Question

Let $S$ be the set of all real values of $k$ which the system of linear equations
$x + y + z = 2$
$2x + y - z = 3$
$3x + 2y + kz = 4$
has a unique solution, then finds the value of $S$ .
A. An empty set.
B. Equal to $R$ - \left\\{ 0 \right\\}
C. Equal to \left\\{ 0 \right\\}
D. Equal to $R$

Answer 1

Suppose a system of linear equations $ax + by + c = 0$ and $dx + ey + g = 0$ , so it will have a unique solution if the two lines represented by the both equations intersect at a point. So, if the two lines are neither parallel nor coincident then the slopes of the two lines should be different.
So, the slopes of the two equations are $- \frac{a}{b}$ and $- \frac{d}{e}$ $\to \left[ { - \frac{a}{b} \ne - \frac{d}{e}} \right]$
If a system of linear equations has a unique solution, therefore the determinant of the coefficient matrix is non-zero.

Complete step-by-step solution:
So, we can write for system of the linear equation which is given in the question,
\left[ A \right] = \left( {\begin{array}{*{20}{c}} 1&1&1 \\\ 2&1&{ - 1} \\\ 3&2&k; \end{array}} \right)
$\left[ B \right] = \left( \begin{gathered} 2 \\\ 3 \\\ 4 \\\ \end{gathered} \right)$
$\left[ X \right] = \left( \begin{gathered} x \\\ y \\\ z \\\ \end{gathered} \right)$
So, here the determination of $\left[ A \right]$ is non-zero because the set of linear equations has a unique solution.
$\left| A \right| \ne 0$
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1&1 \\\ 2&1&{ - 1} \\\ 3&2&k; \end{array}} \right| \ne 0
$\Rightarrow k + 2 - \left( {2k + 3} \right) + 1 \ne 0$
$\Rightarrow k \ne 0$
So, this system has a unique solution for all the real values of $k$ except $0$ .
Thus, we can write k \in R - \left\\{ 0 \right\\} -

Hence, the answer will be option (B).

Note: To find the unique solution to a system of linear equations to a system of linear equations, we must find a numerical value for each variable in the system at the same time. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables.