OCR MEI Further Extra Pure Specimen — Question 5 18 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
SessionSpecimen
Marks18
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Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeEigenvalues and eigenvectors
DifficultyChallenging +1.8 This is a Further Maths question requiring understanding of rotation matrices, eigenvalues, and characteristic equations. While it involves multiple steps (17 marks total), each part follows standard techniques: recognizing rotation properties, finding eigenvectors, computing characteristic polynomials, and using complex eigenvalues. The conceptual demand is moderate for Further Maths students, though the multi-part nature and requirement for detailed reasoning across several connected ideas elevates it above routine exercises.
Spec4.03f Linear transformations 3D: reflections and rotations about axes4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix
In this question you must show detailed reasoning. You are given that the matrix $\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{\sqrt{2}} & \frac{1}{2}
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}
\frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} \end{pmatrix}$ represents a rotation in 3-D space.
  1. Explain why it follows that \(\mathbf{M}\) has 1 as an eigenvalue. [2]
  2. Find a vector equation for the axis of the rotation. [4]
  3. Show that the characteristic equation of \(\mathbf{M}\) can be written as $$\lambda^3 - \lambda^2 + \lambda - 1 = 0.$$ [5]
  4. Find the smallest positive integer \(n\) such that \(\mathbf{M}^n = \mathbf{I}\). [6]
  5. Find the magnitude of the angle of the rotation which \(\mathbf{M}\) represents. Give your reasoning. [1]
Question 5:
AnswerMarks Guidance
5(i) A rotation has an axis …
… and this is a line of invariant points.
AnswerMarks
The direction vector of this line/axis has associated eigenvalue 1.E1
E1
AnswerMarks
[2]1.2
2.2aThe second of these
can be implied
AnswerMarks Guidance
5(ii) DR
 1  1 1 
x x
 2 2 2
   
 1 0  1  y  y
   
 2 2
   
 1 1 1 z z
 2 2 2 
1x 1 y1zx
2 2 2
1 x  1 z y
2 2
1x 1 y1zz
2 2 2
Simplifying algebra which eliminates one variable
1
 
Eigenvector 0 .
 
 
1
1
 
Axis of rotation rt 0
 
 
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1M1
M1
A1
A1
AnswerMarks
[4]3.1a
1.1
1.1
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3.2aFor any two correct
equations
May be awarded for
1
 
starting with y .
 
 
z
Or any multiple
oe
AnswerMarks Guidance
5(iii) DR
1  1 1
2 2 2
1   1 0
2 2
1 1 1
2 2 2
1  1 1  1 1 
2 2 2 2 2
 
 1 2 1
2 4
 1   1 1  1 1  0
2 2 2 2 2
11
2 2
+23
+110
2 2
3+210
AnswerMarks
So 32 10M1
M1
M1
A1
A1
AnswerMarks
[5]1.1
1.1
1.1a
1.1
AnswerMarks
2.1This expands on
middle column; any
other column or row
acceptable.
Condone missing =0.
Correct one line
All correct
Answer given; must
follow from previous
working.
AnswerMarks Guidance
5(iv) DR
 1  1 1  1  1 1 
 2 2 2  2 2 2 
M2  1 0  1  1 0  1 
 2 2 2 2
 1 1 1  1 1 1 
 2 2 2  2 2 2 
 
 
 1 I
 
 
 
M3=M2MI( Cayley-Hamilton theorem)
AnswerMarks
So M3 Ibecause M2 Mfrom aboveB1
M1
AnswerMarks
A13.1a
2.4
AnswerMarks Guidance
2.2aShowing M2 I 0 0 1
 
