2025 Cambridge IAS-Level Mathematics - Further (9231) Practice Paper with Answers

(i) Combining the terms on the right-hand side over a common denominator gives: \( \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} = \frac{(2r+1)^2 - (2r-1)^2}{(2r-1)^2(2r+1)^2} = \frac{(4r^2+4r+1) - (4r^2-4r+1)}{(2r-1)^2(2r+1)^2} = \frac{8r}{(2r-1)^2(2r+1)^2} \). Hence, LHS = RHS. (ii) From part (i), we can write the general term as \( u_r = \frac{1}{8} \left( \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} \right) \). Summing from \( r=1 \) to \( n \): \( \sum_{r=1}^n u_r = \frac{1}{8} \left[ \left(1 - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{25}\right) + \dots + \left(\frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2}\right) \right] \). The intermediate terms cancel out, leaving: \( \sum_{r=1}^n u_r = \frac{1}{8} \left( 1 - \frac{1}{(2n+1)^2} \right) \). (iii) As \( n \to \infty \), the sum to infinity is \( \sum_{r=1}^{\infty} u_r = \frac{1}{8} \). The required sum from \( r=n \) to infinity is \( \sum_{r=n}^{\infty} u_r = \sum_{r=1}^{\infty} u_r - \sum_{r=1}^{n-1} u_r = \frac{1}{8} - \frac{1}{8} \left( 1 - \frac{1}{(2(n-1)+1)^2} \right) = \frac{1}{8(2n-1)^2} \).

M1: For attempting to combine the fractions on the RHS of (i) over a common denominator. A1: For correct algebraic simplification to show the LHS. M1: For writing \( u_r \) as a difference of two terms and attempting to list terms for cancellation. A1: For obtaining the correct cancelled form \( \frac{1}{8} \left( 1 - \frac{1}{(2n+1)^2} \right) \). M1: For using \( \sum_{r=n}^{\infty} u_r = \sum_{r=1}^{\infty} u_r - \sum_{r=1}^{n-1} u_r \) or direct limits on the difference method from \( r=n \). A1: For the correct final expression \( \frac{1}{8(2n-1)^2} \).

Let \( H(n) \) be the statement that \( 4^n + 15n - 1 \) is divisible by 9. Base Case: For \( n = 1 \), \( 4^1 + 15(1) - 1 = 18 \), which is divisible by 9 since \( 18 = 9 \times 2 \). Thus, \( H(1) \) is true. Inductive Step: Assume that \( H(k) \) is true for some positive integer \( k \). That is, \( 4^k + 15k - 1 = 9m \) for some integer \( m \). We want to show that \( H(k+1) \) is true, i.e., \( 4^{k+1} + 15(k+1) - 1 \) is divisible by 9. Consider \( 4^{k+1} + 15(k+1) - 1 = 4 \cdot 4^k + 15k + 14 \). Substituting the induction hypothesis \( 4^k = 9m - 15k + 1 \): \( 4(9m - 15k + 1) + 15k + 14 = 36m - 60k + 4 + 15k + 14 = 36m - 45k + 18 = 9(4m - 5k + 2) \). Since \( m \) and \( k \) are integers, \( 4m - 5k + 2 \) is also an integer, which shows the expression is divisible by 9. Conclusion: Since \( H(1) \) is true, and \( H(k) \implies H(k+1) \), by the principle of mathematical induction, \( H(n) \) is true for all positive integers \( n \).

B1: Verifies the base case \( n=1 \) correctly showing it equals 18, which is divisible by 9. M1: States the inductive hypothesis clearly for some integer \( k \). M1: Attempts to analyze \( H(k+1) \) by writing \( 4^{k+1} \) as \( 4 \cdot 4^k \) or by taking the difference \( f(k+1) - f(k) \). A1: Correct algebraic substitution or expansion to get an expression in terms of \( 9m \). A1: Factorises out 9 to obtain \( 9(4m - 5k + 2) \) or equivalent. A1: Gives a complete and logical conclusion mentioning the induction steps.

(i) From the given equation, we have: \( \sum \alpha = 3 \), \( \sum \alpha\beta = 5 \), and \( \alpha\beta\gamma = 2 \). Then, \( \alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 3^2 - 2(5) = 9 - 10 = -1 \). Also, \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha) = 5^2 - 2(2)(3) = 25 - 12 = 13 \). (ii) The product of the new roots is \( \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 2^2 = 4 \). A cubic equation with roots \( y = \alpha^2, \beta^2, \gamma^2 \) is given by: \( y^3 - (\sum \alpha^2)y^2 + (\sum \alpha^2\beta^2)y - \alpha^2\beta^2\gamma^2 = 0 \). Substituting the values: \( y^3 - (-1)y^2 + 13y - 4 = 0 \implies y^3 + y^2 + 13y - 4 = 0 \). (iii) Since \( \alpha, \beta, \gamma \) are roots of \( x^3 - 3x^2 + 5x - 2 = 0 \), we can sum this relation: \( \sum \alpha^3 - 3\sum \alpha^2 + 5\sum \alpha - 6 = 0 \). Substituting the known values: \( \sum \alpha^3 - 3(-1) + 5(3) - 6 = 0 \implies \sum \alpha^3 + 3 + 15 - 6 = 0 \implies \sum \alpha^3 + 12 = 0 \implies \sum \alpha^3 = -12 \).

B1: Identifies \( \sum \alpha = 3, \sum \alpha\beta = 5, \alpha\beta\gamma = 2 \). M1: Uses the identity for \( \sum \alpha^2 \) and obtains \( -1 \). M1: Uses the identity for \( \sum \alpha^2\beta^2 \) and obtains \( 13 \). A1: Correct values for both sums in part (i). M1: Forms the cubic equation using their sum values and the product of roots \( 4 \). A1: Obtains \( y^3 + y^2 + 13y - 4 = 0 \). M1: Uses the cubic relation to find \( \sum \alpha^3 \) or uses a valid identity. A1: Obtains \( -12 \) correctly.

