| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| 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 structured multi-part question on matrix multiplication and inverse matrices. Part (i) involves routine 3×3 matrix multiplication with algebraic entries. Part (ii) tests understanding that det(A)=0 when inverse doesn't exist and uses AB=(8+a)I to write A^(-1)=B/(8+a). Part (iii) is straightforward application of the inverse to solve equations. Part (iv) requires recognizing that a=-8 makes the system singular. While lengthy, each step follows standard procedures with clear guidance, making it slightly easier than average for Further Maths. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{AB} = \begin{pmatrix} -2 & 1 & -5 \\ 3 & a & 1 \\ 1 & -1 & 2 \end{pmatrix}\begin{pmatrix} 2a+1 & 3 & 1+5a \\ -5 & 1 & -13 \\ -3-a & -1 & -2a-3 \end{pmatrix}\) | M1 | Attempt to find \(\mathbf{AB}\) with some justification of at least two leading diagonal terms and any other |
| \(= \begin{pmatrix} -4a-2-5+15+5a & 0 & 0 \\ 0 & 9+a-1 & 0 \\ 0 & 0 & 1+5a+13-4a-6 \end{pmatrix}\) | ||
| \(= \begin{pmatrix} 8+a & 0 & 0 \\ 0 & 8+a & 0 \\ 0 & 0 & 8+a \end{pmatrix}\) | A1 | Correct |
| \(= (8+a)\mathbf{I}\) | B1 | Relating correct diagonal matrix to \(\mathbf{I}\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{A}^{-1}\) does not exist for \(a = -8\) | B1 | |
| \(\mathbf{A}^{-1} = \dfrac{1}{8+a}\begin{pmatrix} 2a+1 & 3 & 1+5a \\ -5 & 1 & -13 \\ -3-a & -1 & -2a-3 \end{pmatrix}\) | M1 | \(k\mathbf{B}\), \(k\) not equal to 1 |
| A1 | Correct \(\mathbf{A}^{-1}\) as shown | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{A}^{-1} = \dfrac{1}{12}\begin{pmatrix} 9 & 3 & 21 \\ -5 & 1 & -13 \\ -7 & -1 & -11 \end{pmatrix}\) | B1 | |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \dfrac{1}{12}\begin{pmatrix} 9 & 3 & 21 \\ -5 & 1 & -13 \\ -7 & -1 & -11 \end{pmatrix}\begin{pmatrix} -55 \\ -9 \\ 26 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \\ 9 \end{pmatrix}\) | M1 | Correct use of their \(\mathbf{A}^{-1}\) |
| A3 | \(x\), \(y\) and \(z\) cao, \(-1\) each error | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| There is no unique solution. | B1 [1] |
## Question 9:
### Part 9(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{AB} = \begin{pmatrix} -2 & 1 & -5 \\ 3 & a & 1 \\ 1 & -1 & 2 \end{pmatrix}\begin{pmatrix} 2a+1 & 3 & 1+5a \\ -5 & 1 & -13 \\ -3-a & -1 & -2a-3 \end{pmatrix}$ | M1 | Attempt to find $\mathbf{AB}$ with some justification of at least two leading diagonal terms and any other |
| $= \begin{pmatrix} -4a-2-5+15+5a & 0 & 0 \\ 0 & 9+a-1 & 0 \\ 0 & 0 & 1+5a+13-4a-6 \end{pmatrix}$ | | |
| $= \begin{pmatrix} 8+a & 0 & 0 \\ 0 & 8+a & 0 \\ 0 & 0 & 8+a \end{pmatrix}$ | A1 | Correct |
| $= (8+a)\mathbf{I}$ | B1 | Relating correct diagonal matrix to $\mathbf{I}$ |
| | **[3]** | |
### Part 9(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{A}^{-1}$ does not exist for $a = -8$ | B1 | |
| $\mathbf{A}^{-1} = \dfrac{1}{8+a}\begin{pmatrix} 2a+1 & 3 & 1+5a \\ -5 & 1 & -13 \\ -3-a & -1 & -2a-3 \end{pmatrix}$ | M1 | $k\mathbf{B}$, $k$ not equal to 1 |
| | A1 | Correct $\mathbf{A}^{-1}$ as shown |
| | **[3]** | |
### Part 9(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{A}^{-1} = \dfrac{1}{12}\begin{pmatrix} 9 & 3 & 21 \\ -5 & 1 & -13 \\ -7 & -1 & -11 \end{pmatrix}$ | B1 | |
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \dfrac{1}{12}\begin{pmatrix} 9 & 3 & 21 \\ -5 & 1 & -13 \\ -7 & -1 & -11 \end{pmatrix}\begin{pmatrix} -55 \\ -9 \\ 26 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \\ 9 \end{pmatrix}$ | M1 | Correct use of their $\mathbf{A}^{-1}$ |
| | A3 | $x$, $y$ and $z$ cao, $-1$ each error |
| | **[5]** | |
### Part 9(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| There is no unique solution. | B1 **[1]** | |$\mathbf { 9 }$ You are given that $\mathbf { A } = \left( \begin{array} { r r r } - 2 & 1 & - 5 \\ 3 & a & 1 \\ 1 & - 1 & 2 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { c c c } 2 a + 1 & 3 & 1 + 5 a \\ - 5 & 1 & - 13 \\ - 3 - a & - 1 & - 2 a - 3 \end{array} \right)$.\\
(i) Show that $\mathbf { A B } = ( 8 + a ) \mathbf { I }$.\\
(ii) State the value of $a$ for which $\mathbf { A } ^ { - 1 }$ does not exist. Write down $\mathbf { A } ^ { - 1 }$ in terms of $a$, when $\mathbf { A } ^ { - 1 }$ exists.\\
(iii) Use $\mathbf { A } ^ { - 1 }$ to solve the following simultaneous equations.
$$\begin{aligned}
- 2 x + y - 5 z & = - 55 \\
3 x + 4 y + z & = - 9 \\
x - y + 2 z & = 26
\end{aligned}$$
(iv) What can you say about the solutions of the following simultaneous equations?
$$\begin{aligned}
- 2 x + y - 5 z & = p \\
3 x - 8 y + z & = q \\
x - y + 2 z & = r
\end{aligned}$$
\hfill \mbox{\textit{OCR MEI FP1 2011 Q9 [12]}}