Question

Find the value of k for which the simultaneous equation x+y+z=3;x+2y+3z=4 and x+4y+kz=6 will not have a unique solution

Answer 1

$|1 \: 1 \: 1 \\ 1 \: \: 2 \: 3 \\ 1 \: \: 4 \: k| = 0$

$1(2 - 1) - 4(3 - 1) + k(2 - 1) = 0$

$1 - 8 + k = 0$

$k = 7$