}1&0&0&5&-3&2\\0&1&0&-5&4&-3\\0&0&1&1&-1&1\end{array}\right). \] Therefore \[ A^{-1}=\begin{pmatrix}5&-3&2\\-5&4&-3\\1&-1&1\end{pmatrix}. \]

Answer 6. A parameter-dependent system

[15 marks]

d) Use your inverse to solve the system when \(a=5\), leaving the answer in terms of \(b\).

[2 marks]

For \(a=5\), the system is \(A\mathbf{x}=\mathbf{b}\), where \[ \mathbf{b}=\begin{pmatrix}2\\7\\b\end{pmatrix}. \] Using the inverse from part (c), \[ \mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}5&-3&2\\-5&4&-3\\1&-1&1\end{pmatrix}\begin{pmatrix}2\\7\\b\end{pmatrix}. \] Multiply row by column: \[ x=5(2)-3(7)+2b=2b-11, \] \[ y=-5(2)+4(7)-3b=18-3b, \] \[ z=1(2)-1(7)+b=b-5. \] Thus, when \(a=5\), \[ (x,y,z)=(2b-11,18-3b,b-5). \]

Section G: Eigenvalues and Matrix Structure

Answer 7. Diagonalisation and matrix statements

[15 marks]

Let \[ A=\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}. \] You are given eigenvectors \[ \mathbf v_1=\begin{pmatrix}1\\1\\1\end{pmatrix},\qquad \mathbf v_2=\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \mathbf v_3=\begin{pmatrix}1\\0\\-1\end{pmatrix}. \]

a) Verify the eigenvalue belonging to each supplied eigenvector.

[4 marks]

Compute \(A\mathbf v_1\): \[ A\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}2+1+1\\1+2+1\\1+1+2\end{pmatrix}=\begin{pmatrix}4\\4\\4\end{pmatrix}=4\begin{pmatrix}1\\1\\1\end{pmatrix}. \] So \(\mathbf v_1\) has eigenvalue \(4\). Next, \[ A\begin{pmatrix}1\\-1\\0\end{pmatrix}=\begin{pmatrix}2-1+0\\1-2+0\\1-1+0\end{pmatrix}=\begin{pmatrix}1\\-1\\0\end{pmatrix}. \] So \(\mathbf v_2\) has eigenvalue \(1\). Finally, \[ A\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}2+0-1\\1+0-1\\1+0-2\end{pmatrix}=\begin{pmatrix}1\\0\\-1\end{pmatrix}. \] So \(\mathbf v_3\) also has eigenvalue \(1\).

Answer 7. Diagonalisation and matrix statements

[15 marks]

b) Construct matrices \(P\) and \(D\) such that \(A=PDP^{-1}\), and justify that this is a valid diagonalisation.

[4 marks]

Use the supplied eigenvectors as columns of \(P\), in the same order as the eigenvalues: \[ P=\begin{pmatrix}1&1&1\\1&-1&0\\1&0&-1\end{pmatrix}. \] The matching diagonal matrix is \[ D=\begin{pmatrix}4&0&0\\0&1&0\\0&0&1\end{pmatrix}. \] Because the columns of \(P\) are eigenvectors, \[ AP=PD. \] If \(P\) is invertible, this gives \[ A=PDP^{-1}. \] Check that \(P\) is invertible by computing its determinant: \[ \det P=1\det\begin{pmatrix}-1&0\\0&-1\end{pmatrix}-1\det\begin{pmatrix}1&0\\1&-1\end{pmatrix}+1\det\begin{pmatrix}1&-1\\1&0\end{pmatrix}. \] So \[ \det P=1-(-1)+1=3\ne0. \] Therefore \(P\) is invertible and the diagonalisation is valid.

Answer 7. Diagonalisation and matrix statements

[15 marks]

c) Use the diagonalisation to write a formula for \(A^n\), where \(n\) is a positive integer.

[3 marks]

Using \(A=PDP^{-1}\), powers satisfy \[ A^n=PD^nP^{-1} \] for positive integers \(n\). Since \[ D=\begin{pmatrix}4&0&0\\0&1&0\\0&0&1\end{pmatrix}, \] we have \[ D^n=\begin{pmatrix}4^n&0&0\\0&1^n&0\\0&0&1^n\end{pmatrix}=\begin{pmatrix}4^n&0&0\\0&1&0\\0&0&1\end{pmatrix}. \] Thus \[ A^n=P\begin{pmatrix}4^n&0&0\\0&1&0\\0&0&1\end{pmatrix}P^{-1}. \] This form is useful because it replaces repeated matrix multiplication by powers of diagonal entries.

Answer 7. Diagonalisation and matrix statements

[15 marks]

d) Decide whether each statement is true or false, giving a proof or counterexample. (i) Every matrix with a repeated eigenvalue is diagonalizable. (ii) Every \(n imes n\) matrix with \(n\) linearly independent eigenvectors is diagonalizable.

[4 marks]

Statement (i) is false. A counterexample is \[ B=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \] Its only eigenvalue is \(1\), repeated. Solve \((B-I)\mathbf v=\mathbf0\): \[ B-I=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \] The equation gives \(v_2=0\), so the eigenspace consists of vectors \((v_1,0)^T\). This is one-dimensional, so there are not two independent eigenvectors. Hence \(B\) is not diagonalizable. Statement (ii) is true. If a square matrix \(C\) has \(n\) linearly independent eigenvectors \(\mathbf u_1,\ldots,\mathbf u_n\), put these eigenvectors as columns of \[ P=\begin{pmatrix}\mathbf u_1&\cdots&\mathbf u_n\end{pmatrix}. \] Linear independence means \(P\) is invertible. If the corresponding eigenvalues are \(\lambda_1,\ldots,\lambda_n\), then \[ CP=P\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \] Multiplying on the right by \(P^{-1}\) gives \[ C=P\operatorname{diag}(\lambda_1,\ldots,\lambda_n)P^{-1}. \] Therefore \(C\) is diagonalizable.