| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Parameter values for unique solution |
| Difficulty | Challenging +1.2 This is a standard Further Maths linear algebra question requiring determinant calculation to show non-singularity and matrix inversion/row reduction to find the intersection point. While it involves 3×3 matrices and parameter k (making it harder than basic A-level), the techniques are routine for Further Maths students with no novel insight required. The multi-part structure and algebraic manipulation with parameters elevate it slightly above average difficulty. |
| Spec | 4.03j Determinant 3x3: calculation4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | (a) | 𝑘 0 −1 |
| Answer | Marks |
|---|---|
| planes always meet at a point | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | considering correct determinant |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | (b) | 3𝑘−4 −2 𝑘 |
| Answer | Marks |
|---|---|
| 5𝑘(cid:2870)−4𝑘+2 5𝑘(cid:2870)−4𝑘+2 5𝑘(cid:2870)−4𝑘+2 | M1 |
| Answer | Marks |
|---|---|
| [8] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1, 1.1 | at least 5 cofactors correct (need not be in matrix) |
Question 14:
14 | (a) | 𝑘 0 −1
(cid:3629)−1 𝑘 2 (cid:3629)= 𝑘(3𝑘−4)−1(−2−2𝑘(cid:2870))
2𝑘 2 3
= 5k2 – 4k +2
discriminant = 16 – 40 = – 24 < 0
so determinant is never zero
planes always meet at a point | M1
M1
A1
M1
A1
[5] | 1.1
1.1
1.1
3.1a
2.2a | considering correct determinant
finding determinant using any row or column
oe, e.g. completing the square, considering quadratic formula
or showing no real roots
www. Both statements required.
14 | (b) | 3𝑘−4 −2 𝑘
1
𝐌(cid:2879)(cid:2869) = (cid:3437) 4𝑘+3 5𝑘 −2𝑘+1(cid:3441)
5𝑘(cid:2870)−4𝑘+2
−2−2𝑘(cid:2870) −2𝑘 𝑘(cid:2870)
2 6𝑘−10
1
𝐌(cid:2879)(cid:2869)(cid:3437)1(cid:3441)= (cid:3437) 13𝑘+6 (cid:3441)
5𝑘(cid:2870)−4𝑘+2
0 −4𝑘(cid:2870)−2𝑘−4
6𝑘−10 13𝑘+6 −4𝑘(cid:2870)−2𝑘−4
(cid:4678) , , (cid:4679)
5𝑘(cid:2870)−4𝑘+2 5𝑘(cid:2870)−4𝑘+2 5𝑘(cid:2870)−4𝑘+2 | M1
A1
M1
M1
A1
M1
A2,1,0
[8] | 1.1
1.1
1.1
3.1a
1.1
1.1
1.1, 1.1 | at least 5 cofactors correct (need not be in matrix)
all cofactors correct
cofactor matrix transposed
(cid:2869)
Multiplying by
(cid:2930)(cid:2918)(cid:2915)(cid:2919)(cid:2928) (cid:2914)(cid:2915)(cid:2930)(cid:2915)(cid:2928)(cid:2923)(cid:2919)(cid:2924)(cid:2911)(cid:2924)(cid:2930)
correct inverse
2
finding 𝐌(cid:2879)(cid:2869)(cid:3437)1(cid:3441)
0
Award A1 for one of x, y or z coordinates correct or all three
correct FT their determinant14 Three planes have equations
$$\begin{aligned}
k x - z & = 2 \\
- x + k y + 2 z & = 1 \\
2 k x + 2 y + 3 z & = 0
\end{aligned}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item By considering a suitable determinant, show that the three planes meet at a point for all values of $k$.
\item Using a matrix method, find, in terms of $k$, the coordinates of the point of intersection of the planes.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q14 [13]}}