| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix multiplication |
| Difficulty | Standard +0.3 This is a straightforward Further Maths matrices question requiring routine matrix multiplication (2×3 by 3×2 giving 2×2), then finding a 2×2 determinant, and verifying singularity by substitution. All steps are mechanical with no problem-solving insight needed, making it slightly easier than average even for Further Maths content. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks |
|---|---|
| 4(a) | Obtains one correct element in terms of . |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑎𝑎 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept unsimplified. ISW | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 4(b) | Obtains the correct determinant. |
| Answer | Marks | Guidance |
|---|---|---|
| × | 1.1b | B1F |
| Answer | Marks |
|---|---|
| 4(c) | 𝑎𝑎 |
| Answer | Marks | Guidance |
|---|---|---|
| substitutes into their expression for the determinant. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| singular, clearly referring to singular determinant = 0. | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 5 | |
| Q | Marking instructions | AO |
Question 4:
--- 4(a) ---
4(a) | Obtains one correct element in terms of .
Must be a 2 2 matrix.
𝑎𝑎 | 1.1a | M1 | 1 −3
2 𝑎𝑎 3
� � �−2 4𝑎𝑎�
0 −2 1
0 5
2 2
2−2𝑎𝑎+0 −6+4𝑎𝑎 + 15 2−2𝑎𝑎 4𝑎𝑎 +9
×
Obtains the correct product.
Accept unsimplified. ISW | 1.1b | A1
--- 4(b) ---
4(b) | Obtains the correct determinant.
Accept unsimplified. ISW
Follow through their 2 2 matrix with at least one element in
terms of .
× | 1.1b | B1F | = � �= � �
0+4+0 0−8𝑎𝑎+5 4 5−8𝑎𝑎
2
(2−2𝑎𝑎)(5−8𝑎𝑎)−4 ( 4𝑎𝑎 + 9)
2 2
=10−16𝑎𝑎−10𝑎𝑎+ 16𝑎𝑎 −16𝑎𝑎 −36
--- 4(c) ---
4(c) | 𝑎𝑎
Selects a method to show that AB is singular.
e.g.
equates their expression for the determinant to zero
or
substitutes into their expression for the determinant. | 1.1a | M1 | = −26−26𝑎𝑎
AB is singular when AB 0
det =
−26−26𝑎𝑎 = 0
−26𝑎𝑎 = 26
𝑎𝑎 = −1
𝑎𝑎 = −1
Completes a fully correct reasoned argument to show that AB is
singular, clearly referring to singular determinant = 0. | 2.1 | R1
⇔
Total | 5
Q | Marking instructions | AO | Marks | Typical solutionThe matrices $\mathbf{A}$ and $\mathbf{B}$ are such that
$$\mathbf{A} = \begin{bmatrix} 2 & a & 3 \\ 0 & -2 & 1 \end{bmatrix} \quad \text{and} \quad \mathbf{B} = \begin{bmatrix} 1 & -3 \\ -2 & 4a \\ 0 & 5 \end{bmatrix}$$
\begin{enumerate}[label=(\alph*)]
\item Find the product $\mathbf{AB}$ in terms of $a$.
[2 marks]
\item Find the determinant of $\mathbf{AB}$ in terms of $a$.
[1 mark]
\item Show that $\mathbf{AB}$ is singular when $a = -1$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q4 [5]}}