| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | General solution with parameters |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to find when a system has no unique solution (determinant = 0), then solve the resulting dependent system parametrically. It involves matrix theory, determinant calculation with a parameter, and expressing an infinite solution set in vector form—more sophisticated than standard A-level but routine for FP1. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 1 & 3 & k \\ 2 & -1 & -5 \\ 1 & 1 & 2 \end{vmatrix} = 0\) | M1M1 | Using the determinant; Alt method: Uses row operations |
| \(\Rightarrow k = 8\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x + 3y + 8z = 4 \quad (1)\) | M1 | |
| \(4x - 2y - 10z = -5 \quad (2)\) | ||
| From (1) and (2) obtain \(y = -3x\) or other correct expression; \(x + y + 2z = 1 \quad (3)\); Substitute \(x = t\) and \(y = -3t\) in (3) to obtain \(z\) | A1 FT | Alternative: Find REF for augmented matrix and form equations (M1) Correctly (A1) |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \frac{1}{2} \end{pmatrix} + t\begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}\) or \(\begin{pmatrix} \frac{1}{2} \\ \frac{3}{2} \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}\) (OE) | A1 | |
| Total: 3 |
## Question 4:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 1 & 3 & k \\ 2 & -1 & -5 \\ 1 & 1 & 2 \end{vmatrix} = 0$ | M1M1 | Using the determinant; Alt method: Uses row operations |
| $\Rightarrow k = 8$ | A1 | |
| **Total: 3** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + 3y + 8z = 4 \quad (1)$ | M1 | |
| $4x - 2y - 10z = -5 \quad (2)$ | | |
| From (1) and (2) obtain $y = -3x$ or other correct expression; $x + y + 2z = 1 \quad (3)$; Substitute $x = t$ and $y = -3t$ in (3) to obtain $z$ | A1 FT | Alternative: Find REF for augmented matrix and form equations (M1) Correctly (A1) |
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \frac{1}{2} \end{pmatrix} + t\begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} \frac{1}{2} \\ \frac{3}{2} \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -3 \\ 1 \end{pmatrix}$ (OE) | A1 | |
| **Total: 3** | | |
---4 (i) Find the value of $k$ for which the set of linear equations
$$\begin{aligned}
x + 3 y + k z & = 4 \\
4 x - 2 y - 10 z & = - 5 \\
x + y + 2 z & = 1
\end{aligned}$$
has no unique solution.\\
(ii) For this value of $k$, find the set of possible solutions, giving your answer in the form
$$\left( \begin{array} { c }
x \\
y \\
z
\end{array} \right) = \mathbf { a } + t \mathbf { b } ,$$
where $\mathbf { a }$ and $\mathbf { b }$ are vectors and $t$ is a scalar.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q4 [6]}}