| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix multiplication |
| Difficulty | Moderate -0.8 This is a straightforward Further Maths question testing basic matrix operations: routine multiplication, understanding of dimension compatibility, and standard 2×2 matrix inversion formula. While it's Further Maths content, these are foundational techniques requiring only procedural knowledge with no problem-solving or insight needed. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1&2\\3&-1\\4&5\end{pmatrix}\begin{pmatrix}2&-1&4\\1&3&1\end{pmatrix} = \begin{pmatrix}4&5&6\\5&-6&11\\13&11&21\end{pmatrix}\) | M1A2 | M1: 3×3 matrix with a number or numerical expression for each element. A2: cao (-1 each error). Only 1 error award A1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{BA} = \begin{pmatrix}2&-1&4\\1&3&1\end{pmatrix}\begin{pmatrix}1&2\\3&-1\\4&5\end{pmatrix} = \begin{pmatrix}15&25\\14&4\end{pmatrix}\) | B1 | Allow any convincing argument. e.g. BA is a 2×2 matrix (so \(AB \neq BA\)) or dimensionally different. Attempt to evaluate product not required. NB 'Not commutative' only is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\det\mathbf{C} = 2k \times k - 3\times(-2)\) | M1 | Correct attempt at determinant |
| \(\mathbf{C}^{-1} = \frac{1}{2k^2+6}\begin{pmatrix}k&2\\-3&2k\end{pmatrix}\) | M1A1 | M1: \(\frac{1}{\text{their}\det\mathbf{C}}\begin{pmatrix}k&2\\-3&2k\end{pmatrix}\). A1: cao oe |
# Question 4:
$$\mathbf{A} = \begin{pmatrix}1&2\\3&-1\\4&5\end{pmatrix},\quad \mathbf{B} = \begin{pmatrix}2&-1&4\\1&3&1\end{pmatrix}$$
## Part (i)(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&2\\3&-1\\4&5\end{pmatrix}\begin{pmatrix}2&-1&4\\1&3&1\end{pmatrix} = \begin{pmatrix}4&5&6\\5&-6&11\\13&11&21\end{pmatrix}$ | M1A2 | M1: 3×3 matrix with a number or numerical expression for each element. A2: cao (-1 each error). Only 1 error award A1A0 |
## Part (i)(b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{BA} = \begin{pmatrix}2&-1&4\\1&3&1\end{pmatrix}\begin{pmatrix}1&2\\3&-1\\4&5\end{pmatrix} = \begin{pmatrix}15&25\\14&4\end{pmatrix}$ | B1 | Allow any convincing argument. e.g. BA is a 2×2 matrix (so $AB \neq BA$) or dimensionally different. Attempt to evaluate product not required. NB 'Not commutative' only is B0 |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\det\mathbf{C} = 2k \times k - 3\times(-2)$ | M1 | Correct attempt at determinant |
| $\mathbf{C}^{-1} = \frac{1}{2k^2+6}\begin{pmatrix}k&2\\-3&2k\end{pmatrix}$ | M1A1 | M1: $\frac{1}{\text{their}\det\mathbf{C}}\begin{pmatrix}k&2\\-3&2k\end{pmatrix}$. A1: cao oe |
---4. (i) Given that
$$\mathbf { A } = \left( \begin{array} { r r }
1 & 2 \\
3 & - 1 \\
4 & 5
\end{array} \right) \text { and } \mathbf { B } = \left( \begin{array} { r r r }
2 & - 1 & 4 \\
1 & 3 & 1
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item find $\mathbf { A B }$.
\item Explain why $\mathbf { A B } \neq \mathbf { B A }$.\\
(ii) Given that
$$\mathbf { C } = \left( \begin{array} { c r }
2 k & - 2 \\
3 & k
\end{array} \right) \text {, where } k \text { is a real number }$$
find $\mathbf { C } ^ { - 1 }$, giving your answer in terms of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q4 [7]}}