By reducing the augmented matrix to row echelon form,

  (i)     find the rank of the coefficient matrix;

  (ii)     find the value of $k$ for which the system has a solution.

For this value of $k$ , determine the solution.

reducing to row echelon form

$\begin{array}{*{20}{ccc|c}}   1&{ - 1}&2&5 \\   0&4&{ - 5}&{ - 7} \\   0&8&{ - 10}&{ - 14} \\   0&4&{ - 5}&{k - 15} \end{array}$     (M1)(A1)

$\begin{array}{*{20}{ccc|c}}   1&{ - 2}&2&5 \\   0&4&{ - 5}&{ - 7} \\   0&0&0&0 \\   0&0&0&{k - 8} \end{array}$     A1

(i)     this shows that the rank of the matrix is $2$    A1

(ii)     the equations can be solved if $k = 8$     A1

[5 marks]

let $z = \lambda$      A1

then $y = \frac{{5\lambda  - 7}}{4}$     A1

and $x = \left( {5 - 2\lambda  + \frac{{5\lambda  - 7}}{4} = } \right)\frac{{13 - 3\lambda }}{4}$     A1

Note: Accept equivalent expressions.

[3 marks]