$\qquad  = {k^2} - 4k + 3$

$\qquad  = (k - 3)(k - 1)$

${{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}}
{4k}&8\\
{3k - 1}&{k + 3}
\end{array}} \right|$$ = 4{k^2} + 12k - 24k + 8$

$\qquad  = 4{k^2} + 12k - 24k + 8$

$\qquad  = 4({k^2} - 3k + 2)$

$\qquad  = 4(k - 2)(k - 1)$

${{\rm{\Delta }}_2} = \left| {\begin{array}{*{20}{c}}
{k + 1}&{4k}\\
k&{3k - 1}
\end{array}} \right|$$ = 3{k^2} + 2k - 1 - 4{k^2}$

$\qquad  =  - {k^2} + 2k - 1$

$\qquad  =  - {(k - 1)^2}$

As given no solution 

${{\rm{\Delta }}_1}\;\;and\;\;{{\rm{\Delta }}_2} \ne 0$

but  ${\rm{\Delta }} = 0$

$k=3$