Find the value of k such that the following system of equations does not have a unique solution.

\[kx + y + 2z = 4\]

\[ - y + 4z = 5\]

\[3x + 4y + 2z = 1\]

METHOD 1

determinant = 0     M1

\(k( - 2 - 16) - (0 - 12) + 2(0 + 3) = 0\)     (M1)(A1)

\( - 18k + 18 = 0\)     (A1)

\(k = 1\)     A1

METHOD 2

writes in the form

\(\left( {\begin{array}{*{20}{c}}
  k&1&2&4 \\
  0&{ - 1}&4&5 \\
  3&4&2&1
\end{array}} \right)\)     (or attempts to solve simultaneous equations)     (M1)

Having two 0’s in first column (obtaining two equations in the same two variables)     M1

\(\left( {\begin{array}{*{20}{c}}
  k&1&2&4 \\
  0&{ - 1}&4&5 \\
  3&4&{18k - 18}&{21k - 27}
\end{array}} \right)\) (or isolating one variable)     A1

Note: The A1 is to be awarded for the 18k – 18. The final column may not be seen.

k = 1     (M1)A1

[5 marks]

Candidates who used the determinant method usually obtained full marks. Few students used row reduction and of those the success was varied. However, many candidates attempted long algebraic methods, which frequently went wrong at some stage. Of those who did work through to correctly isolate one variable, few were able to interpret the resultant value of \(k\) .