| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix equation solving (AB = C) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on matrix inversion using standard techniques (cofactor method or row reduction). Part (a) requires systematic calculation of the inverse with a parameter, which is routine for F3 students. Part (b) is immediate once the inverse is found: N = M^{-1} × (given matrix). The 9 marks reflect computational length rather than conceptual difficulty—this is a standard textbook exercise testing technique rather than problem-solving. |
| Spec | 4.03a Matrix language: terminology and notation4.03o Inverse 3x3 matrix |
| Answer | Marks |
|---|---|
| 4(a) | (cid:167) 1 k 0(cid:183) |
| Answer | Marks |
|---|---|
| M (cid:32)3(cid:16)k(cid:16)k(cid:11)(cid:16)3(cid:16)1(cid:12)(cid:11)(cid:32)3k(cid:14)3(cid:12) | Correct determinant in any |
| form | B1 |
| Answer | Marks |
|---|---|
| (cid:169) (cid:185) | B1 |
| Answer | Marks |
|---|---|
| (cid:169) (cid:185) | M1: Identifiable full attempt |
| Answer | Marks |
|---|---|
| least 6 correct elements. | M1 |
| Answer | Marks |
|---|---|
| (b) | (cid:167)3 5 6(cid:183) (cid:167)3 5 6(cid:183) |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:169) (cid:185) (cid:169) (cid:185) | Correct statement | B1 |
| Answer | Marks |
|---|---|
| (cid:169) (cid:185)(cid:169) (cid:185) (cid:169) (cid:185) | M1: Multiplies the given |
| Answer | Marks |
|---|---|
| 3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 4:
--- 4(a) ---
4(a) | (cid:167) 1 k 0(cid:183)
(cid:168) (cid:184)
M(cid:32) (cid:16)1 1 1
(cid:168) (cid:184)
(cid:168) (cid:184)
1 k 3
(cid:169) (cid:185)
M (cid:32)3(cid:16)k(cid:16)k(cid:11)(cid:16)3(cid:16)1(cid:12)(cid:11)(cid:32)3k(cid:14)3(cid:12) | Correct determinant in any
form | B1
(cid:167)1 (cid:16)1 1(cid:183) (cid:167)3(cid:16)k (cid:16)4 (cid:16)k(cid:16)1(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
MT (cid:32) k 1 k or minors 3k 3 0
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184)
0 1 3 k 1 1(cid:14)k
(cid:169) (cid:185) (cid:169) (cid:185)
(cid:167)3(cid:16)k 4 (cid:16)k(cid:16)1(cid:183)
(cid:168) (cid:184)
or cofactors (cid:16)3k 3 0
(cid:168) (cid:184)
(cid:168) (cid:184)
k (cid:16)1 1(cid:14)k
(cid:169) (cid:185) | B1
(cid:167) 3(cid:16)k (cid:16)3k k (cid:183)
1 (cid:168) (cid:184)
M(cid:16)1 (cid:32) 4 3 (cid:16)1
(cid:168) (cid:184)
3(cid:14)3k
(cid:168) (cid:184)
(cid:16)k(cid:16)1 0 1(cid:14)k
(cid:169) (cid:185) | M1: Identifiable full attempt
at inverse including
reciprocal of determinant.
Could be indicated by at
least 6 correct elements. | M1
A1ft
A1ft
A1ft: Two rows or two
columns correct (follow
through their determinant
but not incorrect entries in
the matrices used)
A1ft: Fully correct inverse
(follow through as before)
NB: If every element is the negative of the correct element, allow M1A1A0
(5)
(b) | (cid:167)3 5 6(cid:183) (cid:167)3 5 6(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
MN(cid:32) 4 (cid:16)1 1 (cid:159)N(cid:32)M(cid:16)1 4 (cid:16)1 1
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184)
3 2 (cid:16)3 3 2 (cid:16)3
(cid:169) (cid:185) (cid:169) (cid:185) | Correct statement | B1
(cid:167) 3 0 0 (cid:183)(cid:167)3 5 6(cid:183) (cid:167)3 5 6 (cid:183)
1(cid:168) (cid:184)(cid:168) (cid:184) (cid:168) (cid:184)
N(cid:32) 4 3 (cid:16)1 4 (cid:16)1 1 (cid:32) 7 5 10
(cid:168) (cid:184)(cid:168) (cid:184) (cid:168) (cid:184)
3
(cid:168) (cid:184)(cid:168) (cid:184) (cid:168) (cid:184)
(cid:16)1 0 1 3 2 (cid:16)3 0 (cid:16)1 (cid:16)3
(cid:169) (cid:185)(cid:169) (cid:185) (cid:169) (cid:185) | M1: Multiplies the given
matrix by their M(cid:16)1in the
correct order Must include
1
the " "
3 | M1
A(2, 1,
0)
A2: Correct matrix
((cid:16)1 each error). If left with
1
outside the matrix
3
award A0
(4)
(9 marks)
Question | Scheme | Marks$$\mathbf{M} = \begin{pmatrix} 1 & k & 0 \\ -1 & 1 & 1 \\ 1 & k & 3 \end{pmatrix}, \text{ where } k \text{ is a constant}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf{M}^{-1}$ in terms of $k$.
[5]
\end{enumerate}
Hence, given that $k = 0$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the matrix $\mathbf{N}$ such that
$$\mathbf{MN} = \begin{pmatrix} 3 & 5 & 6 \\ 4 & -1 & 1 \\ 3 & 2 & -3 \end{pmatrix}$$
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q4 [9]}}