| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Linear independence and spanning |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question testing standard linear algebra concepts. Part (i) requires computing a 3×3 determinant to verify linear independence (routine calculation), and part (ii) involves solving a 3×3 system of linear equations. Both are textbook exercises with clear methods and no conceptual subtlety, making it slightly easier than average even for Further Maths. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(\begin{vmatrix} 1 & 2 & 3 \\ 2 & 9 & 3 \\ 1 & 0 & 4 \end{vmatrix} = 36 - 10 - 3 \times 9 = -1 \neq 0\) | M1 A1 | Calculates determinant |
| OR: \(\alpha + 2\beta + 3\gamma = 0\), \(2\alpha + 9\beta + 3\gamma = 0 \Rightarrow 2\beta - \gamma = 0\), \(\alpha + 4\gamma = 0\), \(9\beta - 5\gamma = 0\), \(\Rightarrow \beta = 0 \Rightarrow \alpha = \gamma = 0\) | (M1A1) | Solves homogeneous system; eliminates one variable |
| Therefore a, b, c are linearly independent (and span \(\mathbb{R}^3\)) so form a basis for \(\mathbb{R}^3\) | A1 | States either that vectors are linearly independent or that vectors span \(\mathbb{R}^3\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(l\begin{pmatrix}1\\2\\1\end{pmatrix} + m\begin{pmatrix}2\\9\\0\end{pmatrix} + n\begin{pmatrix}3\\3\\4\end{pmatrix} = \begin{pmatrix}0\\-8\\3\end{pmatrix}\), \(\Rightarrow l = m = -1\) and \(n = 1\) | M1 | Sets up system of equations |
| \(\Rightarrow \mathbf{d} = \mathbf{c} - \mathbf{b} - \mathbf{a}\) | A1 | |
| Total: 2 |
## Question 1:
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| EITHER: $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 9 & 3 \\ 1 & 0 & 4 \end{vmatrix} = 36 - 10 - 3 \times 9 = -1 \neq 0$ | M1 A1 | Calculates determinant |
| OR: $\alpha + 2\beta + 3\gamma = 0$, $2\alpha + 9\beta + 3\gamma = 0 \Rightarrow 2\beta - \gamma = 0$, $\alpha + 4\gamma = 0$, $9\beta - 5\gamma = 0$, $\Rightarrow \beta = 0 \Rightarrow \alpha = \gamma = 0$ | (M1A1) | Solves homogeneous system; eliminates one variable |
| Therefore **a**, **b**, **c** are linearly independent (and span $\mathbb{R}^3$) so form a basis for $\mathbb{R}^3$ | A1 | States either that vectors are linearly independent or that vectors span $\mathbb{R}^3$ |
| **Total: 3** | | |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $l\begin{pmatrix}1\\2\\1\end{pmatrix} + m\begin{pmatrix}2\\9\\0\end{pmatrix} + n\begin{pmatrix}3\\3\\4\end{pmatrix} = \begin{pmatrix}0\\-8\\3\end{pmatrix}$, $\Rightarrow l = m = -1$ and $n = 1$ | M1 | Sets up system of equations |
| $\Rightarrow \mathbf{d} = \mathbf{c} - \mathbf{b} - \mathbf{a}$ | A1 | |
| **Total: 2** | | |
---1 The vectors $\mathbf { a } , \mathbf { b } , \mathbf { c }$ and $\mathbf { d }$ in $\mathbb { R } ^ { 3 }$ are given by
$$\mathbf { a } = \left( \begin{array} { l }
1 \\
2 \\
1
\end{array} \right) , \quad \mathbf { b } = \left( \begin{array} { l }
2 \\
9 \\
0
\end{array} \right) , \quad \mathbf { c } = \left( \begin{array} { l }
3 \\
3 \\
4
\end{array} \right) \quad \text { and } \quad \mathbf { d } = \left( \begin{array} { r }
0 \\
- 8 \\
3
\end{array} \right) .$$
(i) Show that $\{ \mathbf { a } , \mathbf { b } , \mathbf { c } \}$ is a basis for $\mathbb { R } ^ { 3 }$.\\
(ii) Express $\mathbf { d }$ in terms of $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q1}}