(a) The Cartesian coordinate \( x \) is given by:
\[ x = r \cos \theta = a(1 - \sin \theta)\cos \theta = a(\cos \theta - \sin \theta \cos \theta) = a\left(\cos \theta - \frac{1}{2}\sin 2\theta\right) \]
To find where the tangent is perpendicular to the initial line, we set \( \frac{dx}{d\theta} = 0 \):
\[ \frac{dx}{d\theta} = a(-\sin \theta - \cos 2\theta) = 0 \]
Using the double-angle identity \( \cos 2\theta = 1 - 2\sin^2 \theta \), we obtain:
\[ -\sin \theta - (1 - 2\sin^2 \theta) = 0 \implies 2\sin^2 \theta - \sin \theta - 1 = 0 \]
Factoring this quadratic equation:
\[ (2\sin \theta + 1)(\sin \theta - 1) = 0 \]
This gives \( \sin \theta = -\frac{1}{2} \) or \( \sin \theta = 1 \).
For \( 0 \le \theta < 2\pi \):
- If \( \sin \theta = 1 \), then \( \theta = \frac{\pi}{2} \). Here, \( r = a(1 - 1) = 0 \). This point lies at the pole.
- If \( \sin \theta = -\frac{1}{2} \), then \( \theta = \frac{7\pi}{6} \) or \( \theta = \frac{11\pi}{6} \).
For both values, the corresponding \( r \) is:
\[ r = a\left(1 - \left(-\frac{1}{2}\right)\right) = \frac{3}{2}a \]
Thus, the polar coordinates of the points are:
\[ \left(0, \frac{\pi}{2}\right), \quad \left(\frac{3}{2}a, \frac{7\pi}{6}\right), \quad \left(\frac{3}{2}a, \frac{11\pi}{6}\right) \]

(b) The area \( A \) of the region enclosed by \( C \) is given by:
\[ A = \frac{1}{2} \int_{0}^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_{0}^{2\pi} (1 - \sin \theta)^2 \, d\theta \]
Expanding the integrand:
\[ A = \frac{1}{2} a^2 \int_{0}^{2\pi} (1 - 2\sin \theta + \sin^2 \theta) \, d\theta \]
Using \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):
\[ A = \frac{1}{2} a^2 \int_{0}^{2\pi} \left(\frac{3}{2} - 2\sin \theta - \frac{1}{2}\cos 2\theta\right) \, d\theta \]
Integrating term-by-term:
\[ A = \frac{1}{2} a^2 \left[ \frac{3}{2}\theta + 2\cos \theta - \frac{1}{4}\sin 2\theta \right]_{0}^{2\pi} \]
Evaluating the limits:
\[ A = \frac{1}{2} a^2 \left( \left(3\pi + 2 - 0\right) - \left(0 + 2 - 0\right) \right) = \frac{3}{2}\pi a^2 \]

(a)
M1: Expresses \( x \) in terms of \( \theta \) using \( x = r\cos\theta \).
A1: Correctly differentiates \( x \) with respect to \( \theta \).
M1: Uses a trigonometric identity to get a quadratic in \( \sin \theta \) and sets \( \frac{dx}{d\theta} = 0 \).
A1: Solves to find \( \sin \theta = -\frac{1}{2} \) and \( \sin \theta = 1 \).
A1: Finds correct values of \( \theta \) as \( \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \).
A1: Correctly states the polar coordinates of the three points.

(b)
M1: Uses the polar area formula \( \frac{1}{2}\int r^2 \, d\theta \) with correct limits.
A1: Correctly expands the integrand.
M1: Uses the double-angle identity for \( \sin^2 \theta \).
A1: Integrates correctly to obtain \( \frac{3}{2}\theta + 2\cos\theta - \frac{1}{4}\sin 2\theta \).
M1: Substitutes the limits \( 0 \) and \( 2\pi \) correctly.
A1.5: Obtains the correct final area \( \frac{3}{2}\pi a^2 \).

(a) The matrix \( \mathbf{M} \) is singular when its determinant is zero:
\[ \det(\mathbf{M}) = 2(3(2) - (-1)(1)) - (-1)((1)(2) - (-1)(a)) + a((1)(1) - 3(a)) \]
\[ = 2(7) + (2 + a) + a(1 - 3a) = 16 + 2a - 3a^2 \]
Setting this to zero:
\[ 3a^2 - 2a - 16 = 0 \implies (3a - 8)(a + 2) = 0 \]
So \( a = \frac{8}{3} \) or \( a = -2 \).

(b) For \( a = 1 \):
\[ \mathbf{M} = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 3 & -1 \\ 1 & 1 & 2 \end{pmatrix} \]
Its determinant is \( \det(\mathbf{M}) = 16 + 2(1) - 3(1)^2 = 15 \).
Next, we find the matrix of cofactors, \( \mathbf{C} \):
- \( C_{11} = +(6 - (-1)) = 7 \)
- \( C_{12} = -(2 - (-1)) = -3 \)
- \( C_{13} = +(1 - 3) = -2 \)
- \( C_{21} = -(-2 - 1) = 3 \)
- \( C_{22} = +(4 - 1) = 3 \)
- \( C_{23} = -(2 - (-1)) = -3 \)
- \( C_{31} = +(1 - 3) = -2 \)
- \( C_{32} = -(-2 - 1) = 3 \)
- \( C_{33} = +(6 - (-1)) = 7 \)
So, the cofactor matrix is:
\[ \mathbf{C} = \begin{pmatrix} 7 & -3 & -2 \\ 3 & 3 & -3 \\ -2 & 3 & 7 \end{pmatrix} \]
Taking the transpose of \( \mathbf{C} \) to get the adjugate matrix:
\[ \mathbf{Adj}(\mathbf{M}) = \mathbf{C}^T = \begin{pmatrix} 7 & 3 & -2 \\ -3 & 3 & 3 \\ -2 & -3 & 7 \end{pmatrix} \]
Thus, the inverse is:
\[ \mathbf{M}^{-1} = \frac{1}{15} \begin{pmatrix} 7 & 3 & -2 \\ -3 & 3 & 3 \\ -2 & -3 & 7 \end{pmatrix} \]

