Question

Let $\left[\lambda \right]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4,3x+2y+5z=3,9x+4y+(28+[\lambda \left]\right)z=\left[\lambda \right]$ has a solution is

• A. $(-\infty ,-9)\cup [-8,\infty )$
• B. $[-9,-8)$
• C. $R$
• D. $(-\infty ,-9)\cup (-9,\infty )$

Answer 1

The correct option is C $R$

$x+y+z=4$
$3x+2y+5z=3$
$9x+4y+(28+[\lambda \left]\right)z=\left[\lambda \right]$

For unique solution $\Delta \ne 0$
$\left|\begin{array}{ccc}1& 1& 1\\ 3& 2& 5\\ 9& 4& 28+\left[\lambda \right]\end{array}\right|\ne 0$
$\Rightarrow (56+2[\lambda ]-20)-1(84+3[\lambda ]-45)+1(-6)\ne 0$
$\Rightarrow 36+2\left[\lambda \right]-39-3\left[\lambda \right]-6\ne 0$
$\Rightarrow \left[\lambda \right]\ne -9$
$\Rightarrow \lambda \in (-\infty ,-9)\cup [-8,\infty )$
and if $\left[\lambda \right]=-9,{\Delta }_x={\Delta }_y={\Delta }_z=0$ gives infinite solution.

$\therefore$ For $\lambda \in R$ set of equations have solution.