M2  0 1 0
 
 
1 0 0
Alternative method
     
     
M3  1  0 I 0 I
     
     
AnswerMarks Guidance
     M1
A1M1 Or convincing
multiplication
AnswerMarks
showing M3 I.Or convincing
A1multiplication
showing M3 I.
M4=M3M2M
M2MM2MI
AnswerMarks
So n4M1
A1
A1
AnswerMarks
[6]1.1
2.1
AnswerMarks
3.2aOr convincing
multiplication.
AnswerMarks Guidance
5(v)
because 4 applications of the transformation gets back to starting point
AnswerMarks Guidance
2E1
[1]2.2a 90°
QuestionAO1 AO2
1i2 1
1ii1 1
1iii2 2 
Question 5:
5 | (i) | A rotation has an axis …
… and this is a line of invariant points.
The direction vector of this line/axis has associated eigenvalue 1. | E1
E1
[2] | 1.2
2.2a | The second of these
can be implied
5 | (ii) | DR
 1  1 1 
x x
 2 2 2
   
 1 0  1  y  y
   
 2 2
   
 1 1 1 z z
 2 2 2 
1x 1 y1zx
2 2 2
1 x  1 z y
2 2
1x 1 y1zz
2 2 2
Simplifying algebra which eliminates one variable
1
 
Eigenvector 0 .
 
 
1
1
 
Axis of rotation rt 0
 
 
1 | M1
M1
A1
A1
[4] | 3.1a
1.1
1.1
3.2a | For any two correct
equations
May be awarded for
1
 
starting with y .
 
 
z
Or any multiple
oe
5 | (iii) | DR
1  1 1
2 2 2
1   1 0
2 2
1 1 1
2 2 2
1  1 1  1 1 
2 2 2 2 2
 
 1 2 1
2 4
 1   1 1  1 1  0
2 2 2 2 2
11
2 2
+23
+110
2 2
3+210
So 32 10 | M1
M1
M1
A1
A1
[5] | 1.1
1.1
1.1a
1.1
2.1 | This expands on
middle column; any
other column or row
acceptable.
Condone missing =0.
Correct one line
All correct
Answer given; must
follow from previous
working.
5 | (iv) | DR
 1  1 1  1  1 1 
 2 2 2  2 2 2 
M2  1 0  1  1 0  1 
 2 2 2 2
 1 1 1  1 1 1 
 2 2 2  2 2 2 
 
 
 1 I
 
 
 
M3=M2MI( Cayley-Hamilton theorem)
So M3 Ibecause M2 Mfrom above | B1
M1
A1 | 3.1a
2.4
2.2a | Showing M2 I | 0 0 1
 
M2  0 1 0
 
 
1 0 0
Alternative method
     
     
M3  1  0 I 0 I
     
     
      | M1
A1 | M1 | Or convincing
multiplication
showing M3 I. | Or convincing
A1 | multiplication
showing M3 I.
M4=M3M2M
M2MM2MI
So n4 | M1
A1
A1
[6] | 1.1
2.1
3.2a | Or convincing
multiplication.
5 | (v) | 
because 4 applications of the transformation gets back to starting point
2 | E1
[1] | 2.2a | 90°
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
1i | 2 | 1 | 0 | 0 | 3
1ii | 1 | 1 | 0 | 0 | 2
1iii | 2 | 2 | 1 | 0 | 5
In this question you must show detailed reasoning.

You are given that the matrix $\mathbf{M} = \begin{pmatrix}
\frac{1}{2} & -\frac{1}{\sqrt{2}} & \frac{1}{2} \\
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\
\frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2}
\end{pmatrix}$ represents a rotation in 3-D space.

\begin{enumerate}[label=(\roman*)]
\item Explain why it follows that $\mathbf{M}$ has 1 as an eigenvalue. [2]

\item Find a vector equation for the axis of the rotation. [4]

\item Show that the characteristic equation of $\mathbf{M}$ can be written as
$$\lambda^3 - \lambda^2 + \lambda - 1 = 0.$$ [5]

\item Find the smallest positive integer $n$ such that $\mathbf{M}^n = \mathbf{I}$. [6]

\item Find the magnitude of the angle of the rotation which $\mathbf{M}$ represents. Give your reasoning. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure  Q5 [18]}}
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