(c) The system of equations can be written as \( \mathbf{M} \mathbf{X} = \mathbf{B} \), where \( \mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( \mathbf{B} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} \).
Using the inverse from part (b):
\[ \mathbf{X} = \mathbf{M}^{-1} \mathbf{B} = \frac{1}{15} \begin{pmatrix} 7 & 3 & -2 \\ -3 & 3 & 3 \\ -2 & -3 & 7 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} \]
\[ = \frac{1}{15} \begin{pmatrix} 7(2) + 3(5) - 2(-1) \\ -3(2) + 3(5) + 3(-1) \\ -2(2) - 3(5) + 7(-1) \end{pmatrix} = \frac{1}{15} \begin{pmatrix} 31 \\ 6 \\ -26 \end{pmatrix} \]
Thus, \( x = \frac{31}{15} \), \( y = \frac{6}{15} = \frac{2}{5} \), and \( z = -\frac{26}{15} \).

(a)
M1: Attempts to find the determinant of \( \mathbf{M} \) in terms of \( a \).
A1: Obtains the correct determinant: \( 16 + 2a - 3a^2 \).
M1: Sets the determinant to 0 and solves the resulting quadratic equation.
A1: Obtains \( a = \frac{8}{3} \) and \( a = -2 \).

(b)
M1: Calculates the determinant for \( a = 1 \) to be 15.
M1: Calculates at least 4 cofactors correctly.
A1: Correctly calculates all cofactors.
M1: Transposes the cofactor matrix and divides by \( \det(\mathbf{M}) \).
A0.5: Obtains the correct inverse matrix.

(c)
M1: Writes the system as \( \mathbf{X} = \mathbf{M}^{-1} \mathbf{B} \).
M1: Attempts matrix multiplication of their inverse by the column matrix \( \mathbf{B} \).
A2: Correctly obtains \( x = \frac{31}{15} \), \( y = \frac{2}{5} \), and \( z = -\frac{26}{15} \) (Award 1 mark for any two correct components).

(a) First, check if the direction vectors are parallel. The directions are \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \). Since \( \mathbf{d}_1 \) is not a scalar multiple of \( \mathbf{d}_2 \), the lines are not parallel.
Next, test for intersection by equating components:
1) \( 1 + 2\lambda = 2 + \mu \implies 2\lambda - \mu = 1 \)
2) \( -2 + \lambda = -\mu \implies \lambda + \mu = 2 \)
3) \( 3 - \lambda = 1 + 2\mu \)
Adding (1) and (2) gives \( 3\lambda = 3 \implies \lambda = 1 \), which yields \( \mu = 1 \).
Substituting these into (3):
LHS: \( 3 - 1 = 2 \)
RHS: \( 1 + 2(1) = 3 \)
Since \( 2 \neq 3 \), the lines do not intersect. Since they are not parallel and do not intersect, they are skew.

(b) The direction of the common perpendicular is:
\[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1(2) - (-1)(-1) \\ -((-1)(1) - (2)(2)) \\ 2(-1) - 1(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} \]
Let \( A = (1, -2, 3) \) be on \( L_1 \) and \( B = (2, 0, 1) \) be on \( L_2 \). Then:
\[ \mathbf{AB} = \begin{pmatrix} 2 - 1 \\ 0 - (-2) \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \]
The shortest distance is the projection of \( \mathbf{AB} \) onto \( \mathbf{n} \):
\[ d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| 1(1) + 2(-5) - 2(-3) \right|}{\sqrt{1^2 + (-5)^2 + (-3)^2}} = \frac{|1 - 10 + 6|}{\sqrt{35}} = \frac{3}{\sqrt{35}} \]

(c) Since \( \Pi \) contains \( L_1 \) and is parallel to \( L_2 \), its normal vector is perpendicular to both lines, which means we can use \( \mathbf{n} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} \).
Using the point \( (1, -2, 3) \) on \( L_1 \):
\[ 1(x - 1) - 5(y + 2) - 3(z - 3) = 0 \implies x - 5y - 3z = 2 \]

(a)
M1: Demonstrates that the direction vectors are not parallel.
M1: Equates components of both lines to set up three equations.
A1: Solves two equations to get unique values for \( \lambda \) and \( \mu \).
A1: Shows that these parameters do not satisfy the third equation, concluding that the lines are skew.

(b)
M1: Computes the cross product of the direction vectors to get \( \mathbf{n} \).
A1: Obtains the correct perpendicular direction \( \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} \).
M1: Obtains a vector between any point on \( L_1 \) and any point on \( L_2 \) and applies the projection formula.
A1.5: Correctly simplifies to get the shortest distance as \( \frac{3}{\sqrt{35}} \).

(c)
M1: Identifies the normal of the plane as \( \mathbf{n} \).
M1: Uses the coordinates of a point on \( L_1 \) to establish the constant of the Cartesian equation.
A2: Correctly writes the final equation as \( x - 5y - 3z = 2 \) (or any non-zero scalar multiple).

