| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Challenging +1.2 This is a Further Maths question on linear algebra requiring understanding of linear dependence and matrix operations. The proof in the first part is straightforward (if αx₁ + βx₂ + γx₃ = 0, then M(αx₁ + βx₂ + γx₃) = 0 by linearity). Part (ii) requires finding a linear combination showing dependence (routine calculation), and part (iii) involves computing three matrix-vector products and identifying which are linearly independent. While this requires multiple steps and FM-level concepts, the techniques are standard and the upper triangular matrix P makes calculations manageable. Slightly above average difficulty due to the abstract proof and multi-part nature, but not requiring deep insight. |
| Spec | 4.03a Matrix language: terminology and notation |
Three $n \times 1$ column vectors are denoted by $\mathbf{x}_1$, $\mathbf{x}_2$, $\mathbf{x}_3$, and $\mathbf{M}$ is an $n \times n$ matrix. Show that if $\mathbf{x}_1$, $\mathbf{x}_2$, $\mathbf{x}_3$ are linearly dependent then the vectors $\mathbf{Mx}_1$, $\mathbf{Mx}_2$, $\mathbf{Mx}_3$ are also linearly dependent. [2]
The vectors $\mathbf{y}_1$, $\mathbf{y}_2$, $\mathbf{y}_3$ and the matrix $\mathbf{P}$ are defined as follows:
$$\mathbf{y}_1 = \begin{pmatrix} 1 \\ 5 \\ 7 \end{pmatrix}, \quad \mathbf{y}_2 = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}, \quad \mathbf{y}_3 = \begin{pmatrix} 5 \\ 51 \\ 55 \end{pmatrix},$$
$$\mathbf{P} = \begin{pmatrix} 1 & -4 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & -7 \end{pmatrix}$$
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathbf{y}_1$, $\mathbf{y}_2$, $\mathbf{y}_3$ are linearly dependent. [2]
\item Find a basis for the linear space spanned by the vectors $\mathbf{Py}_1$, $\mathbf{Py}_2$, $\mathbf{Py}_3$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2003 Q3 [6]}}