# Question 3

Let $M=\begin{pmatrix}3 & 5 & 2\\5 & 3 & -2\\2 & -2 & -4\end{pmatrix}$

## (i) Show that the characteristic equation for M is $\lambda^3 - 2\lambda^2 - 48\lambda = 0$. [4]

M1: Attempt to find $\det(M - \lambda I)$

A1: Correct expansion of determinant

A1: Simplification to correct cubic

A1: Correct factorisation (or equivalent)

## (ii) Find the other two eigenvalues of M, and corresponding eigenvectors. [8]

Given that $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ is an eigenvector of M corresponding to the eigenvalue 0.

M1: Factorise characteristic equation to find $\lambda(\lambda^2 - 2\lambda - 48) = 0$

M1: Solve quadratic to find $\lambda = 8$ and $\lambda = -6$

A1: Both eigenvalues correct

M1: For each eigenvalue, set up $(M - \lambda I)\mathbf{v} = \mathbf{0}$

M1: Reduce to row echelon form (or equivalent)

A1: Eigenvector for $\lambda = 8$ (e.g., $\begin{pmatrix}1\\1\\0\end{pmatrix}$ or scalar multiple)

A1: Eigenvector for $\lambda = -6$ (e.g., $\begin{pmatrix}1\\-1\\-2\end{pmatrix}$ or scalar multiple)

A1: All three eigenvectors correct or appropriately identified

## (iii) Write down a matrix P, and a diagonal matrix D, such that $P^{-1}MP = D$. [3]

B1: $P = \begin{pmatrix}1 & 1 & 1\\-1 & 1 & -1\\1 & 0 & -2\end{pmatrix}$ (or equivalent with eigenvectors in different order)

B1: $D = \begin{pmatrix}0 & 0 & 0\\0 & 8 & 0\\0 & 0 & -6\end{pmatrix}$ (or equivalent corresponding to order of P)

B1: Correct pairing of P and D

## (iv) Use the Cayley-Hamilton theorem to find integers a and b such that $M^4 = aM^2 + bM$. [3]

M1: Apply Cayley-Hamilton theorem: $M^3 - 2M^2 - 48M = 0$, so $M^3 = 2M^2 + 48M$

M1: Multiply both sides by M to find $M^4 = 2M^3 + 48M^2 = 2(2M^2 + 48M) + 48M^2 = 52M^2 + 96M$

A1: $a = 52$ and $b = 96$

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# Section B: Question 4 — Option 1: Hyperbolic functions

## (a) Find $\int_0^1 \frac{1}{9x^2 + 16} dx$, giving your answer in an exact logarithmic form. [5]

M1: Rewrite integrand as $\frac{1}{9(x^2 + \frac{16}{9})}$ or $\frac{1}{9}\cdot\frac{1}{x^2 + (\frac{4}{3})^2}$

M1: Recognise standard form and use $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\arctan\frac{x}{a} + C$

A1: Correct antiderivative: $\frac{1}{12}\arctan\frac{3x}{4}$ (or equivalent)

M1: Apply limits correctly

A1: Answer in exact form, e.g., $\frac{1}{12}\arctan\frac{3}{4}$ or equivalent

## (b)(i) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that $\sinh 2x = 2\sinh x \cosh x$. [2]

M1: Use $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$

A1: Correct algebraic manipulation to obtain $\sinh 2x = 2\sinh x \cosh x$

## (b)(ii) Show that one of the stationary points on the curve $y = 20\cosh x - 3\cosh 2x$ has coordinates $(\ln 3, 59)$, and find the coordinates of the other two stationary points. [7]

M1: Differentiate to find $\frac{dy}{dx} = 20\sinh x - 6\sinh 2x$

M1: Set derivative equal to zero and use $\sinh 2x = 2\sinh x \cosh x$ to get $20\sinh x - 12\sinh x \cosh x = 0$

M1: Factorise as $\sinh x(20 - 12\cosh x) = 0$

M1: From $\sinh x = 0$, get $x = 0$; from $\cosh x = \frac{5}{3}$, use $\cosh x = \frac{e^x + e^{-x}}{2}$ to find $x = \pm\ln 3$

A1: Verify $(\ln 3, 59)$ is a stationary point

A1: Find $y$-coordinate at $x = 0$: $y = 20(1) - 3(1) = 17$, giving $(0, 17)$

A1: Find $y$-coordinate at $x = -\ln 3$: $y = 20\cosh(-\ln 3) - 3\cosh(-2\ln 3) = 59$, giving $(-\ln 3, 59)$

## (b)(iii) Show that $\int_{-\ln 3}^{\ln 3} (20\cosh x - 3\cosh 2x) dx = 40$. [4]

M1: Find antiderivative: $20\sinh x - \frac{3}{2}\sinh 2x$

M1: Recognise that $20\cosh x - 3\cosh 2x$ is an even function, or use symmetry argument

M1: Evaluate at $x = \ln 3$: $20\sinh(\ln 3) - \frac{3}{2}\sinh(2\ln 3) = 20 \cdot \frac{8}{3} - \frac{3}{2} \cdot \frac{80}{9} = \frac{160}{3} - \frac{40}{3} = 40$

A1: Correct final answer: 40