OCR MEI Further Extra Pure 2021 November — Question 3 14 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2021
SessionNovember
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind P and D for diagonalization / matrix powers
DifficultyChallenging +1.3 This is a Further Maths question on matrix diagonalization requiring finding characteristic equation, eigenvalues, eigenvectors, and expressing A^n using diagonalization. While systematic and multi-step (14 marks total), it follows a standard algorithmic procedure without requiring novel insight. The eigenvalues are given for verification, reducing computational burden. Harder than typical A-level due to Further Maths content and the A^n derivation, but still a textbook exercise.
Spec4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix

The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} 3 & 3 & 0 \\ 0 & 2 & 2 \\ 1 & 3 & 4 \end{pmatrix}\).
  1. Determine the characteristic equation of \(\mathbf{A}\). [3]
  2. Hence verify that the eigenvalues of \(\mathbf{A}\) are 1, 2 and 6. [1]
  3. For each eigenvalue of \(\mathbf{A}\) determine an associated eigenvector. [4]
  4. Use the results of parts (b) and (c) to find \(\mathbf{A}^n\) as a single matrix, where \(n\) is a positive integer. [6]

Question 3:
AnswerMarks Guidance
3(a) 3−λ 3 0
det(A−λI)= 0 2−λ 2
1 3 4−λ
=(3−λ)[(2−λ)(4−λ)−2×3]−3(0−2×1) oe
AnswerMarks
= –λ3 + 9λ2 – 20λ + 12 = 0M1
M1
A1
AnswerMarks
[3]1.1a
1.1
AnswerMarks
1.1Formation of appropriate
determinant soi.
Attempt to expand determinant.
Allow one slip.
AnswerMarks
Must be an equation. ISW.May see eg expansion by 1st col:
(3−λ)[(2−λ)(4−λ)−6]+1(6−0)
Or other formulation eg:
((3−λ)(2−λ)(4−λ)+6+0)
−(0+6(3−λ)+0)
AnswerMarks Guidance
3(b) 1, 2 and 6 substituted into (a) equation to verify
[1]1.1 eg checking trace is insufficient.
3(c) 3a + 3b = a or 2a or 6a
and 2b + 2c = b or 2b or 6b
and a + 3b + 4c = c or 2c or 6c
λ = 1: 2a = –3b, b = –2c
or λ = 2: c = 0, a = –3b
or λ = 6: a = b, c = 2b
 3  −3 1
     
−2 or 1 or 1
     
     
 1   0  2
 3  −3 1
     
−2 and 1 and 1
     
     
AnswerMarks
 1   0  2M1
M1
A1
A1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1Correctly forming 3 equations in 3
unknowns for one of their
eigenvalues. May see explicit
choice of eg c = 1 to form 3
equations in 2 unknowns.
Attempt to solve equations for at
least one of their eigenvalues
leading to two unknowns in terms
of 3rd.
or any non-zero multiple.
AnswerMarks
or any non-zero multiple.Or formation of appropriate
i j k
determinant eg −2 4−λ −4 .
0 −1 3−λ
Attempt to expand determinant
(might be in terms of λ) eg
8−7λ+λ2
 
6−2λ . Can be inferred by
 
 2 
2 correct coefficients.
AnswerMarks Guidance
3(d)  3 −3 1
 
−2 1 1
 
 1 0 2
−1
 3 −3 1 −2 −6 4
  1  
−2 1 1 = −5 −5 5 oe
  10 
 1 0 2  1 3 3
1 0 0 n 1 0 0 
 0 2 0  =  0 2n 0 
   
0 0 6  0 0 6n 
 3 −3 11 0 0  −2 −6 4
 −2 1 1  0 2n 0  1  −5 −5 5 
   10 
 1 0 2  0 0 6n   1 3 3
1 0 0 −2 −6 4
 0 2n 0  −5 −5 5  =
  
 0 0 6n  1 3 3
 −2 −6 4 
 
−5×2n −5×2n 5×2n
 
 6n 3×6n 3×6n
 
 3 −3 1 −2 −6 4 
1   
 −2 1 1−5×2n −5×2n 5×2n=
10  
 1 0 2 6n 3×6n 3×6n 
−6+15×2n+6n −18+15×2n+3×6n 12−15×2n+3×6n
 
1
 4−5×2n+6n 12−5×2n+3×6n −8+5×2n+3×6n
10 
 −2+2×6n −6+6n+1 4+6n+1 
AnswerMarks
 M1
A1FT
B1
M1
M1
A1
AnswerMarks
[6]3.1a
3.1a
3.1a
3.1a
1.1
AnswerMarks
1.1Forming matrix of their
eigenvectors, E.
BC. Finding inverse of their
matrix of eigenvectors.
Matrix of eigenvalues must be
consistent with matrix of
eigenvectors. Allow 1n.
Forming EΛnE–1. Can be awarded
if Λn incorrect or uncalculated but
eigenvectors must be in same
order as eigenvalues.
Proper attempt to multiply either
the first two or the last two (of 3)
in the correct order (with or
without 1 ).
10
or
 3 −3×2n 6n  −2 −6 4
 
