3 and 4 .Determinants and Matrices
hard

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:

A

${R}$

B

$(-\infty,-9) \cup(-9, \infty)$

C

$[-9,-8)$

D

$(-\infty,-9) \cup[-8, \infty)$

(JEE MAIN-2021)

Solution

$\left| {\begin{array}{*{20}{r}} 1&1&1\\ 3&2&2\\ 9&4&{28 + [\lambda ]} \end{array}} \right|$ $=-24-[\lambda] +15=-[\lambda]-9$

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 real number.

Standard 12
Mathematics

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