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

Hint: 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.