A system of equations is given below.

     $x + 2y - z = 2$

     $2x + y + z = 1$

     $- x + 4y + az = 4$

(a)     Find the value of a so that the system does not have a unique solution.

(b)     Show that the system has a solution for any value of a.

(a)   $\left[ \begin{array}{l}x + 2y - z = 2\\2x + y + z = 1\\ - x + 4y + az = 4\end{array} \right.$

$\to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z =  - 3\\6y + (a - 1)z = 6\end{array} \right.$     M1A1

$\to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z =  - 3\\(a + 5)z = 0\end{array} \right.$     A1

(or equivalent)

if not a unique solution then $a =  - 5$     A1

Note:     The first M1 is for attempting to eliminate a variable, the first A1 for obtaining two expression in just two variables (plus a), and the second A1 for obtaining an expression in just a and one other variable

[4 marks]

(b)     if $a =  - 5$ there are an infinite number of solutions as last equation always true     R1

and if $a \ne  - 5$ there is a unique solution     R1

hence always a solution     AG

[2 marks]

Total [6 marks]