Question

The number of non-trivial solution of the systems: $x-y+z=0,x+2y-z=0,2x+y+3z=0\text{ }$ is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Hint:For a system of equations, the solutions follow some conditions.
If there are system of equations namely $ax+by+c=0$ and $dx+ey+f=0$ then,
$\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}\Rightarrow $ Infinite solutions
$\dfrac{a}{d}=\dfrac{b}{e}\ne \dfrac{c}{f}\Rightarrow $ No solutions

Complete step-by-step answer:
Definition of system of equations:
If simultaneously we have more than one equation, then the set of those equations is called a system of equations. We can project systems of equations as lines, planes, etc, depending on a number of variables.
If we have 2 variables:
Then the system of equations is analogous to straight lines.
If we have 3 variables:
Then the system of equations is analogous to planes.
Here, we have 3 variables. So, in our case:
Our system of equations is analogous to 3 planes. We have 3 possibilities.
(a) No solution; (b) Infinite solutions ; (c) One solution
(a) No solution:
If 3 planes (infinitely long) have 0 solution then they must not intersect anywhere that means they are parallel planes.
For 3 planes to be parallel they must satisfy:
We use matrix elimination methods.
$ax+by+cz=d$
$ex+fy+gz=h$
$ix+jy+kz=l$
$\Delta =\left| \begin{matrix}
   a & b & c \\
   e & f & g \\
   i & j & k \\
\end{matrix} \right|$
If $\Delta =0$ then no solution is possible.
(b) Infinite solution: If any 2 planes intersect on line or 3 planes coincide in likely of these events we have infinite solutions.
For 3 planes to satisfy this condition we have
\[\left| \begin{matrix}
   a \\
   e \\
   i \\
\end{matrix}\text{ }\begin{matrix}
   b \\
   f \\
   j \\
\end{matrix}\text{ }\begin{matrix}
   c \\
   g \\
   k \\
\end{matrix}\text{ }\begin{matrix}
   d \\
   h \\
   l \\
\end{matrix} \right|\Rightarrow \] By apply row transformations
You must be able to turn at least 1 row into all zeros.
(c) One solution: If all the 3 planes intersect at one single point that is each one perpendicular to each other. We have one solution conditions as
\[\left| \begin{matrix}
   a & b & c \\
   e & f & g \\
   i & j & k \\
\end{matrix} \right|\ne 0\] \[\left| \begin{matrix}
   a & b & d \\
   e & f & h \\
   i & j & l \\
\end{matrix} \right|\ne 0\] \[\left| \begin{matrix}
   a & d & c \\
   e & h & g \\
   i & l & k \\
\end{matrix} \right|\ne 0\] \[{{\Delta }_{z}}\ne 0\]
In our question, we have: $x-y+z=0,x+2y-z=0,2x+y+3z=0\text{ }$
Finding required determinant, we get: \[\left| \begin{matrix}
   1 & -1 & 1 \\
   1 & 2 & -1 \\
   2 & 1 & 3 \\
\end{matrix} \right|=1\left( 6+1 \right)+1\left( 3+2 \right)+1\left( 1-4 \right)\]
By simplifying, we get \[=7+5-3=9\]
As \[\Delta \ne 0\] non trivial solutions are not possible.
Therefore, zero solution. Option (a) is correct.

Note: (1) If $\Delta =0$, then non trivial solutions can be found if the $\Delta $ itself is not zero we can directly say.
(2) Be careful while taking the coefficients into the determinants as you must observe a,b,c,….. all are in the left hand side. If they are in RHS, the sign may change.
(3) Be careful while categorizing solutions with determinant values.