Discussion Forum

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

Answer:

Explanation:

 The given system of equations will be consistent with unique solution , when

$\begin{bmatrix}1 & 1&1 \\1 & 2&3\\1 &4&k \end{bmatrix}\neq0$

 $\Rightarrow  1(2k-12)+1(3-k)+1(4-2)\neq0$

$\Rightarrow  k-12+3+2\neq0\Rightarrow k-7\neq0\Rightarrow k\neq7$