Question 10 - A-Level Maths

CAIE FP1 2010 June — Question 10 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2010
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 Further Maths question on linear systems requiring determinant calculation for uniqueness, then verification of inconsistency and parametric solution for specific cases. While it involves multiple parts and Further Maths content (making it harder than average A-level), the techniques are standard: compute det(A), substitute values, and use row reduction. The steps are clearly signposted with no novel insight required, placing it moderately above average difficulty.
Spec	4.03s Consistent/inconsistent: systems of equations

10 Find the set of values of $a$ for which the system of equations $$\begin{aligned} x + 4 y + 12 z & = 5 \\ 2 x + a y + 12 z & = a - 1 \\ 3 x + 12 y + 2 a z & = 10 \end{aligned}$$ has a unique solution. Show that the system does not have any solution in the case $a = 18$. Given that $a = 8$, show that the number of solutions is infinite and find the solution for which $x + y + z = 1$.

Show mark scheme Show mark scheme source

Answer	Marks
$\begin{pmatrix}1 & 4 & 12 \\ 2 & a & 12 \\ 3 & 12 & 2a\end{pmatrix} \rightarrow ... \rightarrow \begin{pmatrix}1 & 4 & 12 \\ 0 & a-8 & -12 \\ 0 & 0 & 2a-36\end{pmatrix}$	M1A1
$\Rightarrow r(A) = 3$ provided $a \neq 18$ and $a \neq 8$	A1
$\Rightarrow$ unique solution for all values of $a$ except $a = 18$ and $a = 8$	A1
[4]

OR for first 3 marks:

Answer	Marks
$\det A = 0 \Rightarrow ... \Rightarrow a^2 - 26a + 144 = 0$	(M1A1)
$\Rightarrow a = 8$ or $18$	(A1)
$a = 18 \Rightarrow 0z = -5$ which is impossible for any finite $z$, or equivalent contradiction	M1A1
[2]
When $a = 8$ system reduces to 2 equations:	M1
$x + 4y = 2$ and $z = 1/4$
All solutions then of form:	A1
$x = \lambda, y = (2-\lambda)/4, z = 1/4$ where $\lambda$ is real
May be parametrised in any equivalent way	A1
$\Rightarrow$ an infinite number of solutions	A1
$\lambda + (2-\lambda)/4 + 1/4 = 1$	M1
$\Rightarrow \lambda = 1/3 \Rightarrow x = 1/3, y = 5/12, z = 1/4$	A1
[5]

$\begin{pmatrix}1 & 4 & 12 \\ 2 & a & 12 \\ 3 & 12 & 2a\end{pmatrix} \rightarrow ... \rightarrow \begin{pmatrix}1 & 4 & 12 \\ 0 & a-8 & -12 \\ 0 & 0 & 2a-36\end{pmatrix}$ | M1A1 |

$\Rightarrow r(A) = 3$ provided $a \neq 18$ and $a \neq 8$ | A1 |

$\Rightarrow$ unique solution for all values of $a$ except $a = 18$ and $a = 8$ | A1 |

| [4] |

**OR for first 3 marks:**

$\det A = 0 \Rightarrow ... \Rightarrow a^2 - 26a + 144 = 0$ | (M1A1) |

$\Rightarrow a = 8$ or $18$ | (A1) |

$a = 18 \Rightarrow 0z = -5$ which is impossible for any finite $z$, or equivalent contradiction | M1A1 |

| [2] |

When $a = 8$ system reduces to 2 equations: | M1 |

$x + 4y = 2$ and $z = 1/4$ |

All solutions then of form: | A1 |

$x = \lambda, y = (2-\lambda)/4, z = 1/4$ where $\lambda$ is real |

May be parametrised in any equivalent way | A1 |

$\Rightarrow$ an infinite number of solutions | A1 |

$\lambda + (2-\lambda)/4 + 1/4 = 1$ | M1 |

$\Rightarrow \lambda = 1/3 \Rightarrow x = 1/3, y = 5/12, z = 1/4$ | A1 |

| [5] |

Show LaTeX source

10 Find the set of values of $a$ for which the system of equations

$$\begin{aligned}
x + 4 y + 12 z & = 5 \\
2 x + a y + 12 z & = a - 1 \\
3 x + 12 y + 2 a z & = 10
\end{aligned}$$

has a unique solution.

Show that the system does not have any solution in the case $a = 18$.

Given that $a = 8$, show that the number of solutions is infinite and find the solution for which $x + y + z = 1$.

\hfill \mbox{\textit{CAIE FP1 2010 Q10 [11]}}

This paper (12 questions)

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Q1 5 Q2 7 Q3 7 Q4 7 Q5 9 Q6 9 Q7 10 Q8 10 Q9 11 Q10 11 Q11 EITHER Q11 OR

Answer	Marks
\(\begin{pmatrix}1 & 4 & 12 \\ 2 & a & 12 \\ 3 & 12 & 2a\end{pmatrix} \rightarrow ... \rightarrow \begin{pmatrix}1 & 4 & 12 \\ 0 & a-8 & -12 \\ 0 & 0 & 2a-36\end{pmatrix}\)	M1A1
\(\Rightarrow r(A) = 3\) provided \(a \neq 18\) and \(a \neq 8\)	A1
\(\Rightarrow\) unique solution for all values of \(a\) except \(a = 18\) and \(a = 8\)	A1
[4]

Answer	Marks
\(\det A = 0 \Rightarrow ... \Rightarrow a^2 - 26a + 144 = 0\)	(M1A1)
\(\Rightarrow a = 8\) or \(18\)	(A1)
\(a = 18 \Rightarrow 0z = -5\) which is impossible for any finite \(z\), or equivalent contradiction	M1A1
[2]
When \(a = 8\) system reduces to 2 equations:	M1
\(x + 4y = 2\) and \(z = 1/4\)
All solutions then of form:	A1
\(x = \lambda, y = (2-\lambda)/4, z = 1/4\) where \(\lambda\) is real
May be parametrised in any equivalent way	A1
\(\Rightarrow\) an infinite number of solutions	A1
\(\lambda + (2-\lambda)/4 + 1/4 = 1\)	M1
\(\Rightarrow \lambda = 1/3 \Rightarrow x = 1/3, y = 5/12, z = 1/4\)	A1
[5]