| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Rank and null space basis |
| Difficulty | Challenging +1.2 This is a systematic linear algebra question requiring row reduction to find rank, then identifying pivot columns for range space basis and solving the homogeneous system for null space basis. While it involves 4×4 matrices and multiple steps, the procedures are algorithmic and standard for Further Maths students. The computational work is moderate but straightforward, placing it somewhat above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Row reduce \(\mathbf{M}\) to echelon form leading to two zero rows | M1A1 | |
| \(R(\mathbf{M}) = 4 - 2 = 2\) | A1 | |
| Basis for range space: any two independent column vectors from matrix \(\mathbf{M}\) | A1 | |
| Equations: \(2x - y + z + 3t = 0\) and \(y - z + 2t = 0\) | M1 | |
| \(t = \lambda,\ z = \mu,\ y = \mu - 2\lambda,\ x = -\frac{5}{2}\lambda\) | M1 | OE |
| Basis for null space is \(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}\) | A1A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Row reduce \(\mathbf{M}^T\) to echelon form | M1A1 | |
| \(R(\mathbf{M}) = 2\), basis for range space: any two independent column vectors | A1A1 | |
| Apply same row operations to \(\begin{pmatrix}a\\b\\c\\d\end{pmatrix}\) | M1 | |
| \(\rightarrow \begin{pmatrix}a \\ a+2b \\ b+c \\ -5a-4b+2d\end{pmatrix}\) | ||
| Basis for null space is \(\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}\) | M1, A1A1 | OE |
## Question 6:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Row reduce $\mathbf{M}$ to echelon form leading to two zero rows | M1A1 | |
| $R(\mathbf{M}) = 4 - 2 = 2$ | A1 | |
| Basis for range space: any two independent column vectors from matrix $\mathbf{M}$ | A1 | |
| Equations: $2x - y + z + 3t = 0$ and $y - z + 2t = 0$ | M1 | |
| $t = \lambda,\ z = \mu,\ y = \mu - 2\lambda,\ x = -\frac{5}{2}\lambda$ | M1 | OE |
| Basis for null space is $\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}$ | A1A1 | OE |
**Alternative (Transpose Matrix):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Row reduce $\mathbf{M}^T$ to echelon form | M1A1 | |
| $R(\mathbf{M}) = 2$, basis for range space: any two independent column vectors | A1A1 | |
| Apply same row operations to $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$ | M1 | |
| $\rightarrow \begin{pmatrix}a \\ a+2b \\ b+c \\ -5a-4b+2d\end{pmatrix}$ | | |
| Basis for null space is $\left\{\begin{pmatrix}-5\\-4\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}\right\}$ | M1, A1A1 | OE |
**Total: [4]+[4] = [8]**
---6 The linear transformation $\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }$ is represented by the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { r r r r }
2 & - 1 & 1 & 3 \\
2 & 0 & 0 & 5 \\
6 & - 2 & 2 & 11 \\
10 & - 3 & 3 & 19
\end{array} \right)$$
(i) Find the rank of $\mathbf { M }$ and state a basis for the range space of T .\\
(ii) Obtain a basis for the null space of T .
\hfill \mbox{\textit{CAIE FP1 2014 Q6 [8]}}