Question

If $t$ is a real number and $k=\frac{t^2-t+1}{t^2+t+1},$ then the system of equations
$3x-y+4z=3$
$x+2y-3z=-2$
$6x+5y+kz=-3$
For any allowable value of $k$, has

Answer 1

$k=\frac{t^2-t+1}{t^2+t+1}$
$\Rightarrow t^2(k-1)+t(k+1)+k-1=0$
$\because t\in R,$ $\therefore D\ge 0$
$\Rightarrow (k+1{)}^2-4(k-1{)}^2\ge 0$
$\Rightarrow (3k-1)(-k+3)\ge 0$
$\Rightarrow (3k-1)(k-3)\le 0$
$\Rightarrow k\in [\frac{1}{3},3]$

For the given equations,
$\Delta =\left|\begin{array}{ccc}3& -1& 4\\ 1& 2& -3\\ 6& 5& k\end{array}\right|$
$\Delta =7(k+5)>0,\forall k\in [\frac{1}{3},3]$
As $\Delta \ne 0$
$\therefore$ Given system of equations has a unique solution.