CAIE FP1 2018 November — Question 5 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind constant from singularity condition
DifficultyChallenging +1.2 This is a systematic linear algebra question requiring row reduction to find rank, null space basis, and solve a non-homogeneous system. While it involves 4×4 matrices (making calculations longer), the techniques are standard Further Maths fare with no conceptual surprises. The 3-mark allocations suggest routine application of algorithms rather than insight, placing it moderately above average difficulty due to computational demands and being Further Maths content.
Spec4.03s Consistent/inconsistent: systems of equations4.03t Plane intersection: geometric interpretation
The linear transformation \(\mathrm{T} : \mathbb{R}^4 \to \mathbb{R}^4\) is represented by the matrix \(\mathbf{M}\), where $$\mathbf{M} = \begin{pmatrix} 3 & 2 & 0 & 1 \\ 6 & 5 & -1 & 3 \\ 9 & 8 & -2 & 5 \\ -3 & -2 & 0 & -1 \end{pmatrix}.$$
  1. Find the rank of \(\mathbf{M}\). [3]
Let \(K\) be the null space of \(\mathrm{T}\).
  1. Find a basis for \(K\). [3]
  2. Find the general solution of $$\mathbf{M}\mathbf{x} = \begin{pmatrix} 2 \\ 5 \\ 8 \\ -2 \end{pmatrix}.$$ [3]
Question 5:
AnswerMarks
5(i) 3 2 0 1  3 2 0 1
   
6 5 −1 3 0 1 −1 1
   
→
 9 8 −2 5  0 2 −2 2
   
AnswerMarks Guidance
−3 −2 0 −1 0 0 0 0M1 Attempt to row reduction.
3 2 0 1
 
0 1 −1 1
 
→
0 0 0 0
 
AnswerMarks Guidance
0 0 0 0A1 Two correct rows only
( )=4−2=2
AnswerMarks Guidance
r MA1 Obtains rank.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(ii)3x+2y +t =0
y−z+t =0M1 Solves homogeneous system of equations.
2 1
⇒ t =µ, z=λ, y=λ−µ, x=− λ+ µ
AnswerMarks Guidance
3 3M1 Using 2 parameters
−2  1  −1  0 
       
 3 −3   0 −1 
       
A basis is  ,  or  ,  or equivalent
 3   0   3   1 
   
       
AnswerMarks Guidance
   0   3      3   2  A1 AEF
3
AnswerMarks
5(iii)0  2  0
     
1 5 1
M  =  so a particular solution is  
0  8  0
     
AnswerMarks Guidance
0 −2 0B1 Finds a particular solution.
0 −2  1 
     
1 3 −3
General solution: ( x=) +λ  +µ 
0  3   0 
     
AnswerMarks Guidance
0  0   3 M1 Using correct format
A1FT
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(i) ---
5(i) |  3 2 0 1  3 2 0 1
   
6 5 −1 3 0 1 −1 1
   
→
 9 8 −2 5  0 2 −2 2
   
−3 −2 0 −1 0 0 0 0 | M1 | Attempt to row reduction.
3 2 0 1
 
0 1 −1 1
 
→
0 0 0 0
 
0 0 0 0 | A1 | Two correct rows only
( )=4−2=2
r M | A1 | Obtains rank.
3
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | 3x+2y +t =0
y−z+t =0 | M1 | Solves homogeneous system of equations.
2 1
⇒ t =µ, z=λ, y=λ−µ, x=− λ+ µ
3 3 | M1 | Using 2 parameters
−2  1  −1  0 
       
 3 −3   0 −1 
       
A basis is  ,  or  ,  or equivalent
 3   0   3   1 
   
       
   0   3      3   2   | A1 | AEF
3
--- 5(iii) ---
5(iii) | 0  2  0
     
1 5 1
M  =  so a particular solution is  
0  8  0
     
0 −2 0 | B1 | Finds a particular solution.
0 −2  1 
     
1 3 −3
General solution: ( x=) +λ  +µ 
0  3   0 
     
0  0   3  | M1 | Using correct format
A1FT
3
Question | Answer | Marks | Guidance
The linear transformation $\mathrm{T} : \mathbb{R}^4 \to \mathbb{R}^4$ is represented by the matrix $\mathbf{M}$, where
$$\mathbf{M} = \begin{pmatrix} 3 & 2 & 0 & 1 \\ 6 & 5 & -1 & 3 \\ 9 & 8 & -2 & 5 \\ -3 & -2 & 0 & -1 \end{pmatrix}.$$

\begin{enumerate}[label=(\roman*)]
\item Find the rank of $\mathbf{M}$. [3]
\end{enumerate}

Let $K$ be the null space of $\mathrm{T}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find a basis for $K$. [3]

\item Find the general solution of
$$\mathbf{M}\mathbf{x} = \begin{pmatrix} 2 \\ 5 \\ 8 \\ -2 \end{pmatrix}.$$ [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q5 [9]}}
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