1  
−2 2n 6n  −5 −5 5 =
 
10  
 1 0 2×6n 1 3 3
 
etc.
AnswerMarks
Condone 6×6n unsimplified.May be in decimal form:
−0.2 −0.6 0.4
 
−0.5 −0.5 0.5
 
 0.1 0.3 0.3
 3 −3 11 0 0 
 −2 1 1  0 2n 0  =
  
 1 0 2  0 0 6n 
or
 3 −3×2n 6n 
 
−2 2n 6n 
 2×6n
 1 0 
 
Question 3:
3 | (a) | 3−λ 3 0
det(A−λI)= 0 2−λ 2
1 3 4−λ
=(3−λ)[(2−λ)(4−λ)−2×3]−3(0−2×1) oe
= –λ3 + 9λ2 – 20λ + 12 = 0 | M1
M1
A1
[3] | 1.1a
1.1
1.1 | Formation of appropriate
determinant soi.
Attempt to expand determinant.
Allow one slip.
Must be an equation. ISW. | May see eg expansion by 1st col:
(3−λ)[(2−λ)(4−λ)−6]+1(6−0)
Or other formulation eg:
((3−λ)(2−λ)(4−λ)+6+0)
−(0+6(3−λ)+0)
3 | (b) | 1, 2 and 6 substituted into (a) equation to verify | B1
[1] | 1.1 | eg checking trace is insufficient.
3 | (c) | 3a + 3b = a or 2a or 6a
and 2b + 2c = b or 2b or 6b
and a + 3b + 4c = c or 2c or 6c
λ = 1: 2a = –3b, b = –2c
or λ = 2: c = 0, a = –3b
or λ = 6: a = b, c = 2b
 3  −3 1
     
−2 or 1 or 1
     
     
 1   0  2
 3  −3 1
     
−2 and 1 and 1
     
     
 1   0  2 | M1
M1
A1
A1
[4] | 1.1
1.1
1.1
1.1 | Correctly forming 3 equations in 3
unknowns for one of their
eigenvalues. May see explicit
choice of eg c = 1 to form 3
equations in 2 unknowns.
Attempt to solve equations for at
least one of their eigenvalues
leading to two unknowns in terms
of 3rd.
or any non-zero multiple.
or any non-zero multiple. | Or formation of appropriate
i j k
determinant eg −2 4−λ −4 .
0 −1 3−λ
Attempt to expand determinant
(might be in terms of λ) eg
8−7λ+λ2
 
6−2λ . Can be inferred by
 
 2 
2 correct coefficients.
3 | (d) |  3 −3 1
 
−2 1 1
 
 1 0 2
−1
 3 −3 1 −2 −6 4
  1  
−2 1 1 = −5 −5 5 oe
  10 
 1 0 2  1 3 3
1 0 0 n 1 0 0 
 0 2 0  =  0 2n 0 
   
0 0 6  0 0 6n 
 3 −3 11 0 0  −2 −6 4
 −2 1 1  0 2n 0  1  −5 −5 5 
   10 
 1 0 2  0 0 6n   1 3 3
1 0 0 −2 −6 4
 0 2n 0  −5 −5 5  =
  
 0 0 6n  1 3 3
 −2 −6 4 
 
−5×2n −5×2n 5×2n
 
 6n 3×6n 3×6n
 
 3 −3 1 −2 −6 4 
1   
 −2 1 1−5×2n −5×2n 5×2n=
10  
 1 0 2 6n 3×6n 3×6n 
−6+15×2n+6n −18+15×2n+3×6n 12−15×2n+3×6n
 
1
 4−5×2n+6n 12−5×2n+3×6n −8+5×2n+3×6n
10 
 −2+2×6n −6+6n+1 4+6n+1 
  | M1
A1FT
B1
M1
M1
A1
[6] | 3.1a
3.1a
3.1a
3.1a
1.1
1.1 | Forming matrix of their
eigenvectors, E.
BC. Finding inverse of their
matrix of eigenvectors.
Matrix of eigenvalues must be
consistent with matrix of
eigenvectors. Allow 1n.
Forming EΛnE–1. Can be awarded
if Λn incorrect or uncalculated but
eigenvectors must be in same
order as eigenvalues.
Proper attempt to multiply either
the first two or the last two (of 3)
in the correct order (with or
without 1 ).
10
or
 3 −3×2n 6n  −2 −6 4
 
1  
−2 2n 6n  −5 −5 5 =
 
10  
 1 0 2×6n 1 3 3
 
etc.
Condone 6×6n unsimplified. | May be in decimal form:
−0.2 −0.6 0.4
 
−0.5 −0.5 0.5
 
 0.1 0.3 0.3
 3 −3 11 0 0 
 −2 1 1  0 2n 0  =
  
 1 0 2  0 0 6n 
or
 3 −3×2n 6n 
 
−2 2n 6n 
 2×6n
 1 0 
 
The matrix $\mathbf{A}$ is given by $\mathbf{A} = \begin{pmatrix} 3 & 3 & 0 \\ 0 & 2 & 2 \\ 1 & 3 & 4 \end{pmatrix}$.

\begin{enumerate}[label=(\alph*)]
\item Determine the characteristic equation of $\mathbf{A}$. [3]

\item Hence verify that the eigenvalues of $\mathbf{A}$ are 1, 2 and 6. [1]

\item For each eigenvalue of $\mathbf{A}$ determine an associated eigenvector. [4]

\item Use the results of parts (b) and (c) to find $\mathbf{A}^n$ as a single matrix, where $n$ is a positive integer. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2021 Q3 [14]}}