| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question involving routine matrix operations (finding 2×2 inverses using the standard formula, matrix multiplication) and completing a guided proof with scaffolding provided. While it's Further Maths content, the techniques are mechanical and the proof completion requires only basic algebraic manipulation following the given structure. |
| Spec | 4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar4.03n Inverse 2x2 matrix |
Matrices $M$ and $N$ are given by $M = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}$ and $N = \begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix}$.
(i) Find $M^{-1}$ and $N^{-1}$. [3]
(ii) Find $MN$ and $(MN)^{-1}$. Verify that $(MN)^{-1} = N^{-1}M^{-1}$. [6]
(iii) The result $(PQ)^{-1} = Q^{-1}P^{-1}$ is true for any two $2 \times 2$, non-singular matrices $P$ and $Q$. The first two lines of a proof of this general result are given below. Beginning with these two lines, complete the general proof.
$(PQ)^{-1}PQ = I$
$\Rightarrow (PQ)^{-1}PQQ^{-1} = IQ^{-1}$ [4]9 Matrices $\mathbf { M }$ and $\mathbf { N }$ are given by $\mathbf { M } = \left( \begin{array} { l l } 3 & 2 \\ 0 & 1 \end{array} \right)$ and $\mathbf { N } = \left( \begin{array} { r r } 1 & - 3 \\ 1 & 4 \end{array} \right)$.\\
(i) Find $\mathbf { M } ^ { - 1 }$ and $\mathbf { N } ^ { - 1 }$.\\
(ii) Find $\mathbf { M N }$ and $( \mathbf { M N } ) ^ { - \mathbf { 1 } }$. Verify that $( \mathbf { M N } ) ^ { - 1 } = \mathbf { N } ^ { - 1 } \mathbf { M } ^ { - 1 }$.\\
(iii) The result $( \mathbf { P Q } ) ^ { - 1 } = \mathbf { Q } ^ { - 1 } \mathbf { P } ^ { - 1 }$ is true for any two $2 \times 2$, non-singular matrices $\mathbf { P }$ and $\mathbf { Q }$.
The first two lines of a proof of this general result are given below. Beginning with these two lines, complete the general proof.
$$\begin{aligned}
& ( \mathbf { P Q } ) ^ { - 1 } \mathbf { P Q } = \mathbf { I } \\
\Rightarrow & ( \mathbf { P Q } ) ^ { - 1 } \mathbf { P Q Q } \mathbf { Q } ^ { - 1 } = \mathbf { I Q } ^ { - 1 }
\end{aligned}$$
\hfill \mbox{\textit{OCR MEI FP1 2007 Q9 [13]}}