(a) We can factor out a constant \( k > 0 \) from the matrix:
\[ \mathbf{A} = k \begin{pmatrix} \frac{1}{k} & \frac{\sqrt{3}}{k} \\ -\frac{\sqrt{3}}{k} & \frac{1}{k} \end{pmatrix} \]
For the remaining matrix to represent a rotation, its columns must be unit vectors:
\[ \left(\frac{1}{k}\right)^2 + \left(\frac{\sqrt{3}}{k}\right)^2 = 1 \implies \frac{4}{k^2} = 1 \implies k = 2 \; (\text{since } k > 0) \]
This gives:
\[ \mathbf{A} = 2 \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]
The matrix \( \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \) represents a clockwise rotation about the origin through angle \( \theta \).
Here, \( \cos \theta = \frac{1}{2} \) and \( \sin \theta = \frac{\sqrt{3}}{2} \), which implies \( \theta = \frac{\pi}{3} \) (or \( 60^\circ \)).
Thus, the transformation is a clockwise rotation about the origin through \( \frac{\pi}{3} \) radians (or \( 60^\circ \)) followed by (or combined with) an enlargement centered at the origin with scale factor 2.

(b)(i) The standard matrix for a shear parallel to the \( y \)-axis is:
\[ \mathbf{B} = \begin{pmatrix} 1 & 0 \\ s & 1 \end{pmatrix} \]
Given that the image of \( (1,0) \) is \( (1,2) \):
\[ \begin{pmatrix} 1 & 0 \\ s & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ s \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \implies s = 2 \]
Thus, \( \mathbf{B} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \).

(b)(ii) We calculate \( \mathbf{C} = \mathbf{AB} \):
\[ \mathbf{C} = \begin{pmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1(1) + 2\sqrt{3} & 1(0) + \sqrt{3}(1) \\ -\sqrt{3}(1) + 2(1) & -\sqrt{3}(0) + 1(1) \end{pmatrix} = \begin{pmatrix} 1 + 2\sqrt{3} & \sqrt{3} \\ 2 - \sqrt{3} & 1 \end{pmatrix} \]
The area of the image of the unit square is given by \( |\det(\mathbf{C})| \).
Using the property \( \det(\mathbf{C}) = \det(\mathbf{A}) \det(\mathbf{B}) \):
\[ \det(\mathbf{A}) = 1(1) - (-\sqrt{3})(\sqrt{3}) = 1 + 3 = 4 \]
\[ \det(\mathbf{B}) = 1(1) - 0(2) = 1 \]
Thus, \( \det(\mathbf{C}) = 4 \times 1 = 4 \).
Therefore, the area of the image of the unit square is 4.

(a)
M1: Sets up the equation to find the scale factor \( k \).
A1: Correctly identifies the scale factor as 2.
M1: Equates the factored matrix to the trigonometric form of a rotation matrix.
A1: Finds the angle \( \theta = 60^\circ \) (or \( \frac{\pi}{3} \) radians).
A1: States the direction of rotation correctly (clockwise).

(b)(i)
M1: Recalls the structure of a shear matrix parallel to the \( y \)-axis.
A1: Correctly identifies \( s = 2 \) and writes down the matrix \( \mathbf{B} \).

(b)(ii)
M1: Sets up and attempts the matrix multiplication \( \mathbf{AB} \).
A2: Correctly obtains the matrix \( \mathbf{C} \) (A1 if only one error).
M1: Recognizes that the area scale factor is given by the determinant of \( \mathbf{C} \).
A1.5: Calculates the correct area of 4.

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 2m + 5 = 0 \]
Using the quadratic formula:
\[ m = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2\mathrm{i} \]
Thus, the CF is:
\[ y_{\text{CF}} = \mathrm{e}^{-x}(A \cos 2x + B \sin 2x) \]

Next, find the particular integral (PI) by assuming a form:
\[ y_{\text{PI}} = p \cos x + q \sin x \]
Differentiating:
\[ y' = -p \sin x + q \cos x \]
\[ y'' = -p \cos x - q \sin x \]
Substitute these into the original differential equation:
\[ (-p \cos x - q \sin x) + 2(-p \sin x + q \cos x) + 5(p \cos x + q \sin x) = 10 \cos x \]
Grouping terms in \( \cos x \) and \( \sin x \):
\[ (4p + 2q) \cos x + (4q - 2p) \sin x = 10 \cos x \]
By equating coefficients:
\[ 4p + 2q = 10 \implies 2p + q = 5 \]
\[ 4q - 2p = 0 \implies p = 2q \]
Substituting \( p = 2q \) into the first equation:
\[ 5q = 5 \implies q = 1 \implies p = 2 \]
Thus, the PI is:
\[ y_{\text{PI}} = 2 \cos x + \sin x \]

The general solution is:
\[ y = \mathrm{e}^{-x}(A \cos 2x + B \sin 2x) + 2 \cos x + \sin x \]

Now, apply the initial conditions to find the constants \( A \) and \( B \):
Using \( y = 3 \) when \( x = 0 \):
\[ 3 = 1(A \cdot 1 + 0) + 2 \cdot 1 + 0 \implies 3 = A + 2 \implies A = 1 \]

Differentiate the general solution to apply the boundary condition for \( y' \):
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{-x}(A \cos 2x + B \sin 2x) + \mathrm{e}^{-x}(-2A \sin 2x + 2B \cos 2x) - 2\sin x + \cos x \]
Using \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) when \( x = 0 \):
\[ 0 = -1(A) + 1(2B) + 1 \implies 2B - A + 1 = 0 \]
Since \( A = 1 \):
\[ 2B - 1 + 1 = 0 \implies 2B = 0 \implies B = 0 \]

Therefore, the particular solution is:
\[ y = \mathrm{e}^{-x} \cos 2x + 2 \cos x + \sin x \]

M1: Set up and solve auxiliary equation to find complex roots.
A1: State correct complementary function with arbitrary constants.
M1: Formulate particular integral of the form \( p \cos x + q \sin x \) and substitute into the DE.
A1: Correctly calculate \( p = 2 \) and \( q = 1 \).
A1: State the correct general solution.
M1: Use initial condition \( y(0) = 3 \) to find \( A \).
M1: Correctly differentiate the general solution and use \( y'(0) = 0 \) to find \( B \).
A1.375: State the correct particular solution.

