Question

Let $S$ be the set of all real values of $k$ for which the system of linear 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 $R-\left\{0\right\}$
• C. Equal to $\left\{0\right\}$
• D. Equal to $R$

Answer 1

Answer: (B)

Equal to $R-\left\{0\right\}$

Given are the system of linear equations.

$x+y+z=2$
$2x+y-z=3$
$3x+2y+kz=4$

We know that, this system has a unique solution.

Therefore, coefficient determinant is non-zero.

$\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 3& 2& k\end{array}\right|\ne 0$

$\Rightarrow k+2-(2k+3)+1\ne 0$

$\Rightarrow k\ne 0$

Therefore, $k\in R-\left\{0\right\}\equiv S$