Question 8 - A-Level Maths

CAIE Further Paper 2 2024 June — Question 8 16 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2024
Session	June
Marks	16
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹
Difficulty	Challenging +1.8 This is a substantial Further Maths question combining linear systems, eigenvalues, and matrix diagonalization. Parts (a)-(b) require understanding of consistency conditions (moderate difficulty). Part (c) requires finding eigenvalues of a transformed matrix and constructing P and D (standard but lengthy). Part (d) uses the Cayley-Hamilton theorem creatively to establish a non-trivial matrix identity, requiring algebraic manipulation and insight beyond routine application. The multi-part structure and need to connect different concepts elevates this above typical questions.
Spec	4.03l Singular/non-singular matrices 4.03o Inverse 3x3 matrix 4.03r Solve simultaneous equations: using inverse matrix 4.03t Plane intersection: geometric interpretation

Find the set of values of $a$ for which the system of equations $$\begin{array} { c l } 6 x + a y & = 3 \\ 2 x - y & = 1 \\ x + 5 y + 4 z & = 2 \end{array}$$ has a unique solution.
Show that the system of equations in part (a) is consistent for all values of $a$.
The matrix $\mathbf { A }$ is given by $$\mathbf { A } = \left( \begin{array} { r r r } 6 & 0 & 0 \\ 2 & - 1 & 0 \\ 1 & 5 & 4 \end{array} \right)$$
Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $( 14 \mathbf { A } + 24 \mathbf { I } ) ^ { 2 } = \mathbf { P D P } ^ { - 1 }$.
Use the characteristic equation of $\mathbf { A }$ to show that $$( 14 \mathbf { A } + 24 \mathbf { I } ) ^ { 2 } = \mathbf { A } ^ { 4 } ( \mathbf { A } + b \mathbf { I } ) ^ { 2 }$$ where $b$ is an integer to be determined.
If you use the following page to complete the answer to any question, the question number must be clearly shown.

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Question 8(a):

Answer	Marks	Guidance
$\begin{vmatrix}6 & a & 0\\2 & -1 & 0\\1 & 5 & 4\end{vmatrix} = -8a - 24$	M1	Finds determinant or solves equations.
$a \neq -3$	A1

Question 8(b):

Answer	Marks	Guidance
If $a \neq -3$ then system has unique solution (so consistent).	B1 FT
If $a=-3$ then: $6x - 3y = 3$, $2x - y = 1 \Rightarrow 11x + 4z = 7$, $x + 5y + 4z = 2$	M1	Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with $a = -3$.
So the system has infinitely many solutions (so consistent).	A1
Alternative: Sets $y = 0$	B1
$\left(\frac{1}{2}, 0, \frac{3}{8}\right)$ is a solution so system is consistent for all values of $a$.	M1 A1	Finds a solution with $y=0$ and states conclusion.

Question 8(c):

Answer	Marks	Guidance
Eigenvalues of A are $6,\,-1$ and $4$.	B1	Lower diagonal matrix or characteristic equation.
$\lambda=6$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-7&0\\1&5&-2\end{vmatrix} = \begin{pmatrix}14\\4\\17\end{pmatrix}$	M1 A1	Uses vector product (or equations) to find corresponding eigenvectors.
$\lambda=-1$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&0&0\\1&5&5\end{vmatrix} = \begin{pmatrix}0\\-10\\10\end{pmatrix}\sim\begin{pmatrix}0\\-1\\1\end{pmatrix}$	A1
$\lambda=4$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-5&0\\1&5&0\end{vmatrix} = \begin{pmatrix}0\\0\\15\end{pmatrix}\sim\begin{pmatrix}0\\0\\1\end{pmatrix}$	A1
$\mathbf{P}=\begin{pmatrix}14&0&0\\4&-1&0\\17&1&1\end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix}11664&0&0\\0&100&0\\0&0&6400\end{pmatrix}$	M1 A1	Or correctly matched permutations of columns. Column of zeros in P gets M0.

Question 8(d):

Answer	Marks	Guidance
$(\lambda-6)(\lambda+1)(\lambda-4) = \lambda^3 - 9\lambda^2 + 14\lambda + 24 = 0$	B1	Characteristic equation.
$14\mathbf{A} + 24\mathbf{I} = -\mathbf{A}^3 + 9\mathbf{A}^2$	M1	Substitutes for A and makes $14\mathbf{A}+24\mathbf{I}$ the subject.
$(14\mathbf{A}+24\mathbf{I})^2 = \left(-\mathbf{A}^3+9\mathbf{A}^2\right)^2 = \mathbf{A}^4(\mathbf{A}-9\mathbf{I})^2$	M1 A1	Squares and factorises. CWO.

## Question 8(a):

$\begin{vmatrix}6 & a & 0\\2 & -1 & 0\\1 & 5 & 4\end{vmatrix} = -8a - 24$ | **M1** | Finds determinant or solves equations.

$a \neq -3$ | **A1** |

---

## Question 8(b):

If $a \neq -3$ then system has unique solution (so consistent). | **B1 FT** |

If $a=-3$ then: $6x - 3y = 3$, $2x - y = 1 \Rightarrow 11x + 4z = 7$, $x + 5y + 4z = 2$ | **M1** | Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with $a = -3$.

So the system has infinitely many solutions (so consistent). | **A1** |

**Alternative:** Sets $y = 0$ | **B1** |

$\left(\frac{1}{2}, 0, \frac{3}{8}\right)$ is a solution so system is consistent for all values of $a$. | **M1 A1** | Finds a solution with $y=0$ and states conclusion.

---

## Question 8(c):

Eigenvalues of **A** are $6,\,-1$ and $4$. | **B1** | Lower diagonal matrix or characteristic equation.

