| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | December |
| Marks | 6 |
| Topic | 3x3 Matrices |
| Type | Inverse given/derived then solve system |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard matrix operations: multiply two given 3×3 matrices, recognize the result as the identity to deduce the inverse, then apply it to solve a system. All steps are routine procedures with no conceptual challenges—students are even given matrix B explicitly rather than computing the inverse themselves. While 3×3 matrices involve more arithmetic than 2×2, the question requires only mechanical application of learned techniques. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}\) | B1 | |
| (b) \(\frac{1}{4}\begin{pmatrix} 1 & 0 & 1 \\ -8 & 4 & 0 \\ 19 & -8 & -1 \end{pmatrix}\) oe | B1 | |
| (c)(i) E.g. Because expressed in matrix form the system of equations is \(AX = C\) where \(A\) is the matrix is \(A\) from above; E.g. And it has been established that an inverse exists. | B1, B1 | |
| (c)(ii) \(\begin{pmatrix} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 3 & -2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}\); \(\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 1 & 0 & 1 \\ -8 & 4 & 0 \\ 19 & -8 & -1 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}\); \(\Rightarrow x=1, y=1, z=-3\) | M1, A1 | Use B and multiply; BC; AG |
**(a)** $\begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}$ | B1 |
**(b)** $\frac{1}{4}\begin{pmatrix} 1 & 0 & 1 \\ -8 & 4 & 0 \\ 19 & -8 & -1 \end{pmatrix}$ oe | B1 |
**(c)(i)** E.g. Because expressed in matrix form the system of equations is $AX = C$ where $A$ is the matrix is $A$ from above; E.g. And it has been established that an inverse exists. | B1, B1 |
**(c)(ii)** $\begin{pmatrix} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 3 & -2 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$; $\Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 1 & 0 & 1 \\ -8 & 4 & 0 \\ 19 & -8 & -1 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -3 \end{pmatrix}$; $\Rightarrow x=1, y=1, z=-3$ | M1, A1 | Use B and multiply; BC; AG
---5 You are given that $\mathbf { A } = \left( \begin{array} { c c c } 1 & 2 & 1 \\ 2 & 5 & 2 \\ 3 & - 2 & - 1 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { c c c } 1 & 0 & 1 \\ - 8 & 4 & 0 \\ 19 & - 8 & - 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { A B }$.
\item Hence write down $\mathbf { A } ^ { - 1 }$.
\item You are given three simultaneous equations
$$\begin{array} { r }
x + 2 y + z = 0 \\
2 x + 5 y + 2 z = 1 \\
3 x - 2 y - z = 4
\end{array}$$
\begin{enumerate}[label=(\roman*)]
\item Explain how you can tell, without solving them, that there is a unique solution to these equations.
\item Find this unique solution.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q5 [6]}}