CAIE Further Paper 2 2023 November — Question 7 11 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionNovember
Marks11
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Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹
DifficultyChallenging +1.2 This is a Further Maths question on diagonalization and Cayley-Hamilton theorem. Part (a) requires finding eigenvalues (trivial from upper triangular form), eigenvectors, and constructing P and D—standard procedure with 7 marks. Part (b) uses Cayley-Hamilton to find the inverse, which is a bookwork application once the characteristic equation is known. While this requires multiple techniques, both parts follow well-established algorithms without requiring novel insight or complex problem-solving.
Spec4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
The matrix A is given by $$\mathbf{A} = \begin{pmatrix} -6 & 2 & 13 \\ 0 & -2 & 5 \\ 0 & 0 & 8 \end{pmatrix}.$$
  1. Find a matrix P and a diagonal matrix D such that \(\mathbf{A}^{-1} = \mathbf{PDP}^{-1}\). [7]
  2. Use the characteristic equation of A to find \(\mathbf{A}^{-1}\). [4]
Question 7:
AnswerMarks Guidance
7(a)=−6, =−2, =8 B1
i j k 8 1
   
=−6: 0 4 5 = 0 0
   
   
AnswerMarks Guidance
0 0 2 0 0M1 A1 Uses vector product (or equations) to find corresponding
eigenvectors.
i j k 10 1
   
=−2: −4 2 13 = 20 2
   
   
AnswerMarks
0 0 5  0  0A1
i j k 140 2
   
=8: −14 2 13 = 70 1
   
   
AnswerMarks
0 −10 5 140 2A1
1 1 2 −1 0 0
   6 
Thus P=  0 2 1  and D=  0 −1 2 0
   1
0 0 2   0 0 
AnswerMarks Guidance
8M1 A1 Or correctly matched permutations of columns.
M1 for their (non-zero) eigenvectors matched to their
eigenvalues.
7
AnswerMarks Guidance
7(b)A3−52A−96I=0 M1
96A−1 =A2 −52IM1 Multiples through by A−1.
36 −16 36
 
A2 = 0 4 30
 
 
AnswerMarks Guidance
 0 0 64B1
QuestionAnswer Marks
7(b)−1 −1 3 
 6 6 8 
A−1 =  0 −1 5 
2 16
  0 0 1  
AnswerMarks
8A1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 7:
--- 7(a) ---
7(a) | =−6, =−2, =8 | B1
i j k 8 1
   
=−6: 0 4 5 = 0 0
   
   
0 0 2 0 0 | M1 A1 | Uses vector product (or equations) to find corresponding
eigenvectors.
i j k 10 1
   
=−2: −4 2 13 = 20 2
   
   
0 0 5  0  0 | A1
i j k 140 2
   
=8: −14 2 13 = 70 1
   
   
0 −10 5 140 2 | A1
1 1 2 −1 0 0
   6 
Thus P=  0 2 1  and D=  0 −1 2 0
   1
0 0 2   0 0 
8 | M1 A1 | Or correctly matched permutations of columns.
M1 for their (non-zero) eigenvectors matched to their
eigenvalues.
7
--- 7(b) ---
7(b) | A3−52A−96I=0 | M1 | Substitutes A into characteristic equation.
96A−1 =A2 −52I | M1 | Multiples through by A−1.
36 −16 36
 
A2 = 0 4 30
 
 
 0 0 64 | B1
Question | Answer | Marks | Guidance
7(b) | −1 −1 3 
 6 6 8 
A−1 =  0 −1 5 
2 16
  0 0 1  
8 | A1
4
Question | Answer | Marks | Guidance
The matrix A is given by
$$\mathbf{A} = \begin{pmatrix} -6 & 2 & 13 \\ 0 & -2 & 5 \\ 0 & 0 & 8 \end{pmatrix}.$$

\begin{enumerate}[label=(\alph*)]
\item Find a matrix P and a diagonal matrix D such that $\mathbf{A}^{-1} = \mathbf{PDP}^{-1}$. [7]

\item Use the characteristic equation of A to find $\mathbf{A}^{-1}$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q7 [11]}}
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