OCR MEI Further Pure Core 2022 June — Question 4 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeVolume/area scale factors
DifficultyStandard +0.8 Part (a) requires understanding that mapping a 3D cube to a plane means the transformation has determinant zero, then computing a 3Γ—3 determinant and solving. Part (b) involves combining knowledge of determinants, composition of transformations, and the relationship between determinant sign/magnitude and orientation/area scale factor. While these are standard Further Maths topics, the question requires connecting multiple concepts rather than routine calculation, placing it moderately above average difficulty.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03j Determinant 3x3: calculation
4
  1. A transformation with associated matrix \(\left( \begin{array} { r r r } m & 2 & 1 \\ 0 & 1 & - 2 \\ 2 & 0 & 3 \end{array} \right)\), where \(m\) is a constant, maps the vertices of a cube to points that all lie in a plane. Find \(m\).
  2. The transformations S and T of the plane have associated matrices \(\mathbf { M }\) and \(\mathbf { N }\) respectively, where \(\mathbf { M } = \left( \begin{array} { r r } k & 1 \\ - 3 & 4 \end{array} \right)\) and the determinant of \(\mathbf { N }\) is \(3 k + 1\). The transformation \(U\) is equivalent to the combined transformation consisting of S followed by T . Given that U preserves orientation and has an area scale factor 2, find the possible values of \(k\).
Question 4:
AnswerMarks Guidance
4(a) π‘š 2 1
0 1 βˆ’2= 3π‘šβˆ’2Γ—4+1Γ—(βˆ’2) = 3π‘šβˆ’10
2 0 3
10
so 3π‘šβˆ’10 = 0 β‡’ π‘š =
AnswerMarks
3M1
A1
AnswerMarks
A13.1a
1.1
AnswerMarks
1.1finding determinant
Alternative solution
π‘š 2 1 π‘š 3
(0).(1)Γ—(βˆ’2) = (0).(βˆ’6)
2 0 3 2 βˆ’5
= 3π‘šβˆ’10 [= 0]
10
β‡’ π‘š =
AnswerMarks
33.1a
1.1
1.1
M1
A1
A1
[3]
AnswerMarks Guidance
4(b) det𝑴 = 4π‘˜+3
det(𝑡𝑴) = det𝑡×det𝑴
β‡’ (3π‘˜+1)(4π‘˜+3) = 2
1
π‘˜ = βˆ’1 or βˆ’
AnswerMarks
12B1
M1
M1
AnswerMarks
A11.2
3.1a
1.1
AnswerMarks
1.1Soi
Equating to 2
AnswerMarks
Alternative solution1.2
3.1a
1.1
1.1
AnswerMarks
det(𝑁𝑀) = (π‘Žπ‘˜βˆ’3𝑏)(𝑐+4𝑑)βˆ’(π‘Ž+4𝑏)(π‘π‘˜βˆ’3𝑑)B1
(π‘Žπ‘˜βˆ’3𝑏)(𝑐+4𝑑)βˆ’(π‘Ž+4𝑏)(π‘π‘˜βˆ’3𝑑) = 2M1
(3π‘˜+1)(4π‘˜+3) = 2M1
1
π‘˜ = βˆ’1 or βˆ’
AnswerMarks
12A1
[4]
Alternative solution
π‘š 2 1 π‘š 3
(0).(1)Γ—(βˆ’2) = (0).(βˆ’6)
2 0 3 2 βˆ’5
= 3π‘šβˆ’10 [= 0]
10
β‡’ π‘š =
3
3.1a
1.1
1.1
1.2
3.1a
1.1
1.1
Question 4:
4 | (a) | π‘š 2 1
|0 1 βˆ’2| = 3π‘šβˆ’2Γ—4+1Γ—(βˆ’2) = 3π‘šβˆ’10
2 0 3
10
so 3π‘šβˆ’10 = 0 β‡’ π‘š =
3 | M1
A1
A1 | 3.1a
1.1
1.1 | finding determinant
Alternative solution
π‘š 2 1 π‘š 3
(0).(1)Γ—(βˆ’2) = (0).(βˆ’6)
2 0 3 2 βˆ’5
= 3π‘šβˆ’10 [= 0]
10
β‡’ π‘š =
3 | 3.1a
1.1
1.1
M1
A1
A1
[3]
4 | (b) | det𝑴 = 4π‘˜+3
det(𝑡𝑴) = det𝑡×det𝑴
β‡’ (3π‘˜+1)(4π‘˜+3) = 2
1
π‘˜ = βˆ’1 or βˆ’
12 | B1
M1
M1
A1 | 1.2
3.1a
1.1
1.1 | Soi
Equating to 2
Alternative solution | 1.2
3.1a
1.1
1.1
det(𝑁𝑀) = (π‘Žπ‘˜βˆ’3𝑏)(𝑐+4𝑑)βˆ’(π‘Ž+4𝑏)(π‘π‘˜βˆ’3𝑑) | B1
(π‘Žπ‘˜βˆ’3𝑏)(𝑐+4𝑑)βˆ’(π‘Ž+4𝑏)(π‘π‘˜βˆ’3𝑑) = 2 | M1
(3π‘˜+1)(4π‘˜+3) = 2 | M1
1
π‘˜ = βˆ’1 or βˆ’
12 | A1
[4]
Alternative solution
π‘š 2 1 π‘š 3
(0).(1)Γ—(βˆ’2) = (0).(βˆ’6)
2 0 3 2 βˆ’5
= 3π‘šβˆ’10 [= 0]
10
β‡’ π‘š =
3
3.1a
1.1
1.1
1.2
3.1a
1.1
1.1
4
\begin{enumerate}[label=(\alph*)]
\item A transformation with associated matrix $\left( \begin{array} { r r r } m & 2 & 1 \\ 0 & 1 & - 2 \\ 2 & 0 & 3 \end{array} \right)$, where $m$ is a constant, maps the vertices of a cube to points that all lie in a plane.

Find $m$.
\item The transformations S and T of the plane have associated matrices $\mathbf { M }$ and $\mathbf { N }$ respectively, where $\mathbf { M } = \left( \begin{array} { r r } k & 1 \\ - 3 & 4 \end{array} \right)$ and the determinant of $\mathbf { N }$ is $3 k + 1$. The transformation $U$ is equivalent to the combined transformation consisting of S followed by T .

Given that U preserves orientation and has an area scale factor 2, find the possible values of $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q4 [7]}}
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