CAIE FP1 2014 November — Question 5 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeGeneral solution with parameters
DifficultyChallenging +1.2 This is a standard Further Maths linear systems question requiring determinant calculation to find when no unique solution exists, then finding the general solution with a parameter. While it involves multiple parts and requires understanding of rank/consistency, the techniques are routine for FP1 students: set det=0, solve for a, then use row reduction. The geometric interpretation is straightforward recall. More mechanical than insightful compared to average A-level questions.
Spec4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix4.03s Consistent/inconsistent: systems of equations4.03t Plane intersection: geometric interpretation

5 Find the value of \(a\) for which the system of equations $$\begin{aligned} & x - y + 2 z = 4 \\ & x + a y - 3 z = b \\ & x - y + 7 z = 13 \end{aligned}$$ where \(a\) and \(b\) are constants, has no unique solution. Taking \(a\) as the value just found,
  1. find the general solution in the case \(b = - 5\),
  2. interpret the situation geometrically in the case \(b \neq - 5\).

Question 5:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\begin{vmatrix}1&-1&2\\1&-1&-3\\1&-1&7\end{vmatrix}=0\)M1
\(\Rightarrow 5a+5=0\Rightarrow a=-1\)A1A1 (3)
\(\begin{pmatrix}1&-1&2&\mid&4\\1&-1&-3&\mid&-5\\1&-1&7&\mid&13\end{pmatrix}\rightarrow\ldots\rightarrow\begin{pmatrix}1&-1&2&\mid&4\\0&0&5&\mid&9\\0&0&0&\mid&0\end{pmatrix}\), or by eliminationM1
\(x-y+2z=4,\quad 5z=9\)A1
\(\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0.4\\0\\1.8\end{pmatrix}+t\begin{pmatrix}1\\1\\0\end{pmatrix}\) (OE)M1A1 (4)
Planes form a prism, or words to that effect.B1 (1)
## Question 5:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}1&-1&2\\1&-1&-3\\1&-1&7\end{vmatrix}=0$ | M1 | |
| $\Rightarrow 5a+5=0\Rightarrow a=-1$ | A1A1 (3) | |
| $\begin{pmatrix}1&-1&2&\mid&4\\1&-1&-3&\mid&-5\\1&-1&7&\mid&13\end{pmatrix}\rightarrow\ldots\rightarrow\begin{pmatrix}1&-1&2&\mid&4\\0&0&5&\mid&9\\0&0&0&\mid&0\end{pmatrix}$, or by elimination | M1 | |
| $x-y+2z=4,\quad 5z=9$ | A1 | |
| $\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0.4\\0\\1.8\end{pmatrix}+t\begin{pmatrix}1\\1\\0\end{pmatrix}$ (OE) | M1A1 (4) | |
| Planes form a prism, or words to that effect. | B1 (1) | |

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5 Find the value of $a$ for which the system of equations

$$\begin{aligned}
& x - y + 2 z = 4 \\
& x + a y - 3 z = b \\
& x - y + 7 z = 13
\end{aligned}$$

where $a$ and $b$ are constants, has no unique solution.

Taking $a$ as the value just found,\\
(i) find the general solution in the case $b = - 5$,\\
(ii) interpret the situation geometrically in the case $b \neq - 5$.

\hfill \mbox{\textit{CAIE FP1 2014 Q5 [8]}}