Question

If the equations $kx+4y+1=0,x+ky+1=0$ and $2x-3y+1=0$ are consistent, then show that the value of $k$ is not a real number.

Answer 1

Finding the value of k by using the determinant of given equations by arranging them in a matrix format.

$det\left(\begin{array}{ccc}k& 4& 1\\ 1& k& 1\\ 2& -3& 1\end{array}\right)$

$=kdet\left(\begin{array}{cc}k& 1\\ -3& 1\end{array}\right)-4\cdot det\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)+1\cdot det\left(\begin{array}{cc}1& k\\ 2& -3\end{array}\right)$

$=k(k+3)-4(-1)+1\cdot (-3-2k)$

$=k^2+k+1$

$k=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(1\right)}}{2}=\frac{-1\pm i\sqrt{3}}{2}$

$k=-\frac{1}{2}+i\frac{\sqrt{3}}{2},k=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Hence it is proved that, k is not a real value.