| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 4 |
| Topic | Matrices |
| Type | Solving linear systems using matrices |
| Difficulty | Standard +0.3 This is a straightforward FP1 matrix question requiring standard techniques: finding the inverse of a 3×3 matrix (likely given a formula or using cofactors) and solving simultaneous equations via matrix multiplication. While it involves more computation than basic A-level questions, it requires only routine application of learned methods with no problem-solving insight, making it slightly easier than average overall but typical for Further Maths. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(A^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ -1 & 0 & 3 \end{pmatrix}\) | B1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} -3 & 3 & 2 \\ 5 & -4 & -3 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}\) | [1] M1 | |
| Answer: \(\begin{pmatrix} -3 & 3 & 2 \\ 5 & -4 & -3 \\ -1 & 1 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}\) or \(\begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ -1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}\) | M1ft | |
| Answer: \(x = 12a + 1, y = 24a - 13\) and \(z = 21 - 12a\) | A1 | All three |
## (i)
Answer: $A^{-1} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ -1 & 0 & 3 \end{pmatrix}$ | B1 | BC
## (ii)
Answer: Converting to matrix form soi
$\begin{pmatrix} -3 & 3 & 2 \\ 5 & -4 & -3 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}$ | [1] M1 |
Answer: $\begin{pmatrix} -3 & 3 & 2 \\ 5 & -4 & -3 \\ -1 & 1 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}$ or $\begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ -1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 12a \\ -6 \\ 7 \end{pmatrix}$ | M1ft |
Answer: $x = 12a + 1, y = 24a - 13$ and $z = 21 - 12a$ | A1 | All three
---The matrix $\mathbf{A}$ is given by $\mathbf{A} = \begin{pmatrix} -3 & 3 & 2 \\ 5 & -4 & -3 \\ -1 & 1 & 1 \end{pmatrix}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathbf{A}^{-1}$. [1]
\item Solve the simultaneous equations
$$-3x + 3y + 2z = 12a$$
$$5x - 4y - 3z = -6$$
$$-x + y + z = 7$$
giving your solution in terms of $a$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 AS 2017 Q1 [4]}}