| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | General solution with parameters |
| Difficulty | Challenging +1.2 This is a systematic application of standard matrix techniques (determinant for uniqueness, row reduction for special cases) with multiple parts guiding students through the process. While it requires understanding of when systems have unique/no/infinite solutions and involves parameter work, the question structure is scaffolded and uses routine Further Maths methods without requiring novel insight or complex algebraic manipulation. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1&-2&-2&-7\\2&a-9&-10&-11\\3&-6&2a&-29\end{pmatrix} \sim \begin{pmatrix}1&-2&-2&-7\\0&a-5&-6&3\\0&0&2a+6&-8\end{pmatrix}\) | M1A1 | Reduces augmented matrix |
| Unique solution for all real \(a\) except \(a=-3\) or \(5\) | A1A1 | 4 marks total |
| Alternative: \(\begin{vmatrix}1&-2&-2\\2&(a-9)&-10\\3&-6&2a\end{vmatrix} \neq 0 \Rightarrow (a-5)(a+3)\neq 0\) | ||
| (i) \(a=-3\): \(2a+6=0 \Rightarrow 0z=-8 \Rightarrow\) no solutions | M1A1\(\checkmark\) | 2 marks |
| (ii) \(a=5\): \(\Rightarrow z=-\frac{1}{2}\) and \(x-2y=-8\) (*); \(\therefore\) infinite number of solutions | M1A1\(\checkmark\) | 2 marks |
| \(z=-\frac{1}{2}\) and \(x+y+z=2 \Rightarrow x+y=\frac{5}{2}\) | B1 | Obtains particular solution |
| Solving simultaneously with (*): \(x=-1\), \(y=\frac{7}{2}\), \(z=-\frac{1}{2}\) | M1A1 | 3 marks total |
## Question 10:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&-2&-2&-7\\2&a-9&-10&-11\\3&-6&2a&-29\end{pmatrix} \sim \begin{pmatrix}1&-2&-2&-7\\0&a-5&-6&3\\0&0&2a+6&-8\end{pmatrix}$ | M1A1 | Reduces augmented matrix |
| Unique solution for all real $a$ except $a=-3$ or $5$ | A1A1 | 4 marks total |
| **Alternative:** $\begin{vmatrix}1&-2&-2\\2&(a-9)&-10\\3&-6&2a\end{vmatrix} \neq 0 \Rightarrow (a-5)(a+3)\neq 0$ | | |
**(i)** $a=-3$: $2a+6=0 \Rightarrow 0z=-8 \Rightarrow$ no solutions | M1A1$\checkmark$ | 2 marks |
**(ii)** $a=5$: $\Rightarrow z=-\frac{1}{2}$ and $x-2y=-8$ (*); $\therefore$ infinite number of solutions | M1A1$\checkmark$ | 2 marks |
| $z=-\frac{1}{2}$ and $x+y+z=2 \Rightarrow x+y=\frac{5}{2}$ | B1 | Obtains particular solution |
| Solving simultaneously with (*): $x=-1$, $y=\frac{7}{2}$, $z=-\frac{1}{2}$ | M1A1 | 3 marks total |
---10 Find the set of values of $a$ for which the system of equations
$$\begin{aligned}
x - 2 y - 2 z & = - 7 \\
2 x + ( a - 9 ) y - 10 z & = - 11 \\
3 x - 6 y + 2 a z & = - 29
\end{aligned}$$
has a unique solution.
Show that the system has no solution in the case $a = - 3$.
Given that $a = 5$,\\
(i) show that the number of solutions is infinite,\\
(ii) find the solution for which $x + y + z = 2$.
\hfill \mbox{\textit{CAIE FP1 2012 Q10 [11]}}