| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | General solution with parameters |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on matrices with clear scaffolding through parts (i)-(iii). Finding determinant and inverse with parameter 'a' is standard FP2 technique. Part (ii) applies the inverse directly. Part (iii) requires recognizing singular case and finding general solution with geometric interpretation, which adds modest challenge but follows predictable patterns for this topic. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det(M) = 1(2a + 8) - 2(-2 - 12) + 3(2 - 3a) = 42 - 7a\) | M1 A1 | Obtaining \(\det(M)\) in terms of \(a\); Accept unsimplified |
| \(\Rightarrow\) no inverse if \(a = 6\) | A1 | Accept \(a \neq 6\) after correct det M0 if more than 1 is multiplied by the corresponding element |
| \(M^{-1} = \frac{1}{42 - 7a}\begin{pmatrix} 2a + 8 & -10 & 8 - 3a \\ 14 & -7 & -7 \\ 2 - 3a & 8 & a + 2 \end{pmatrix}\) | M1 A1 M1 A1 [7] | At least 4 cofactors correct (including one involving \(a\)); Six signed cofactors correct; Transposing and \(\div\) by \(\det(M)\). Dependent on previous M1 M1; Mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{42}\begin{pmatrix} 8 & -10 & 8 \\ 14 & -7 & -7 \\ 2 & 8 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\) | M1 | Substituting \(a = 0\) |
| \(\Rightarrow x = \frac{6}{7}, y = \frac{1}{2}, z = -\frac{2}{7}\) | M1 A2 [4] | Correct use of inverse; Dependent on both M marks. Give A1 for one correct; SC1 for \(x = 6\), \(y = 3.5\), \(z = -2\); After M0, give SC2 for correct solution and SC1 for one correct; Answers unsupported score 0 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. \(7x - 10y = 10, 7x - 10y = 3b - 2\) (or e.g. \(4x + 5z = 5, 4x + 5z = \frac{b+1}{2}\)) (or e.g. \(8y + 7z = -1, 8y + 7z = 3 - b\)) | M1 A1 M1 | Eliminating one variable in two different ways; Two correct equations; Validly obtaining a value of \(b\) |
| For solutions, \(10 = 3b - 2 \Rightarrow b = 4\) | A1 | |
| OR | M2 A1 A1 [7] | A method leading to an equation from which \(b\) could be found; A correct equation |
| \(b = 4\) | ||
| Obtaining general soln. by e.g. setting one unknown = \(\lambda\) and finding other two in terms of \(\lambda\) | M1 A1 B1 [7] | Obtaining general soln. by e.g. setting one unknown = \(\lambda\) and finding other two in terms of \(\lambda\); Any correct form; Accept "sheaf", "pages of a book", etc.; Accept unknown instead of \(\lambda\); \(x = \frac{10}{7}\lambda + \frac{10}{7}\), \(y = \lambda\), \(z = -\frac{3}{4}\lambda - \frac{1}{8}\); \(x = \frac{3}{8} - \frac{5}{4}\lambda\), \(y = -\frac{7}{8}\lambda - \frac{1}{8}\), \(z = \lambda\); Independent of all previous marks. Ignore other comments |
| Straight line |
### (i)
$\det(M) = 1(2a + 8) - 2(-2 - 12) + 3(2 - 3a) = 42 - 7a$ | M1 A1 | Obtaining $\det(M)$ in terms of $a$; Accept unsimplified
$\Rightarrow$ no inverse if $a = 6$ | A1 | Accept $a \neq 6$ after correct det M0 if more than 1 is multiplied by the corresponding element
$M^{-1} = \frac{1}{42 - 7a}\begin{pmatrix} 2a + 8 & -10 & 8 - 3a \\ 14 & -7 & -7 \\ 2 - 3a & 8 & a + 2 \end{pmatrix}$ | M1 A1 M1 A1 [7] | At least 4 cofactors correct (including one involving $a$); Six signed cofactors correct; Transposing and $\div$ by $\det(M)$. Dependent on previous M1 M1; Mark final answer
### (ii)
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{42}\begin{pmatrix} 8 & -10 & 8 \\ 14 & -7 & -7 \\ 2 & 8 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ | M1 | Substituting $a = 0$
$\Rightarrow x = \frac{6}{7}, y = \frac{1}{2}, z = -\frac{2}{7}$ | M1 A2 [4] | Correct use of inverse; Dependent on both M marks. Give A1 for one correct; SC1 for $x = 6$, $y = 3.5$, $z = -2$; After M0, give SC2 for correct solution and SC1 for one correct; Answers unsupported score 0
### (iii)
e.g. $7x - 10y = 10, 7x - 10y = 3b - 2$ (or e.g. $4x + 5z = 5, 4x + 5z = \frac{b+1}{2}$) (or e.g. $8y + 7z = -1, 8y + 7z = 3 - b$) | M1 A1 M1 | Eliminating one variable in two different ways; Two correct equations; Validly obtaining a value of $b$
For solutions, $10 = 3b - 2 \Rightarrow b = 4$ | A1 |
**OR** | M2 A1 A1 [7] | A method leading to an equation from which $b$ could be found; A correct equation
$b = 4$ | |
**Obtaining general soln. by e.g. setting one unknown = $\lambda$ and finding other two in terms of $\lambda$** | M1 A1 B1 [7] | Obtaining general soln. by e.g. setting one unknown = $\lambda$ and finding other two in terms of $\lambda$; Any correct form; Accept "sheaf", "pages of a book", etc.; Accept unknown instead of $\lambda$; $x = \frac{10}{7}\lambda + \frac{10}{7}$, $y = \lambda$, $z = -\frac{3}{4}\lambda - \frac{1}{8}$; $x = \frac{3}{8} - \frac{5}{4}\lambda$, $y = -\frac{7}{8}\lambda - \frac{1}{8}$, $z = \lambda$; Independent of all previous marks. Ignore other comments
Straight line | |
---3 (i) Find the value of $a$ for which the matrix
$$\mathbf { M } = \left( \begin{array} { r r r }
1 & 2 & 3 \\
- 1 & a & 4 \\
3 & - 2 & 2
\end{array} \right)$$
does not have an inverse.\\
Assuming that $a$ does not have this value, find the inverse of $\mathbf { M }$ in terms of $a$.\\
(ii) Hence solve the following system of equations.
$$\begin{aligned}
x + 2 y + 3 z & = 1 \\
- x + 4 z & = - 2 \\
3 x - 2 y + 2 z & = 1
\end{aligned}$$
(iii) Find the value of $b$ for which the following system of equations has a solution.
$$\begin{aligned}
x + 2 y + 3 z & = 1 \\
- x + 6 y + 4 z & = - 2 \\
3 x - 2 y + 2 z & = b
\end{aligned}$$
Find the general solution in this case and describe the solution geometrically.
\hfill \mbox{\textit{OCR MEI FP2 2012 Q3 [18]}}