CAIE FP1 2019 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks10
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Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A = PDP⁻¹
DifficultyStandard +0.8 This is a Further Maths diagonalization problem with an upper triangular matrix. Part (i) requires finding eigenvalues (straightforward from diagonal of triangular matrix), eigenvectors (moderate algebra with parameter m), and constructing P and D (7 marks indicates substantial work). Part (ii) involves a non-standard calculation with the transpose. While diagonalization is a core FM topic, the multi-step nature, parameter handling, and the somewhat unusual second part place this above average difficulty but not exceptionally hard.
Spec4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
The matrix \(\mathbf{M}\) is defined by $$\mathbf{M} = \begin{pmatrix} 2 & m & 1 \\ 0 & m & 7 \\ 0 & 0 & 1 \end{pmatrix},$$ where \(m \neq 0, 1, 2\).
  1. Find a matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{M} = \mathbf{PDP}^{-1}\). [7]
  2. Find \(\mathbf{M}^T\mathbf{P}\). [3]
Question 8:
AnswerMarks
8(i)Eigenvalues of (upper diagonal matrix) A are 2, m and 1.
( )( )( )=0)
AnswerMarks
(Or from characteristic equation: λ−2 λ−m λ−1B1
i j k 2−m 1
   
λ=2: e = 0 m−2 1 = 0 =t 0
1    
   
0 0 −1 0 0
AnswerMarks Guidance
   M1 A1 Uses vector product (or equations) to find corresponding
eigenvectors.
i j k  7m   m 
λ=m: e = 2−m m 1 =  7 ( m−2 ) =t  m−2 
2    
   
0 0 7 0 0
AnswerMarks
   A1
i j k 6m+1 6m+1
   
λ=1: e = 1 m 1 = −7 =t −7
3    
   
0 m−1 7 m−1 m−1
AnswerMarks
   A1
1 m 6m+1 2 0 0
   
Thus P= 0 m−2 −7 and D= 0 m 0
   
   
0 0 m−1 0 0 1
AnswerMarks
   M1 A1
FTOr correctly matched permutations of columns. No follow
through on two or more zero eigenvectors.
7
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(ii)1 m 6m+127 0 0
  
M7P = PD7P−1P = PD7= 0 m−2 −7 0 m7 0
  
 0 0 m−1  0 0 1
AnswerMarks
  M1 A1
FTM7 = PD7P−1
Applies .
27 m8 6m+1
 
= 0 m8 −2m7 −7
 
 0 0 m−1 
AnswerMarks Guidance
 A1 Order of columns might be swapped depending on P.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(i) ---
8(i) | Eigenvalues of (upper diagonal matrix) A are 2, m and 1.
( )( )( )=0)
(Or from characteristic equation: λ−2 λ−m λ−1 | B1
i j k 2−m 1
   
λ=2: e = 0 m−2 1 = 0 =t 0
1    
   
0 0 −1 0 0
    | M1 A1 | Uses vector product (or equations) to find corresponding
eigenvectors.
i j k  7m   m 
λ=m: e = 2−m m 1 =  7 ( m−2 ) =t  m−2 
2    
   
0 0 7 0 0
    | A1
i j k 6m+1 6m+1
   
λ=1: e = 1 m 1 = −7 =t −7
3    
   
0 m−1 7 m−1 m−1
    | A1
1 m 6m+1 2 0 0
   
Thus P= 0 m−2 −7 and D= 0 m 0
   
   
0 0 m−1 0 0 1
    | M1 A1
FT | Or correctly matched permutations of columns. No follow
through on two or more zero eigenvectors.
7
Question | Answer | Marks | Guidance
--- 8(ii) ---
8(ii) | 1 m 6m+127 0 0
  
M7P = PD7P−1P = PD7= 0 m−2 −7 0 m7 0
  
 0 0 m−1  0 0 1
   | M1 A1
FT | M7 = PD7P−1
Applies .
27 m8 6m+1
 
= 0 m8 −2m7 −7
 
 0 0 m−1 
  | A1 | Order of columns might be swapped depending on P.
3
Question | Answer | Marks | Guidance
The matrix $\mathbf{M}$ is defined by
$$\mathbf{M} = \begin{pmatrix} 2 & m & 1 \\ 0 & m & 7 \\ 0 & 0 & 1 \end{pmatrix},$$
where $m \neq 0, 1, 2$.

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

\item Find $\mathbf{M}^T\mathbf{P}$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q8 [10]}}
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