(i) Solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \):
\[ \begin{vmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ -1 & -1 & 1-\lambda \end{vmatrix} = 0 \]
Adding Row 3 to Row 1:
\[ \begin{vmatrix} 2-\lambda & 0 & 2-\lambda \\ 2 & 4-\lambda & 2 \\ -1 & -1 & 1-\lambda \end{vmatrix} = 0 \]
Factoring out \( 2-\lambda \) from Row 1:
\[ (2-\lambda) \begin{vmatrix} 1 & 0 & 1 \\ 2 & 4-\lambda & 2 \\ -1 & -1 & 1-\lambda \end{vmatrix} = 0 \]
Subtracting Column 1 from Column 3:
\[ (2-\lambda) \begin{vmatrix} 1 & 0 & 0 \\ 2 & 4-\lambda & 0 \\ -1 & -1 & 2-\lambda \end{vmatrix} = 0 \]
This upper triangular form gives the determinant:
\[ (2-\lambda)(4-\lambda)(2-\lambda) = 0 \implies (2-\lambda)^2(4-\lambda) = 0 \]
Thus, the eigenvalues are \( \lambda = 2 \) (repeated twice) and \( \lambda = 4 \).

(ii) For \( \lambda = 4 \):
\[ \mathbf{A} - 4\mathbf{I} = \begin{pmatrix} -1 & 1 & 1 \\ 2 & 0 & 2 \\ -1 & -1 & -3 \end{pmatrix} \implies \begin{pmatrix} -1 & 1 & 1 \\ 2 & 0 & 2 \\ -1 & -1 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
From row 2: \( 2x + 2z = 0 \implies z = -x \).
From row 1: \( -x + y + z = 0 \implies y = x - z = 2x \).
Thus, an eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \).

For \( \lambda = 2 \):
\[ \mathbf{A} - 2\mathbf{I} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ -1 & -1 & -1 \end{pmatrix} \implies \begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ -1 & -1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This system collapses to the single equation:
\[ x + y + z = 0 \]
We can choose any two linearly independent vectors satisfying this equation, for instance:
\[ \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{v}_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]

(iii) The matrix \( \mathbf{P} \) is formed by placing the eigenvectors as columns, and \( \mathbf{D} \) is the diagonal matrix of corresponding eigenvalues:
\[ \mathbf{P} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & -1 \end{pmatrix} \quad \text{and} \quad \mathbf{D} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \]

(i)
M1: Formulate the characteristic determinant equation.
M1: Use appropriate row/column operations or algebraic expansion to simplify the determinant.
A1: Successfully factorize the polynomial to find \( (2-\lambda)^2(4-\lambda) = 0 \).
A1: State eigenvalues are 2 and 4.
(ii)
M1: Set up system for \( \lambda = 4 \) and solve for a relation between variables.
A1: State a correct eigenvector for \( \lambda = 4 \).
M1: Set up system for \( \lambda = 2 \) and express the resulting linear constraint.
A1.375: State two correct linearly independent eigenvectors for \( \lambda = 2 \).
(iii)
B1: State a correct matrix \( \mathbf{P} \) using their eigenvectors.
B1: State a correct corresponding diagonal matrix \( \mathbf{D} \).

(i) To find a reduction formula, we integrate by parts:
Let \( u = x^n \implies \mathrm{d}u = n x^{n-1} \mathrm{d}x \).
Let \( \mathrm{d}v = \mathrm{e}^{-2x} \mathrm{d}x \implies v = -\frac{1}{2} \mathrm{e}^{-2x} \).
Using \( \int u \mathrm{d}v = u v - \int v \mathrm{d}u \):
\[ I_n = \left[ -\frac{1}{2} x^n \mathrm{e}^{-2x} \right]_0^1 - \int_0^1 \left( -\frac{1}{2} \mathrm{e}^{-2x} \right) n x^{n-1} \mathrm{d}x \]
\[ I_n = -\frac{1}{2}(1)^n \mathrm{e}^{-2} - 0 + \frac{n}{2} \int_0^1 x^{n-1} \mathrm{e}^{-2x} \mathrm{d}x \]
\[ I_n = -\frac{1}{2}\mathrm{e}^{-2} + \frac{n}{2} I_{n-1} \]
Multiply both sides of the equation by 2:
\[ 2I_n = n I_{n-1} - \mathrm{e}^{-2} \]
which is the desired result.

(ii) First, calculate \( I_0 \):
\[ I_0 = \int_0^1 \mathrm{e}^{-2x} \mathrm{d}x = \left[ -\frac{1}{2}\mathrm{e}^{-2x} \right]_0^1 = -\frac{1}{2}\mathrm{e}^{-2} + \frac{1}{2} = \frac{1}{2} - \frac{1}{2}\mathrm{e}^{-2} \]

Now use the reduction formula successively:
For \( n = 1 \):
\[ 2I_1 = 1 I_0 - \mathrm{e}^{-2} = \left( \frac{1}{2} - \frac{1}{2}\mathrm{e}^{-2} \right) - \mathrm{e}^{-2} = \frac{1}{2} - \frac{3}{2}\mathrm{e}^{-2} \implies I_1 = \frac{1}{4} - \frac{3}{4}\mathrm{e}^{-2} \]

For \( n = 2 \):
\[ 2I_2 = 2 I_1 - \mathrm{e}^{-2} = 2\left( \frac{1}{4} - \frac{3}{4}\mathrm{e}^{-2} \right) - \mathrm{e}^{-2} = \frac{1}{2} - \frac{5}{2}\mathrm{e}^{-2} \implies I_2 = \frac{1}{4} - \frac{5}{4}\mathrm{e}^{-2} \]

For \( n = 3 \):
\[ 2I_3 = 3 I_2 - \mathrm{e}^{-2} = 3\left( \frac{1}{4} - \frac{5}{4}\mathrm{e}^{-2} \right) - \mathrm{e}^{-2} = \frac{3}{4} - \frac{19}{4}\mathrm{e}^{-2} \implies I_3 = \frac{3}{8} - \frac{19}{8}\mathrm{e}^{-2} \]

(i)
M1: Set up integration by parts by identifying suitable functions \( u \) and \( v' \).
A1: Correctly differentiate \( u \) and integrate \( v' \).
A1: Substitute the limits correctly into the boundary term.
M1: Express the integral term in terms of \( I_{n-1} \).
A1: Multiply by 2 and obtain the correct reduction formula.
(ii)
B1: Calculate the correct value of \( I_0 \).
M1: Apply the reduction formula to calculate \( I_1 \).
A1: Find correct values for \( I_1 \) and \( I_2 \).
A1.375: State the correct final value for \( I_3 \).

