Question:
Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x+y+z=4,3 x+2 y+5 z=3$, $9 x+4 y+(28+[\lambda]) z=[\lambda]$ has a solution is:
Correct Option: 1
Solution:
$\mathrm{D}=\left|\begin{array}{llc}1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28+[\lambda]\end{array}\right|=-24-[\lambda]+15=-[\lambda]-9$
if $[\lambda]+9 \neq 0$ then unique solution
if $[\lambda]+9=0$ then $\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$
so infinite solutions
Hence $\lambda$ can be any red number.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.