| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix equation solving (AB = C) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard matrix operations: finding a scalar multiple using AA^{-1}=I, solving simultaneous equations via matrix inversion, finding unknown entries using the inverse property, and applying the reverse order law (AB)^{-1}=B^{-1}A^{-1}. All parts are routine applications of well-practiced techniques with no novel problem-solving required. While it's Further Maths content (making it slightly above average A-level difficulty), the question is entirely procedural and computational. |
| Spec | 4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((8 \times 4) - (7 \times 5) - (12 \times 1) = -15\) | M1 | Any valid method soi |
| \(\Rightarrow k = -\dfrac{1}{15}\) | A1 | No working or wrong working SC B1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = -\dfrac{1}{15}\begin{pmatrix} 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & -1 & 2 \end{pmatrix}\begin{pmatrix} 14 \\ -25 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix}\) | B1 | Use of \(\mathbf{A}^{-1}\) in correct position(s) |
| M1 | Attempt to multiply matrices to obtain column vector | |
| \(x = -1,\ y = 2,\ z = -3\) | A2 | \(-1\) each error |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1 \times a) + (-8 \times -4) + (-21 \times 2) = 0 \Rightarrow a = 10\) | M1 | Attempt to multiply \(\mathbf{BB}^{-1}\) matrices to find \(a\) or \(b\) soi |
| \((-7 \times 5) + (5 \times 1) + (15 \times b) = 0 \Rightarrow b = 2\) | A1 | For both |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\) | B1 | By notation or explicitly |
| \(= \dfrac{1}{3}\begin{pmatrix} 1 & 0 & 5 \\ -4 & -3 & 1 \\ 2 & 1 & 2 \end{pmatrix} \times -\dfrac{1}{15}\begin{pmatrix} 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & -1 & 2 \end{pmatrix}\) | M1 | Attempt to multiply in correct sequence, may be implied by the answer (at least 7 elements correct) |
| \(= -\dfrac{1}{45}\begin{pmatrix} 9 & -3 & 13 \\ -30 & -21 & -10 \\ 15 & 6 & 10 \end{pmatrix}\) | A2 | \(-1\) each error FT their value of \(b\) |
| [4] |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(8 \times 4) - (7 \times 5) - (12 \times 1) = -15$ | M1 | Any valid method soi |
| $\Rightarrow k = -\dfrac{1}{15}$ | A1 | No working or wrong working SC B1 |
| **[2]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = -\dfrac{1}{15}\begin{pmatrix} 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & -1 & 2 \end{pmatrix}\begin{pmatrix} 14 \\ -25 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix}$ | B1 | Use of $\mathbf{A}^{-1}$ in correct position(s) | Condone missing $k$ |
| | M1 | Attempt to multiply matrices to obtain column vector | |
| $x = -1,\ y = 2,\ z = -3$ | A2 | $-1$ each error |
| **[4]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1 \times a) + (-8 \times -4) + (-21 \times 2) = 0 \Rightarrow a = 10$ | M1 | Attempt to multiply $\mathbf{BB}^{-1}$ matrices to find $a$ or $b$ soi |
| $(-7 \times 5) + (5 \times 1) + (15 \times b) = 0 \Rightarrow b = 2$ | A1 | For both |
| **[2]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$ | B1 | By notation or explicitly |
| $= \dfrac{1}{3}\begin{pmatrix} 1 & 0 & 5 \\ -4 & -3 & 1 \\ 2 & 1 & 2 \end{pmatrix} \times -\dfrac{1}{15}\begin{pmatrix} 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & -1 & 2 \end{pmatrix}$ | M1 | Attempt to multiply in correct sequence, may be implied by the answer (at least 7 elements correct) | Must include $k$ |
| $= -\dfrac{1}{45}\begin{pmatrix} 9 & -3 & 13 \\ -30 & -21 & -10 \\ 15 & 6 & 10 \end{pmatrix}$ | A2 | $-1$ each error FT their value of $b$ |
| **[4]** | | |9 You are given that $\mathbf { A } = \left( \begin{array} { r r r } 8 & - 7 & - 12 \\ - 10 & 5 & 15 \\ - 9 & 6 & 6 \end{array} \right)$ and $\mathbf { A } ^ { - 1 } = k \left( \begin{array} { r r r } 4 & 2 & 3 \\ 5 & 4 & 0 \\ 1 & - 1 & 2 \end{array} \right)$.\\
(i) Find the exact value of $k$.\\
(ii) Using your answer to part (i), solve the following simultaneous equations.
$$\begin{aligned}
8 x - 7 y - 12 z & = 14 \\
- 10 x + 5 y + 15 z & = - 25 \\
- 9 x + 6 y + 6 z & = 3
\end{aligned}$$
You are also given that $\mathbf { B } = \left( \begin{array} { r r r } - 7 & 5 & 15 \\ a & - 8 & - 21 \\ 2 & - 1 & - 3 \end{array} \right)$ and $\mathbf { B } ^ { - 1 } = \frac { 1 } { 3 } \left( \begin{array} { r r r } 1 & 0 & 5 \\ - 4 & - 3 & 1 \\ 2 & 1 & b \end{array} \right)$.\\
(iii) Find the values of $a$ and $b$.\\
(iv) Write down an expression for $( \mathbf { A B } ) ^ { - 1 }$ in terms of $\mathbf { A } ^ { - 1 }$ and $\mathbf { B } ^ { - 1 }$. Hence find $( \mathbf { A B } ) ^ { - 1 }$.
\hfill \mbox{\textit{OCR MEI FP1 2013 Q9 [12]}}