CAIE FP1 2013 November — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeRank and null space basis
DifficultyChallenging +1.2 This is a structured Further Maths question on linear algebra requiring row reduction to find rank and null space basis, followed by verification and application of the rank-nullity theorem. While it involves multiple steps and FM content, the techniques are algorithmic (Gaussian elimination) with clear guidance on what to show, making it moderately above average difficulty but not requiring deep conceptual insight.
Spec4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix
6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { r r r r }
M1 Reduces to echelon form.
A1 Obtains rank \(r(M) = 4 - 2 = 2\)
A1 Basis for null space: \(\begin{pmatrix} -5 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 7 \\ 3 \\ 0 \\ 1 \end{pmatrix}\) (OE)
M1 Evaluates matrix product \(M = \begin{pmatrix} -2 & 16 \\ -3 & 10 \\ -4 & 22 \end{pmatrix}\)
A1 Finds general solution of equations: \(x - 3y - z + 2t = 0\) and \(y + 2z - 3t = 0\)
A1 Obtains \(t = \mu, z = \lambda, y = 3\mu - 2\lambda, x = 7\mu - 5\lambda\)
B1 Expresses particular solution \(x = \begin{pmatrix} 1 \\ -2 \\ -3 \\ -4 \end{pmatrix} \in K\)
M1 Writes general solution in form \(x = \begin{pmatrix} 1 \\ -2 \\ -3 \\ -4 \end{pmatrix} + \lambda e_1 + \mu e_2\)
A1 (AG) Completes solution correctly.
M1 Reduces to echelon form.

A1 Obtains rank $r(M) = 4 - 2 = 2$

A1 Basis for null space: $\begin{pmatrix} -5 \\ -2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 7 \\ 3 \\ 0 \\ 1 \end{pmatrix}$ (OE)

M1 Evaluates matrix product $M = \begin{pmatrix} -2 & 16 \\ -3 & 10 \\ -4 & 22 \end{pmatrix}$

A1 Finds general solution of equations: $x - 3y - z + 2t = 0$ and $y + 2z - 3t = 0$

A1 Obtains $t = \mu, z = \lambda, y = 3\mu - 2\lambda, x = 7\mu - 5\lambda$

B1 Expresses particular solution $x = \begin{pmatrix} 1 \\ -2 \\ -3 \\ -4 \end{pmatrix} \in K$

M1 Writes general solution in form $x = \begin{pmatrix} 1 \\ -2 \\ -3 \\ -4 \end{pmatrix} + \lambda e_1 + \mu e_2$

A1 (AG) Completes solution correctly.
6 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where

$$\mathbf { M } = \left( \begin{array} { r r r r }

\hfill \mbox{\textit{CAIE FP1 2013 Q6 [9]}}
This paper (8 questions)
View full paper
Q6 9 Q7 Q9 Q10 12 Q11 Q16 EITHER Q16 OR Q22