OCR MEI Further Pure Core Specimen — Question 13 13 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeGeometric interpretation of systems
DifficultyChallenging +1.2 This is a Further Maths question testing multiple matrix/3D geometry concepts (determinant calculation, quadratic surface intersection, plane arrangement, transformation scale factors). While it requires connecting several ideas and has 13 marks total, each individual part uses standard techniques: determinant expansion is routine, part (ii) cleverly uses the matrix structure but follows from (i), plane verification is trivial, and the geometric interpretation requires understanding of determinant=0. The novel connection between parts elevates it slightly above average Further Maths difficulty, but no individual step requires exceptional insight.
Spec4.03j Determinant 3x3: calculation4.03k Determinant 3x3: volume scale factor4.03t Plane intersection: geometric interpretation
Matrix M is given by \(\mathbf{M} = \begin{pmatrix} k & 1 & -5 \\ 2 & 3 & -3 \\ -1 & 2 & 2 \end{pmatrix}\), where \(k\) is a constant.
  1. Show that \(\det \mathbf{M} = 12(k - 3)\). [2]
  2. Find a solution of the following simultaneous equations for which \(x \neq z\). $$4x^2 + y^2 - 5z^2 = 6$$ $$2x^2 + 3y^2 - 3z^2 = 6$$ $$-x^2 + 2y^2 + 2z^2 = -6$$ [3]
    1. Verify that the point \((2, 0, 1)\) lies on each of the following three planes. $$3x + y - 5z = 1$$ $$2x + 3y - 3z = 1$$ $$-x + 2y + 2z = 0$$ [1]
    2. Describe how the three planes in part (iii) (A) are arranged in 3-D space. Give reasons for your answer. [4]
  4. Find the values of \(k\) for which the transformation represented by M has a volume scale factor of 6. [3]
Question 13:
AnswerMarks Guidance
13(i) detM(cid:32)k(cid:11)6(cid:14)6(cid:12)–(cid:11)4–3(cid:12)–5(cid:11)4(cid:14)3(cid:12) Simplify to
12(cid:11)k –3(cid:12) AGM1
A1
AnswerMarks
[2]1.1
1.1Answer given so method
must be clear
AnswerMarks Guidance
13(ii) (cid:167)x2(cid:183) (cid:167)1 (cid:16)1 1(cid:183)(cid:167) 6 (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)(cid:168) (cid:184)
y2 (cid:32) (cid:16) 1 1 1 6
(cid:168) (cid:184) (cid:168)1 2 4 6(cid:184)(cid:168) (cid:184)
(cid:168)z2(cid:184) (cid:168) 7 (cid:16)3 5 (cid:184)(cid:168) (cid:16)6 (cid:184)
(cid:169) (cid:185) (cid:169) 12 4 6 (cid:185)(cid:169) (cid:185)
(cid:167)(cid:16)6(cid:183)
(cid:168) (cid:184)
(cid:32) 0
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)(cid:16)6(cid:185)
One of
x(cid:32) 6i, y(cid:32)0,z(cid:32)(cid:16) 6i
e
AnswerMarks
or x(cid:32)(cid:16) 6i, y(cid:32)0,z(cid:32) 6iM1
A1
A1
c
AnswerMarks
[3]3.1a
1.1
m
3.2a
AnswerMarks
iUse of inverse matrix with
kn = 4.
BC
e
A0 if any solution given as
final answer with x(cid:32)z.
AnswerMarks Guidance
13(iii) (A)
3(cid:117)2(cid:14) 0–5(cid:117)1(cid:32)1
S
2(cid:117)2(cid:14)3(cid:117)0–3(cid:117)1(cid:32)1
AnswerMarks Guidance
–2(cid:14)2(cid:117)0(cid:14)2(cid:117)1(cid:32)0B1
[1]1.1 Convincing substitution of
the point into all three
equations
AnswerMarks Guidance
