CAIE Further Paper 2 2024 November — Question 8 14 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionNovember
Marks14
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TopicInvariant lines and eigenvalues and vectors
TypeFind eigenvalues of 3×3 matrix
DifficultyChallenging +1.3 This is a structured Further Maths linear algebra question with three parts: (a) finding eigenvalues via characteristic equation (routine but computational), (b) using Cayley-Hamilton theorem to express A^4 (standard technique once learned), and (c) diagonalizing a transformed matrix (requires understanding of eigenvalue shifts). While it involves multiple Further Maths concepts, each part follows standard procedures without requiring novel insight. The computational load and multi-step nature place it above average difficulty, but it remains a textbook-style question testing technique rather than problem-solving creativity.
Spec4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices
The matrix \(\mathbf{A}\) is given by $$\mathbf{A} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 7 & 9 \\ 4 & 1 & 7 \end{pmatrix}.$$
  1. Show that the characteristic equation of \(\mathbf{A}\) is \(\lambda^3 - 12\lambda^2 + 124 + 80 = 0\) and find the eigenvalues of \(\mathbf{A}\). [4]
  2. Use the characteristic equation of \(\mathbf{A}\) to show that $$\mathbf{A}^4 = p\mathbf{A}^2 + q\mathbf{A} + r\mathbf{I},$$ where \(p\), \(q\) and \(r\) are integers to be determined. [4]
  3. Find a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \((\mathbf{A} - 3\mathbf{I})^4 = \mathbf{PDP}^{-1}\). [6]
Question 8:
AnswerMarks
8(a)−2− 0 0
0 7− 9 =0
AnswerMarks Guidance
4 1 7−B1 Sets determinant equal to zero.
(−2−) ( (7−)2 −9 ) =0(−2−)( 49−14+2 −9 )
AnswerMarks Guidance
3−122 +12+80=0M1A1 Expands, AG
=−2,4,10B1
4
AnswerMarks Guidance
8(b)A3−12A2 +12A+80I=0A4 −12A3+12A2 +80A=0 M1A1
A4 =12 ( 12A2 −12A−80I ) −12A2 −80A=132A2 −224A−960IM1A1 Substitutes A3.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(c)i j k  72   2 
   
=−2: 0 9 9 = 36 1
   
   
AnswerMarks Guidance
4 1 9 −36 −1*M1A1 Uses vector product (or equations) to find
corresponding eigenvectors.
i j k  0   0  i j k  0  0
       
=4: 0 3 9 = 36 ~ 3 =10: −12 0 0 = 108 ~ 3
       
       
AnswerMarks
4 1 3 −12 −1 0 −3 9  36  1A1A1
 2 0 0 625 0 0 
   
Thus P= 1 3 3 and D= 0 1 0
   
   
AnswerMarks Guidance
−1 −1 1  0 0 2401DM1A1 Or correctly matched permutations of
columns.
Correctly matched but eigenvalues not
changed scores M1 A0.
M0 if a column of zeros appears in P.
6
Question 8:
--- 8(a) ---
8(a) | −2− 0 0
0 7− 9 =0
4 1 7− | B1 | Sets determinant equal to zero.
(−2−) ( (7−)2 −9 ) =0(−2−)( 49−14+2 −9 )
3−122 +12+80=0 | M1A1 | Expands, AG
=−2,4,10 | B1
4
--- 8(b) ---
8(b) | A3−12A2 +12A+80I=0A4 −12A3+12A2 +80A=0 | M1A1 | Substitutes A and multiplies through by A.
A4 =12 ( 12A2 −12A−80I ) −12A2 −80A=132A2 −224A−960I | M1A1 | Substitutes A3.
4
Question | Answer | Marks | Guidance
--- 8(c) ---
8(c) | i j k  72   2 
   
=−2: 0 9 9 = 36 1
   
   
4 1 9 −36 −1 | *M1A1 | Uses vector product (or equations) to find
corresponding eigenvectors.
i j k  0   0  i j k  0  0
       
=4: 0 3 9 = 36 ~ 3 =10: −12 0 0 = 108 ~ 3
       
       
4 1 3 −12 −1 0 −3 9  36  1 | A1A1
 2 0 0 625 0 0 
   
Thus P= 1 3 3 and D= 0 1 0
   
   
−1 −1 1  0 0 2401 | DM1A1 | Or correctly matched permutations of
columns.
Correctly matched but eigenvalues not
changed scores M1 A0.
M0 if a column of zeros appears in P.
6
The matrix $\mathbf{A}$ is given by
$$\mathbf{A} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 7 & 9 \\ 4 & 1 & 7 \end{pmatrix}.$$

\begin{enumerate}[label=(\alph*)]
\item Show that the characteristic equation of $\mathbf{A}$ is $\lambda^3 - 12\lambda^2 + 124 + 80 = 0$ and find the eigenvalues of $\mathbf{A}$. [4]

\item Use the characteristic equation of $\mathbf{A}$ to show that
$$\mathbf{A}^4 = p\mathbf{A}^2 + q\mathbf{A} + r\mathbf{I},$$
where $p$, $q$ and $r$ are integers to be determined. [4]

\item Find a matrix $\mathbf{P}$ and a diagonal matrix $\mathbf{D}$ such that $(\mathbf{A} - 3\mathbf{I})^4 = \mathbf{PDP}^{-1}$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q8 [14]}}
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