3 and 4 .Determinants and Matrices
hard

Let $S$ be the set of all real values of $k$ for which the system oflinear equations $x +y + z = 2$ ; $2x +y - z = 3$ ; $3x + 2y + kz = 4$ has a unique solution. Then $S$ is

A

an empty set

B

equal to $R- \{0\}$

C

equal to $\{0\}$

D

equal to $R$

(JEE MAIN-2018)

Solution

The system of linear equations is:

$x+y+z=2$

$2x+y-z=3$

$3x+2y+kz=4$

As, system as unique solution.

So, $\begin{array}{*{20}{c}}
1&1&1\\
2&1&{ – 1}\\
3&2&k
\end{array} \ne 0$

$ \Rightarrow k + 2 – \left( {2k + 3} \right) + 1 \ne 0$

$ \Rightarrow k \ne 0$

Hence, $k \in R – \left\{ 0 \right\} \equiv S$

Standard 12
Mathematics

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