13(iii) (B)
and determinant of M is zero when k (cid:32)3
Hence no unique solution
AnswerMarks
The planes are distinct, so the planes form a sheafB1
B1
E1
E1
AnswerMarks
[4]2.4
2.4
2.1
AnswerMarks Guidance
2.2aThis may be implied
13(iv) 12(cid:11)k –3(cid:12)(cid:32)6 …
… or –6
k = 21 or k (cid:32)31
AnswerMarks
2 2M1
M1
A1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
1.1or statement relating
determinant to volume scale
factor
Question 13:
13 | (i) | detM(cid:32)k(cid:11)6(cid:14)6(cid:12)–(cid:11)4–3(cid:12)–5(cid:11)4(cid:14)3(cid:12) Simplify to
12(cid:11)k –3(cid:12) AG | M1
A1
[2] | 1.1
1.1 | Answer given so method
must be clear
13 | (ii) | (cid:167)x2(cid:183) (cid:167)1 (cid:16)1 1(cid:183)(cid:167) 6 (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)(cid:168) (cid:184)
y2 (cid:32) (cid:16) 1 1 1 6
(cid:168) (cid:184) (cid:168)1 2 4 6(cid:184)(cid:168) (cid:184)
(cid:168)z2(cid:184) (cid:168) 7 (cid:16)3 5 (cid:184)(cid:168) (cid:16)6 (cid:184)
(cid:169) (cid:185) (cid:169) 12 4 6 (cid:185)(cid:169) (cid:185)
(cid:167)(cid:16)6(cid:183)
(cid:168) (cid:184)
(cid:32) 0
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)(cid:16)6(cid:185)
One of
x(cid:32) 6i, y(cid:32)0,z(cid:32)(cid:16) 6i
e
or x(cid:32)(cid:16) 6i, y(cid:32)0,z(cid:32) 6i | M1
A1
A1
c
[3] | 3.1a
1.1
m
3.2a
i | Use of inverse matrix with
kn = 4.
BC
e
A0 if any solution given as
final answer with x(cid:32)z.
13 | (iii) | (A) | p
3(cid:117)2(cid:14) 0–5(cid:117)1(cid:32)1
S
2(cid:117)2(cid:14)3(cid:117)0–3(cid:117)1(cid:32)1
–2(cid:14)2(cid:117)0(cid:14)2(cid:117)1(cid:32)0 | B1
[1] | 1.1 | Convincing substitution of
the point into all three
equations
13 | (iii) | (B) | The coefficients are the matrix M with k (cid:32)3
and determinant of M is zero when k (cid:32)3
Hence no unique solution
The planes are distinct, so the planes form a sheaf | B1
B1
E1
E1
[4] | 2.4
2.4
2.1
2.2a | This may be implied
13 | (iv) | 12(cid:11)k –3(cid:12)(cid:32)6 …
… or –6
k = 21 or k (cid:32)31
2 2 | M1
M1
A1
[3] | 1.1
3.1a
1.1 | or statement relating
determinant to volume scale
factor
Matrix M is given by $\mathbf{M} = \begin{pmatrix} k & 1 & -5 \\ 2 & 3 & -3 \\ -1 & 2 & 2 \end{pmatrix}$, where $k$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Show that $\det \mathbf{M} = 12(k - 3)$. [2]

\item Find a solution of the following simultaneous equations for which $x \neq z$.
$$4x^2 + y^2 - 5z^2 = 6$$
$$2x^2 + 3y^2 - 3z^2 = 6$$
$$-x^2 + 2y^2 + 2z^2 = -6$$ [3]

\item \begin{enumerate}[label=(\Alph*)]
\item Verify that the point $(2, 0, 1)$ lies on each of the following three planes.
$$3x + y - 5z = 1$$
$$2x + 3y - 3z = 1$$
$$-x + 2y + 2z = 0$$ [1]

\item Describe how the three planes in part (iii) (A) are arranged in 3-D space. Give reasons for your answer. [4]
\end{enumerate}

\item Find the values of $k$ for which the transformation represented by M has a volume scale factor of 6. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q13 [13]}}
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