(i) By de Moivre’s theorem, we have:
\[ \cos(5\theta) + \mathrm{i}\sin(5\theta) = (\cos \theta + \mathrm{i}\sin \theta)^5 \]
Expanding the right-hand side using the binomial theorem:
\[ (\cos \theta + \mathrm{i}\sin \theta)^5 = \cos^5 \theta + 5\mathrm{i}\cos^4 \theta \sin \theta - 10\cos^3 \theta \sin^2 \theta - 10\mathrm{i}\cos^2 \theta \sin^3 \theta + 5\cos \theta \sin^4 \theta + \mathrm{i}\sin^5 \theta \]
Equating imaginary parts from both sides:
\[ \sin(5\theta) = 5\cos^4 \theta \sin \theta - 10\cos^2 \theta \sin^3 \theta + \sin^5 \theta \]
We use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to convert the expression into terms of \( \sin \theta \):
\[ \sin(5\theta) = 5(1 - \sin^2 \theta)^2 \sin \theta - 10(1 - \sin^2 \theta)\sin^3 \theta + \sin^5 \theta \]
\[ \sin(5\theta) = 5(1 - 2\sin^2 \theta + \sin^4 \theta)\sin \theta - 10\sin^3 \theta + 10\sin^5 \theta + \sin^5 \theta \]
\[ \sin(5\theta) = 5\sin \theta - 10\sin^3 \theta + 5\sin^5 \theta - 10\sin^3 \theta + 11\sin^5 \theta \]
\[ \sin(5\theta) = 16\sin^5 \theta - 20\sin^3 \theta + 5\sin \theta \]

(ii) Let \( x = \sin \theta \). The equation \( 16x^5 - 20x^3 + 5x = \frac{1}{2} \) then becomes:
\[ \sin(5\theta) = \frac{1}{2} \]
For \( 5\theta \), the solutions are:
\[ 5\theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 5\theta = \frac{5\pi}{6} + 2k\pi \]
where \( k \in \mathbb{Z} \).
This gives:
\[ \theta = \frac{\pi}{30} + \frac{2k\pi}{5} \quad \text{or} \quad \theta = \frac{\pi}{6} + \frac{2k\pi}{5} \]
To obtain 5 distinct roots of the quintic equation within the interval \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), we evaluate \( \theta \) for appropriate values of \( k \):
- For \( 5\theta = -\frac{11\pi}{6} \implies \theta = -\frac{11\pi}{30} \)
- For \( 5\theta = -\frac{7\pi}{6} \implies \theta = -\frac{7\pi}{30} \)
- For \( 5\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{30} \)
- For \( 5\theta = \frac{5\pi}{6} \implies \theta = \frac{\pi}{6} \)
- For \( 5\theta = \frac{13\pi}{6} \implies \theta = \frac{13\pi}{30} \)

Thus, the five exact roots in the required form are:
\[ x = \sin\left(-\frac{11\pi}{30}\right), \quad x = \sin\left(-\frac{7\pi}{30}\right), \quad x = \sin\left(\frac{\pi}{30}\right), \quad x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad x = \sin\left(\frac{13\pi}{30}\right) \]

(i)
M1: Apply de Moivre's theorem to express \( \cos(5\theta) + \mathrm{i}\sin(5\theta) \).
A1: Correctly write down the binomial expansion of the right-hand side.
M1: Equate imaginary parts to find the expression for \( \sin(5\theta) \).
M1: Substitute \( \cos^2\theta = 1 - \sin^2\theta \).
A1: Expand and simplify to obtain the given identity.
(ii)
M1: Use substitution \( x = \sin \theta \) to obtain \( \sin(5\theta) = \frac{1}{2} \).
A1: Solve to find values for \( 5\theta \).
M1: Find five distinct values of \( \theta \) within the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
A1.375: State all five exact solutions in the required form.

(i) We use the hyperbolic identity \( \cosh^2 x = 1 + \sinh^2 x \) to rewrite the equation:
\[ 2(1 + \sinh^2 x) + \sinh x = 5 \implies 2\sinh^2 x + \sinh x - 3 = 0 \]
Let \( u = \sinh x \). The equation is now quadratic:
\[ 2u^2 + u - 3 = 0 \implies (2u + 3)(u - 1) = 0 \]
Thus, the solutions are:
\[ \sinh x = 1 \quad \text{or} \quad \sinh x = -\frac{3}{2} \]

Using the logarithmic form for inverse hyperbolic sine, \( \sinh^{-1} u = \ln(u + \sqrt{u^2 + 1}) \):
For \( \sinh x = 1 \):
\[ x = \ln(1 + \sqrt{1 + 1}) = \ln(1 + \sqrt{2}) \]
For \( \sinh x = -\frac{3}{2} \):
\[ x = \ln\left(-\frac{3}{2} + \sqrt{\frac{9}{4} + 1}\right) = \ln\left(-\frac{3}{2} + \sqrt{\frac{13}{4}}\right) = \ln\left(\frac{-3 + \sqrt{13}}{2}\right) \]

(ii) To compute the integral, rewrite the integrand:
\[ \cosh^3 x = \cosh^2 x \cosh x = (1 + \sinh^2 x) \cosh x \]
Let \( u = \sinh x \implies \mathrm{d}u = \cosh x \mathrm{d}x \).
Now change the limits of integration:
- When \( x = 0 \), \( u = \sinh 0 = 0 \).
- When \( x = \ln 3 \), \( u = \sinh(\ln 3) = \frac{\mathrm{e}^{\ln 3} - \mathrm{e}^{-\ln 3}}{2} = \frac{3 - 1/3}{2} = \frac{8/3}{2} = \frac{4}{3} \).