$\lambda=6$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-7&0\\1&5&-2\end{vmatrix} = \begin{pmatrix}14\\4\\17\end{pmatrix}$ | **M1 A1** | Uses vector product (or equations) to find corresponding eigenvectors.

$\lambda=-1$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&0&0\\1&5&5\end{vmatrix} = \begin{pmatrix}0\\-10\\10\end{pmatrix}\sim\begin{pmatrix}0\\-1\\1\end{pmatrix}$ | **A1** |

$\lambda=4$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-5&0\\1&5&0\end{vmatrix} = \begin{pmatrix}0\\0\\15\end{pmatrix}\sim\begin{pmatrix}0\\0\\1\end{pmatrix}$ | **A1** |

$\mathbf{P}=\begin{pmatrix}14&0&0\\4&-1&0\\17&1&1\end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix}11664&0&0\\0&100&0\\0&0&6400\end{pmatrix}$ | **M1 A1** | Or correctly matched permutations of columns. Column of zeros in **P** gets M0.

---

## Question 8(d):

$(\lambda-6)(\lambda+1)(\lambda-4) = \lambda^3 - 9\lambda^2 + 14\lambda + 24 = 0$ | **B1** | Characteristic equation.

$14\mathbf{A} + 24\mathbf{I} = -\mathbf{A}^3 + 9\mathbf{A}^2$ | **M1** | Substitutes for **A** and makes $14\mathbf{A}+24\mathbf{I}$ the subject.

$(14\mathbf{A}+24\mathbf{I})^2 = \left(-\mathbf{A}^3+9\mathbf{A}^2\right)^2 = \mathbf{A}^4(\mathbf{A}-9\mathbf{I})^2$ | **M1 A1** | Squares and factorises. CWO.

Show LaTeX source

8
\begin{enumerate}[label=(\alph*)]
\item Find the set of values of $a$ for which the system of equations

$$\begin{array} { c l } 
6 x + a y & = 3 \\
2 x - y & = 1 \\
x + 5 y + 4 z & = 2
\end{array}$$

has a unique solution.
\item Show that the system of equations in part (a) is consistent for all values of $a$.\\

The matrix $\mathbf { A }$ is given by

$$\mathbf { A } = \left( \begin{array} { r r r } 
6 & 0 & 0 \\
2 & - 1 & 0 \\
1 & 5 & 4
\end{array} \right)$$
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $( 14 \mathbf { A } + 24 \mathbf { I } ) ^ { 2 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to show that

$$( 14 \mathbf { A } + 24 \mathbf { I } ) ^ { 2 } = \mathbf { A } ^ { 4 } ( \mathbf { A } + b \mathbf { I } ) ^ { 2 }$$

where $b$ is an integer to be determined.\\

If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q8 [16]}}

This paper (8 questions)

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Q1 5 Q2 4 Q3 7 Q4 8 Q5 11 Q6 12 Q7 12 Q8 16

Answer	Marks	Guidance
\(\begin{vmatrix}6 & a & 0\\2 & -1 & 0\\1 & 5 & 4\end{vmatrix} = -8a - 24\)	M1	Finds determinant or solves equations.
\(a \neq -3\)	A1

Answer	Marks	Guidance
If \(a \neq -3\) then system has unique solution (so consistent).	B1 FT
If \(a=-3\) then: \(6x - 3y = 3\), \(2x - y = 1 \Rightarrow 11x + 4z = 7\), \(x + 5y + 4z = 2\)	M1	Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with \(a = -3\).
So the system has infinitely many solutions (so consistent).	A1
Alternative: Sets \(y = 0\)	B1
\(\left(\frac{1}{2}, 0, \frac{3}{8}\right)\) is a solution so system is consistent for all values of \(a\).	M1 A1	Finds a solution with \(y=0\) and states conclusion.

Answer	Marks	Guidance
Eigenvalues of A are \(6,\,-1\) and \(4\).	B1	Lower diagonal matrix or characteristic equation.
\(\lambda=6\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-7&0\\1&5&-2\end{vmatrix} = \begin{pmatrix}14\\4\\17\end{pmatrix}\)	M1 A1	Uses vector product (or equations) to find corresponding eigenvectors.
\(\lambda=-1\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&0&0\\1&5&5\end{vmatrix} = \begin{pmatrix}0\\-10\\10\end{pmatrix}\sim\begin{pmatrix}0\\-1\\1\end{pmatrix}\)	A1
\(\lambda=4\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-5&0\\1&5&0\end{vmatrix} = \begin{pmatrix}0\\0\\15\end{pmatrix}\sim\begin{pmatrix}0\\0\\1\end{pmatrix}\)	A1
\(\mathbf{P}=\begin{pmatrix}14&0&0\\4&-1&0\\17&1&1\end{pmatrix}\) and \(\mathbf{D}=\begin{pmatrix}11664&0&0\\0&100&0\\0&0&6400\end{pmatrix}\)	M1 A1	Or correctly matched permutations of columns. Column of zeros in P gets M0.

Answer	Marks	Guidance
\((\lambda-6)(\lambda+1)(\lambda-4) = \lambda^3 - 9\lambda^2 + 14\lambda + 24 = 0\)	B1	Characteristic equation.
\(14\mathbf{A} + 24\mathbf{I} = -\mathbf{A}^3 + 9\mathbf{A}^2\)	M1	Substitutes for A and makes \(14\mathbf{A}+24\mathbf{I}\) the subject.
\((14\mathbf{A}+24\mathbf{I})^2 = \left(-\mathbf{A}^3+9\mathbf{A}^2\right)^2 = \mathbf{A}^4(\mathbf{A}-9\mathbf{I})^2\)	M1 A1	Squares and factorises. CWO.