| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | December |
| Marks | 9 |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing basic matrix operations and eigenvalue concepts. Part (a) requires routine matrix multiplication to verify associativity, part (b) is a simple demonstration that AC ≠ CA, and part (c) involves finding an eigenvector relationship which reduces to solving simple simultaneous equations. All parts are mechanical with no problem-solving insight required, making it easier than average even for Further Maths. |
| Spec | 4.03c Matrix multiplication: properties (associative, not commutative) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(AB = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 1 \\ 2a - 4 & -a - 1 \end{pmatrix}\) | M1 | Finding AB (or BC) |
| \((AB)C = \begin{pmatrix} 10 & 1 \\ 2a - 4 & -a - 1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 48 & 2 \\ 12a - 18 & -2a - 2 \end{pmatrix}\) | A1 | Finding (AB)C (or A(BC)) |
| \(BC = \begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 12 & -2 \\ 18 & 2 \end{pmatrix}\) | M1 | Finding BC (or AB) |
| \(A(BC) = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 12 & -2 \\ 18 & 2 \end{pmatrix} = \begin{pmatrix} 48 & 2 \\ 12a - 18 & -2a - 2 \end{pmatrix} = (AB)C\) (demonstrates associativity of matrix multiplication) | A1 | Correct final matrix and statement of equality |
| (b) \(AC = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 5a + 2 & -2 \end{pmatrix}\) | M1 | Finding AC (or CA) |
| \(CA = \begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 2a - 2 & -6 \end{pmatrix} \neq AC\) (so matrix multiplication is not commutative) | A1 | Finding the other and statement of non-equality |
| (c) \(\begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ ax - y \end{pmatrix}\) | M1 | Multiplying the vector into the matrix using the correct procedure |
| \(x + 2y = 3x \Rightarrow y = x\) | A1 | |
| \(ax - y = 3y\) and \(y = x \Rightarrow a = 4\) | A1 |
**(a)** $AB = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 1 \\ 2a - 4 & -a - 1 \end{pmatrix}$ | M1 | Finding AB (or BC)
$(AB)C = \begin{pmatrix} 10 & 1 \\ 2a - 4 & -a - 1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 48 & 2 \\ 12a - 18 & -2a - 2 \end{pmatrix}$ | A1 | Finding (AB)C (or A(BC))
$BC = \begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 12 & -2 \\ 18 & 2 \end{pmatrix}$ | M1 | Finding BC (or AB)
$A(BC) = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 12 & -2 \\ 18 & 2 \end{pmatrix} = \begin{pmatrix} 48 & 2 \\ 12a - 18 & -2a - 2 \end{pmatrix} = (AB)C$ (demonstrates associativity of matrix multiplication) | A1 | Correct final matrix and statement of equality
**(b)** $AC = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 5a + 2 & -2 \end{pmatrix}$ | M1 | Finding AC (or CA)
$CA = \begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix} = \begin{pmatrix} 5 & 10 \\ 2a - 2 & -6 \end{pmatrix} \neq AC$ (so matrix multiplication is not commutative) | A1 | Finding the other and statement of non-equality
**(c)** $\begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ ax - y \end{pmatrix}$ | M1 | Multiplying the vector into the matrix using the correct procedure
$x + 2y = 3x \Rightarrow y = x$ | A1 |
$ax - y = 3y$ and $y = x \Rightarrow a = 4$ | A1 |
---Three matrices, A, B and C, are given by $\mathbf{A} = \begin{pmatrix} 1 & 2 \\ a & -1 \end{pmatrix}$, $\mathbf{B} = \begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix}$ and $\mathbf{C} = \begin{pmatrix} 5 & 0 \\ -2 & 2 \end{pmatrix}$ where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Using A, B and C in that order demonstrate explicitly the associativity property of matrix multiplication. [4]
\item Use A and C to disprove by counterexample the proposition 'Matrix multiplication is commutative'. [2]
\end{enumerate}
For a certain value of $a$, $\mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix} = 3\begin{pmatrix} x \\ y \end{pmatrix}$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find
\begin{itemize}
\item $y$ in terms of $x$,
\item the value of $a$.
\end{itemize} [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q6 [9]}}