CAIE FP1 2015 November — Question 6 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A = PDP⁻¹
DifficultyStandard +0.3 This is a standard Further Maths eigenvalue/eigenvector question with straightforward calculations. Finding eigenvectors from given eigenvalues is routine (solve (A-λI)v=0), verifying an eigenvector requires simple matrix multiplication, and constructing the diagonalization P^{-1}AP=D follows a standard algorithm. The 3×3 matrix adds minor computational complexity but no conceptual challenge beyond typical FP1 level.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
The matrix \(\mathbf{A}\), where $$\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 10 & -7 & 10 \\ 7 & -5 & 8 \end{pmatrix},$$ has eigenvalues 1 and 3. Find corresponding eigenvectors. [3] It is given that \(\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\) is an eigenvector of \(\mathbf{A}\). Find the corresponding eigenvalue. [2] Find a diagonal matrix \(\mathbf{D}\) and matrices \(\mathbf{P}\) and \(\mathbf{P}^{-1}\) such that \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}\). [5]
Question 6:
AnswerMarks
6i j k −6  1 
   
λ =1:10 −8 10 = 0 ~ 0  oe
   
7 −5 7 6 −1
   
i j k  0  0
   
λ =3: −2 0 0 =20~1 oe
   
10 −10 10 20 1
   
 1 0 0 0  0  0
      
10 −7 102=−4=−22⇒λ =−2
      
7 −5 8 1 −2 1
      
−2 0 0 0 1 0
   
D= 0 1 0, P=2 0 1 (or other multiples or permutations).
   
0 0 3 1 −1 1
   
Det P = –1 (or 1 depending on permutation).
 1 −1 1 −1 1 −1
   
Adj P = −1 0 0 ⇒ P–1 =  1 0 0  ( or other permutations).
   
−2 1 2 2 −1 2
AnswerMarks
   M1A1
A1
[3]
M1A1
[2]
B1 B1
B1
M1A1
[5]

Total

10
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Cambridge International A Level – October/November 20159231 11
Qn &
AnswerMarks Guidance
PartSolution Marks 
Question 6:
6 | i j k −6  1 
   
λ =1:10 −8 10 = 0 ~ 0  oe
   
7 −5 7 6 −1
   
i j k  0  0
   
λ =3: −2 0 0 =20~1 oe
   
10 −10 10 20 1
   
 1 0 0 0  0  0
      
10 −7 102=−4=−22⇒λ =−2
      
7 −5 8 1 −2 1
      
−2 0 0 0 1 0
   
D= 0 1 0, P=2 0 1 (or other multiples or permutations).
   
0 0 3 1 −1 1
   
Det P = –1 (or 1 depending on permutation).
 1 −1 1 −1 1 −1
   
Adj P = −1 0 0 ⇒ P–1 =  1 0 0  ( or other permutations).
   
−2 1 2 2 −1 2
    | M1A1
A1
[3]
M1A1
[2]
B1 B1
B1
M1A1
[5]
Total
10
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 11
Qn &
Part | Solution | Marks
The matrix $\mathbf{A}$, where
$$\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 \\ 10 & -7 & 10 \\ 7 & -5 & 8 \end{pmatrix},$$

has eigenvalues 1 and 3. Find corresponding eigenvectors. [3]

It is given that $\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ is an eigenvector of $\mathbf{A}$. Find the corresponding eigenvalue. [2]

Find a diagonal matrix $\mathbf{D}$ and matrices $\mathbf{P}$ and $\mathbf{P}^{-1}$ such that $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}$. [5]

\hfill \mbox{\textit{CAIE FP1 2015 Q6 [10]}}
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