CAIE Further Paper 2 2021 November — Question 6 11 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeUse Cayley-Hamilton for inverse
DifficultyChallenging +1.2 This is a structured Further Maths linear algebra question requiring the Cayley-Hamilton theorem for matrix inversion, similarity transformations, and properties of eigenvalues under powers. While it involves multiple techniques, each part follows standard procedures without requiring novel insight—the Cayley-Hamilton approach is a taught method, the similarity transformation is direct matrix multiplication, and eigenvalue/eigenvector properties under powers are standard results. The upper-triangular structure simplifies the characteristic equation significantly. Moderately above average difficulty due to the multi-step nature and Further Maths content, but well within expected exam techniques.
Spec4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
The matrix \(\mathbf{P}\) is given by $$\mathbf{P} = \begin{pmatrix} 1 & 6 & 6 \\ 0 & 2 & 6 \\ 0 & 0 & -3 \end{pmatrix}.$$
  1. Use the characteristic equation of \(\mathbf{P}\) to find \(\mathbf{P}^{-1}\). [5]
  2. Find the matrix \(\mathbf{A}\) such that $$\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{pmatrix}.$$ [4]
  3. State the eigenvalues and corresponding eigenvectors of \(\mathbf{A}^3\). [2]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/22 Cambridge International AS & A Level – Mark Scheme October/November 2021
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2021 Page 5 of 14
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(a)(λ−1 ) (λ−2)(λ+3)=λ3−7λ+6[=0] M1 A1
P2 −7I+6P −1=0M1 −1.
Substitutes P and multiplies through by P
1 18 24 1 −3 −4
P2 =  0 4 −6  leading toP −1 =  0 1 1 
   2 
AnswerMarks
 0 0 9     0 0 −1 3  M1 A1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)4 0 0
A=P  0 5 0  P −1
 
 
0 0 6
AnswerMarks Guidance
 M1 Applies A=PDP −1.
4 30 36 1 −3 −4 1 6 6 4 −12 −16
     
0 10 36 0 1 1 or 0 2 6 0 5 5
  2    2 
AnswerMarks Guidance
 0 0 −18     0 0 −1 3     0 0 −3     0 0 −2  M1 A1 Multiplies two adjacent matrices.
4 3 2 
 
0 5 −2
 
 
0 0 6
AnswerMarks Guidance
 A1 A1 linked to first M1.
4
AnswerMarks
6(c)1 6  6 
     
Eigenvectors: 0 , 2 , 6
     
AnswerMarks Guidance
 0    0     −3  B1 Accept any non-zero multiples of these.
Eigenvalues: 64, 125, 216B1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/22 Cambridge International AS & A Level – Mark Scheme October/November 2021
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2021 Page 5 of 14
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | (λ−1 ) (λ−2)(λ+3)=λ3−7λ+6[=0] | M1 A1 | Finds characteristic equation.
P2 −7I+6P −1=0 | M1 | −1.
Substitutes P and multiplies through by P
1 18 24 1 −3 −4
P2 =  0 4 −6  leading toP −1 =  0 1 1 
   2 
 0 0 9     0 0 −1 3   | M1 A1
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 4 0 0
A=P  0 5 0  P −1
 
 
0 0 6
  | M1 | Applies A=PDP −1.
4 30 36 1 −3 −4 1 6 6 4 −12 −16
     
0 10 36 0 1 1 or 0 2 6 0 5 5
  2    2 
 0 0 −18     0 0 −1 3     0 0 −3     0 0 −2   | M1 A1 | Multiplies two adjacent matrices.
4 3 2 
 
0 5 −2
 
 
0 0 6
  | A1 | A1 linked to first M1.
4
--- 6(c) ---
6(c) | 1 6  6 
     
Eigenvectors: 0 , 2 , 6
     
 0    0     −3   | B1 | Accept any non-zero multiples of these.
Eigenvalues: 64, 125, 216 | B1
2
Question | Answer | Marks | Guidance
The matrix $\mathbf{P}$ is given by

$$\mathbf{P} = \begin{pmatrix} 1 & 6 & 6 \\ 0 & 2 & 6 \\ 0 & 0 & -3 \end{pmatrix}.$$

\begin{enumerate}[label=(\alph*)]
\item Use the characteristic equation of $\mathbf{P}$ to find $\mathbf{P}^{-1}$. [5]

\item Find the matrix $\mathbf{A}$ such that

$$\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 6 \end{pmatrix}.$$ [4]

\item State the eigenvalues and corresponding eigenvectors of $\mathbf{A}^3$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q6 [11]}}
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