Substituting these into the integral:
\[ \int_0^{\ln 3} \cosh^3 x \mathrm{d}x = \int_0^{4/3} (1 + u^2) \mathrm{d}u = \left[ u + \frac{u^3}{3} \right]_0^{4/3} \]
\[ = \frac{4}{3} + \frac{1}{3} \left( \frac{64}{27} \right) = \frac{4}{3} + \frac{64}{81} = \frac{108 + 64}{81} = \frac{172}{81} \]

(i)
M1: Substitute \( \cosh^2 x = 1 + \sinh^2 x \) to form a quadratic in \( \sinh x \).
A1: Solve the quadratic correctly to obtain \( \sinh x = 1 \) and \( \sinh x = -1.5 \).
M1: Apply the inverse hyperbolic sine formula in logarithmic form.
A1: Obtain the solution \( x = \ln(1+\sqrt{2}) \).
A1: Obtain the solution \( x = \ln\left(\frac{-3+\sqrt{13}}{2}\right) \).
(ii)
M1: Rewrite \( \cosh^3 x \) as \( (1+\sinh^2 x)\cosh x \).
M1: Apply substitution \( u = \sinh x \) and compute the new integration limits.
A1: Integrate \( 1+u^2 \) correctly.
A1.375: Obtain the final exact fraction \( \frac{172}{81} \).

(i) Let \( y = \tan^{-1}(x+1) \implies \tan y = x+1 \).
Differentiating both sides with respect to \( x \):
\[ \sec^2 y \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sec^2 y} = \frac{1}{1+\tan^2 y} = \frac{1}{1+(x+1)^2} = \frac{1}{x^2 + 2x + 2} \] (Q.E.D)

(ii) We compute the values at \( x = 0 \):
\[ f(0) = \tan^{-1}(1) = \frac{\pi}{4} \]
\[ f'(x) = (x^2+2x+2)^{-1} \implies f'(0) = 2^{-1} = \frac{1}{2} \]

Differentiate to find higher derivatives:
\[ f''(x) = -(x^2+2x+2)^{-2} (2x+2) \implies f''(0) = -(2)^{-2}(2) = -\frac{2}{4} = -\frac{1}{2} \]

Using the product rule / quotient rule on \( f''(x) = -2(x+1)(x^2+2x+2)^{-2} \):
\[ f'''(x) = -2(x^2+2x+2)^{-2} + 4(x+1)(x^2+2x+2)^{-3}(2x+2) \]
\[ f'''(x) = -2(x^2+2x+2)^{-2} + 8(x+1)^2 (x^2+2x+2)^{-3} \]
At \( x = 0 \):
\[ f'''(0) = -2(2)^{-2} + 8(1)^2 (2)^{-3} = -\frac{2}{4} + \frac{8}{8} = -\frac{1}{2} + 1 = \frac{1}{2} \]

(iii) The Maclaurin series expansion of \( f(x) \) is given by:
\[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \]
Substituting our derived values:
\[ f(x) \approx \frac{\pi}{4} + \frac{1}{2}x + \frac{-1/2}{2}x^2 + \frac{1/2}{6}x^3 \]
\[ f(x) \approx \frac{\pi}{4} + \frac{1}{2}x - \frac{1}{4}x^2 + \frac{1}{12}x^3 \]

(i)
M1: Apply implicit differentiation or standard inverse tangent derivative formula.
A1: Simplify algebra to obtain \( \frac{1}{x^2+2x+2} \).
(ii)
B1: State \( f(0) = \frac{\pi}{4} \).
B1: State \( f'(0) = \frac{1}{2} \).
M1: Differentiate \( f'(x) \) and find \( f''(0) \).
A1: Obtain \( f''(0) = -\frac{1}{2} \).
M1: Differentiate \( f''(x) \) using standard rules and find \( f'''(0) \).
A1.375: Obtain \( f'''(0) = \frac{1}{2} \).
(iii)
M1: State the general formula for Maclaurin's series.
A1: Substitute calculated coefficients to obtain \( \frac{\pi}{4} + \frac{1}{2}x - \frac{1}{4}x^2 + \frac{1}{12}x^3 \).

(i) This is a first-order linear differential equation of the form \( \frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x) \) where \( P(x) = \cot x \).
The integrating factor is:
\[ I(x) = \mathrm{e}^{\int \cot x \mathrm{d}x} = \mathrm{e}^{\ln|\sin x|} = \sin x \]
(since \( 0 < x < \pi \), we have \( \sin x > 0 \)).

