CAIE FP1 2018 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind eigenvectors given eigenvalue
DifficultyModerate -0.3 This is a straightforward Further Maths question testing basic 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 standard textbook exercises with no conceptual difficulty or novel insight required, though the calculations are slightly more involved than typical A-level questions due to the Further Maths context.
Spec4.04b Plane equations: cartesian and vector forms

The vectors \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\) and \(\mathbf{d}\) in \(\mathbb{R}^3\) are given by $$\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \\ 9 \\ 0 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 3 \\ 5 \\ 4 \end{pmatrix} \quad \text{and} \quad \mathbf{d} = \begin{pmatrix} 0 \\ -8 \\ -3 \end{pmatrix}.$$
  1. Show that \(\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}\) is a basis for \(\mathbb{R}^3\). [3]
  2. Express \(\mathbf{d}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\). [2]

Question 1:

AnswerMarks
1(i)EITHER:
1 2 3
2 9 3 =36−10−3×9=−1≠0
AnswerMarks Guidance
1 0 4M1 A1 Calculates determinant
OR:
α +2β+3γ =0
2α+9β+3γ =0 ⇒2β−γ=0
α +4γ=0
9β−5γ=0
AnswerMarks Guidance
⇒β=0⇒α=γ=0(M1A1) Solves homogeneous system of equations.
Eliminates one variable.
Therefore a,b,c are linearly independent (and span R3) so form a basis for
AnswerMarks Guidance
R3.A1 States either that vectors are linearly independent or that
vectors span R3.
3

AnswerMarks
1(ii)1 2 3  0 
       
l 2 +m 9 +n 3 = −8
       
       
1 0 4  3 
AnswerMarks Guidance
⇒l=m=−1 and n=1.M1 Sets up system of equations.
⇒d=c−b−aA1
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
--- 1(i) ---
1(i) | EITHER:
1 2 3
2 9 3 =36−10−3×9=−1≠0
1 0 4 | M1 A1 | Calculates determinant
OR:
α +2β+3γ =0
2α+9β+3γ =0 ⇒2β−γ=0
α +4γ=0
9β−5γ=0
⇒β=0⇒α=γ=0 | (M1A1) | Solves homogeneous system of equations.
Eliminates one variable.
Therefore a,b,c are linearly independent (and span R3) so form a basis for
R3. | A1 | States either that vectors are linearly independent or that
vectors span R3.
3
--- 1(ii) ---
1(ii) | 1 2 3  0 
       
l 2 +m 9 +n 3 = −8
       
       
1 0 4  3 
⇒l=m=−1 and n=1. | M1 | Sets up system of equations.
⇒d=c−b−a | A1
2
Question | Answer | Marks | Guidance
The vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ and $\mathbf{d}$ in $\mathbb{R}^3$ are given by

$$\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \\ 9 \\ 0 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 3 \\ 5 \\ 4 \end{pmatrix} \quad \text{and} \quad \mathbf{d} = \begin{pmatrix} 0 \\ -8 \\ -3 \end{pmatrix}.$$

\begin{enumerate}[label=(\roman*)]
\item Show that $\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$ is a basis for $\mathbb{R}^3$. [3]

\item Express $\mathbf{d}$ in terms of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q1 [5]}}