| Date | May 2018 | Marks available | 4 | Reference code | 18M.1.hl.TZ0.8 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Consider the simultaneous linear equations
\(x + z = - 1\)
\(3x + y + 2z = 1\)
\(2x + ay - z = b\)
where \(a\) and \(b\) are constants.
Using row reduction, find the solutions in terms of \(a\) and \(b\) when \(a\) ≠ 3 .
Explain why the equations have no unique solution when \(a\) = 3.
Find all the solutions to the equations when \(a\) = 3, \(b\) = 10 in the form r = s + \(\lambda \)t.
Markscheme
\(\left({\begin{array}{*{20}{c}} 1 \\ 3 \\ 2 \end{array}\,\, \begin{array}{*{20}{c}} 0 \\ 1 \\ a \end{array}\,\, \begin{array}{*{20}{c}} 1 \\ 2 \\ {-1} \end{array}\,\, \begin{array}{*{20}{c}} {-1} \\ 1 \\ b \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} 1 \\ {3a - 2} \\ 2 \end{array}\,\, \begin{array}{*{20}{c}} 0 \\ 0 \\ a \end{array}\,\, \begin{array}{*{20}{c}} 1 \\ {2a+1} \\ {-1} \end{array}\,\, \begin{array}{*{20}{c}} {-1} \\ {a-b} \\ b \end{array}} \right)\) or equivalent M1A1
\(\left( {\begin{array}{*{20}{c}}
1 \\
{3a - 2} \\
2
\end{array}\,\,\begin{array}{*{20}{c}}
0 \\
0 \\
a
\end{array}\,\,\begin{array}{*{20}{c}}
{a - 3} \\
{2a + 1} \\
{ - 1}
\end{array}\,\,\begin{array}{*{20}{c}}
{ - 4a + 2 + b} \\
{a - b} \\
b
\end{array}} \right)\)
\(z = \frac{{ - 4a + b + 2}}{{a - 3}}\) M1A1
\(x = - 1 - z\) M1
\(x = - 1 - \left( {\frac{{ - 4a + b + 2}}{{a - 3}}} \right)\)
\(x = \frac{{ - a + 3 + 4a - b - 2}}{{a - 3}}\)
\(x = \frac{{3a - b + 1}}{{a - 3}}\) A1
\(y = 1 - 3x - 2z\) M1
\(y = 1 - 3\left( {\frac{{3a - b + 1}}{{a - 3}}} \right) - 2\left( {\frac{{ - 4a + b + 2}}{{a - 3}}} \right)\)
\( = \frac{{a - 3 - 9a + 3b - 3 + 8a - 2b - 4}}{{a - 3}}\)
\( = \frac{{b - 10}}{{a - 3}}\) A1
[8 marks]
when \(a\) = 3 the denominator of \(x\), \(y\) and \(z\) = 0 R1
Note: Accept any valid reason.
hence no unique solutions AG
[1 mark]
For example let \(z = \lambda \) (M1)
\(x = - 1 - \lambda \) (A1)
\(y = 1 - 3\left( { - 1 - \lambda } \right) - 2\lambda \)
\(y = 4 + \lambda \) (A1)
r = \(\left( \begin{gathered}
- 1 \hfill \\
4 \hfill \\
0 \hfill \\
\end{gathered} \right) + \lambda \left( \begin{gathered}
- 1 \hfill \\
1 \hfill \\
1 \hfill \\
\end{gathered} \right)\) A1
[4 marks]
Note: Accept answers which let \(x = \lambda \) or \(y = \lambda \).