Multiply both sides of the differential equation by \( \sin x \):
\[ \sin x \frac{\mathrm{d}y}{\mathrm{d}x} + y \cos x = \sin x \cos^2 x \]
\[ \frac{\mathrm{d}}{\mathrm{d}x} (y \sin x) = \sin x \cos^2 x \]

Now integrate both sides with respect to \( x \):
\[ y \sin x = \int \sin x \cos^2 x \mathrm{d}x \]
Let \( u = \cos x \implies \mathrm{d}u = -\sin x \mathrm{d}x \):
\[ y \sin x = \int -u^2 \mathrm{d}u = -\frac{1}{3}u^3 + C = -\frac{1}{3}\cos^3 x + C \]
Dividing by \( \sin x \), we get the general solution:
\[ y = -\frac{\cos^3 x}{3\sin x} + \frac{C}{\sin x} \]

(ii) Using the initial condition \( y = 0 \) when \( x = \frac{\pi}{2} \):
\[ 0 \cdot \sin\left(\frac{\pi}{2}\right) = -\frac{1}{3}\cos^3\left(\frac{\pi}{2}\right) + C \]
\[ 0 = 0 + C \implies C = 0 \]
Substituting \( C = 0 \) into the general solution:
\[ y = -\frac{\cos^3 x}{3\sin x} \]

(i)
M1: Calculate integrating factor using \( \mathrm{e}^{\int \cot x \mathrm{d}x} \).
A1: Correctly simplify integrating factor to \( \sin x \).
M1: Write the differential equation in the form \( \frac{\mathrm{d}}{\mathrm{d}x}(y \sin x) = \sin x \cos^2 x \).
M1: Integrate the right-hand side using substitution.
A1: Correct integration to get \( -\frac{1}{3}\cos^3 x + C \).
A1.375: State correct general solution.
(ii)
M1: Substitute \( x = \frac{\pi}{2} \) and \( y = 0 \) to solve for \( C \).
A1: Find \( C = 0 \).
A1: Express the particular solution as \( y = -\frac{\cos^3 x}{3\sin x} \).

(i) Solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \):
\[ \begin{vmatrix} 1-\lambda & 0 & 2 \\ 0 & 2-\lambda & 1 \\ 1 & -1 & -\lambda \end{vmatrix} = 0 \]
Expand along the first row:
\[ (1-\lambda) [ (2-\lambda)(-\lambda) - (-1)(1) ] + 2 [ 0( -1) - 1(2-\lambda) ] = 0 \]
\[ (1-\lambda) [ \lambda^2 - 2\lambda + 1 ] + 2[ \lambda - 2 ] = 0 \]
\[ (1-\lambda)(\lambda-1)^2 + 2\lambda - 4 = 0 \]
\[ -(\lambda-1)^3 + 2\lambda - 4 = 0 \]
\[ -(\lambda^3 - 3\lambda^2 + 3\lambda - 1) + 2\lambda - 4 = 0 \]
\[ -\lambda^3 + 3\lambda^2 - \lambda - 3 = 0 \]
Thus, the characteristic equation is:
\[ \lambda^3 - 3\lambda^2 + \lambda + 3 = 0 \]

(ii) By the Cayley-Hamilton theorem, \( \mathbf{A} \) satisfies its own characteristic equation:
\[ \mathbf{A}^3 - 3\mathbf{A}^2 + \mathbf{A} + 3\mathbf{I} = \mathbf{0} \]
Multiplying through by \( \mathbf{A}^{-1} \):
\[ \mathbf{A}^2 - 3\mathbf{A} + \mathbf{I} + 3\mathbf{A}^{-1} = \mathbf{0} \implies 3\mathbf{A}^{-1} = -(\mathbf{A}^2 - 3\mathbf{A} + \mathbf{I}) \]
\[ \mathbf{A}^{-1} = -\frac{1}{3} (\mathbf{A}^2 - 3\mathbf{A} + \mathbf{I}) \] (Q.E.D)

Now calculate \( \mathbf{A}^2 \):
\[ \mathbf{A}^2 = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & 2 \\ 1 & -2 & 1 \end{pmatrix} \]
Substitute this into the expression for \( \mathbf{A}^{-1} \):
\[ \mathbf{A}^{-1} = -\frac{1}{3} \left[ \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & 2 \\ 1 & -2 & 1 \end{pmatrix} - \begin{pmatrix} 3 & 0 & 6 \\ 0 & 6 & 3 \\ 3 & -3 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right] \]
\[ \mathbf{A}^{-1} = -\frac{1}{3} \begin{pmatrix} 1 & -2 & -4 \\ 1 & -2 & -1 \\ -2 & 1 & 2 \end{pmatrix} = \begin{pmatrix} -1/3 & 2/3 & 4/3 \\ -1/3 & 2/3 & 1/3 \\ 2/3 & -1/3 & -2/3 \end{pmatrix} \]

(iii) From the Cayley-Hamilton equation:
\[ \mathbf{A}^3 = 3\mathbf{A}^2 - \mathbf{A} - 3\mathbf{I} \]
Multiply by \( \mathbf{A} \):
\[ \mathbf{A}^4 = 3\mathbf{A}^3 - \mathbf{A}^2 - 3\mathbf{A} \]
Substitute \( \mathbf{A}^3 \) back into the equation:
\[ \mathbf{A}^4 = 3(3\mathbf{A}^2 - \mathbf{A} - 3\mathbf{I}) - \mathbf{A}^2 - 3\mathbf{A} \]
\[ \mathbf{A}^4 = 9\mathbf{A}^2 - 3\mathbf{A} - 9\mathbf{I} - \mathbf{A}^2 - 3\mathbf{A} \]
\[ \mathbf{A}^4 = 8\mathbf{A}^2 - 6\mathbf{A} - 9\mathbf{I} \]
Thus, \( a = 8 \), \( b = -6 \), and \( c = -9 \).

(i)
M1: Set up the characteristic determinant.
M1: Expand the determinant.
A1: Obtain correct characteristic equation \( \lambda^3 - 3\lambda^2 + \lambda + 3 = 0 \).
(ii)
M1: Use Cayley-Hamilton theorem and multiply by \( \mathbf{A}^{-1} \) to verify the algebraic identity.
A1: Correctly calculate the matrix \( \mathbf{A}^2 \).
A1: Obtain correct matrix form of \( \mathbf{A}^{-1} \).
(iii)
M1: Write expression for \( \mathbf{A}^4 \) using \( \mathbf{A}^3 \).
M1: Substitute the expression for \( \mathbf{A}^3 \).
A1.375: Obtain final correct constants \( a = 8 \), \( b = -6 \), \( c = -9 \).