Question

In a system of three linear homogeneous equations with three unknowns, if $\Delta =0$, ${{\Delta }_{x}}=0$, ${{\Delta }_{y}}\ne 0$ and ${{\Delta }_{z}}=0$, then the system has
(a) Unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution

Answer 1

We start solving the problem by assuming three linear homogeneous equations given in the problem. We then recall the matrix representation of these equations and definition of each matrix. We then recall the definitions of ${{\Delta }_{x}}$, ${{\Delta }_{y}}$, ${{\Delta }_{z}}$ and find the absolute values of them to check the compatibility of the linear homogeneous equations which leads us to the required answer.

Complete step-by-step answer:
According to the problem, we have a system of three linear homogeneous equations with three unknowns and it is given $\Delta =0$, ${{\Delta }_{x}}=0$, ${{\Delta }_{y}}\ne 0$ and ${{\Delta }_{z}}=0$. We need to find the number of solutions that this system of equations has.
Let us assume the three linear homogeneous equations be
$\Rightarrow {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=0$.
$\Rightarrow {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=0$.
$\Rightarrow {{a}_{2}}x+{{b}_{2}}y+{{c}_{3}}z=0$.
We know that the matrix representation of the linear homogeneous equation is $AX=0$, where A is known as coefficient matrix (as mentioned below) and X is known as variable matrix (as mentioned below).
So, we have $A=\left[ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right]$ and $X=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]$.
We know that $\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|$.
We know that the ${{\Delta }_{x}}$ is the determinant of the matrix formed by replacing every element in the 1st column of the coefficient matrix with zero. We know that the determinant of the matrix with row or column full of zeroes is zero. So, we always get ${{\Delta }_{x}}=0$.
Similarly, we also get ${{\Delta }_{y}}$, ${{\Delta }_{z}}$ are the determinants of the matrices formed by replacing every element in 2nd and 3rd column of coefficient matrix with zero. We know that the determinant of the matrix with row or column full of zeroes is zero. So, we always get ${{\Delta }_{y}}=0$, ${{\Delta }_{z}}=0$.
We know that if these determinants are not equal to zero, then the linear homogeneous equations are not compatible. This means that the given set of linear homogeneous equations don’t have a solution.

So, the correct answer is “Option d”.

Note: We should always remember the properties about the linear homogeneous equation which tells us about its solutions:
(i) If the value of $\Delta$ is not zero, then the system of equations will have the trivial solutions $x=y=z=0$.
(ii) If the value of $\Delta$ is zero and ${{\Delta }_{x}}={{\Delta }_{y}}={{\Delta }_{z}}=0$, then the system of equations will have infinite number of solutions.
(iii) If the value of $\Delta$ is zero and any of ${{\Delta }_{x}}$, ${{\Delta }_{y}}$, ${{\Delta }_{z}}$ is not zero, then the system of equations will not have a solution.
All these results can be verified by using Gauss Jordan method or Cramer’s rule.