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.

$\left({\begin{array}{*{20}{c}} 1 \\ 3 \\ 2 \end{array}\,\, \begin{array}{*{20}{c}} 0 \\ 1 \\ a \end{array}\,\, \begin{array}{*{20}{c}} 1 \\ 2 \\ {-1} \end{array}\,\, \begin{array}{*{20}{c}} {-1} \\ 1 \\ b \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} 1 \\ {3a - 2} \\ 2 \end{array}\,\, \begin{array}{*{20}{c}} 0 \\ 0 \\ a \end{array}\,\, \begin{array}{*{20}{c}} 1 \\ {2a+1} \\ {-1} \end{array}\,\, \begin{array}{*{20}{c}} {-1} \\ {a-b} \\ b \end{array}} \right)$ 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 $y = \lambda$.