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

• A. a unique solution
• B. infinite solutions
• C. no solution
• D. $2$ solutions

Answer 1

Answer: (A)

$k=\frac{t^{2} - t + 1}{t^{2} + t + 1}\Rightarrow t^{2}\left(k - 1\right)+t\left(k + 1\right)+k-1=0$
as $t\in R$
$D\geq 0\Rightarrow \left(k + 1\right)^{2}-4\left(k - 1\right)^{2}\geq 0$
$\Rightarrow k\in \left[\frac{1}{3} , 3\right]$
for the given equation,
$\Delta =\begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix}=7\left(k + 5\right)>0 \, \forall k\in \left[\frac{1}{3} , 3\right]$
hence, unique solution