Question

For what values of k will the pair of equations kx + 3y = k-3 , 12x + ky = k have no solution

Answer 1

$\left\{\begin{array}{ccc}kx+3y=k-3\\12x+ky=k\end{array}\right\\\\ W=\left|\begin{array}{ccc}k&3\\12&k\end{array}\right|=k^2-3\cdot12=k^2-36\\\\W_x=\left|\begin{array}{ccc}k-3&3\\k&k\end{array}\right|=k(k-3)-3k=k^2-6k\\\\W_y=\left|\begin{array}{ccc}k&k-3\\12&k\end{array}\right|=k^2-12(k-3)=k^2-12k+36\\\\System\ of\ equals\ hasn't\ solution\ when\:\\W=0\ and\ (W_x=0\ or\ W_y=0)$

$W=0\iff k^2-36=0\to k^2=36\to k=-6\ or\ k=6\\\\W_x=0\iff k^2-6k=0\to k(k-6)=0\to k=0\ or\ k=6\\\\W_y=0\iff k^2-12k+36=0\to(k-6)^2=0\to k=6$

$Answer:\\\\for\ k=6\ the\ system\ has\ infinity\ solutions\\\\for\ k=-6\ system\ has\ no\ solution.$