Question

The system of linear equations, $x+y+z=2$, $2x+y-z=3$,$3x+2y+kz=4$ has a unique solution, if
(a) $k\ne 0$
(b) $-1 < k < 1$
(c) $-2 < k < 2$
(d) $k=0$

Answer 1

We are given the equations $x+y+z=2$, $2x+y-z=3$, $3x+2y+kz=4$ which has a unique solution. So use condition for unique solution i.e. $D\ne 0$. Try it and you will get the answer.

Complete step-by-step answer:
In geometry, a line can be defined as a straight one- dimensional figure that has no thickness and extends endlessly in both directions.
It is often described as the shortest distance between any two points.
A line is one-dimensional. It has zero width. If you draw a line with a pencil, examination with a microscope would show that the pencil mark has a measurable width. The pencil line is just a way to illustrate the idea on paper. In geometry however, a line has no width.
A straight line is the shortest distance between any two points on a plane.
Line, Basic element of Euclidean geometry. Euclid defined a line as an interval between two points and claimed it could be extended indefinitely in either direction. Such an extension in both directions is now thought of as a line, while Euclid’s original definition is considered a line segment. A ray is part of a line extending infinitely from a point on the line in only one direction. In a coordinate system on a plane, a line can be represented by the linear equation $ax+by+c=0$. This is often written in the slope-intercept form as $y=mx+b$, in which $m$ is the slope and $b$ is the value where the line crosses the $y$-axis. Because geometrical objects whose edges are line segments are completely understood, mathematicians frequently try to reduce more complex structures into simpler ones made up of connected line segments.
In question we have given equations $x+y+z=2$, $2x+y-z=3$,$3x+2y+kz=4$ which has a unique solution.
We know condition for unique solution,
$D\ne 0$
That determinant should not be equal to $0$.
So determinant of above equations becomes,
$\left| \begin{matrix} 1 & 1 & 1 \\\ 2 & 1 & -1 \\\ 3 & 2 & k \\\ \end{matrix} \right|\ne 0$
Simplifying we get,
$\begin{aligned} & 1(k+2)-1(2k+3)+1(4-3)\ne 0 \\\ & \Rightarrow k+2-2k-3+4-3\ne 0 \\\ & \Rightarrow -k\ne 0 \\\ & \Rightarrow k\ne 0 \\\ \end{aligned}$
So from the above option, option (A) is correct.

Note: Read the question carefully. You should know the concepts behind the determinant. Also, you must know the conditions for a unique solution. Do not miss any term while solving. While simplifying do not confuse regarding signs.