HOMOGENEOUS EQUATION: Each equation of the system of following linear equations
[katex]a_{11}x_1+a_{12}x_2+a_{13}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(1)[/katex].
[katex]a_{21}x_1+a_{22}x_2+a_{23}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(2)[/katex].
[katex]a_{31}x_1+a_{32}x_2+a_{33}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(3)[/katex].
is always satisfied by [katex]x_1= 0,\;\; x_2= 0\;\; and \;\;x_3 = 0[/katex], so such a system is always consistent. The solution [katex](0, 0, 0)[/katex] of the above homogeneous equations [katex](i),\;\; (ii),\;\; and\;\; (iii)[/katex] is called the trivial solution. Any other solution of equations [katex](i),\;\; (ii)\;\; and\;\; (iii)[/katex] other than the trivial solution is called a non-trivial solution.
The above system can be written as
[katex]AX = O\;\;[/katex] where [katex]\;\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}[/katex]
If [katex]\left|A\right|\neq0[/katex], then [katex]A[/katex] is non-singular and [katex]A^{-1}[/katex] exists, that is
[katex]A^{-1}(AX)=(A^{-1}A)X=OX=O[/katex]
[katex]\;\;(A^{-1}A)X=O[/katex]
[katex]⇒X=O[/katex]
[katex]i.e,\;\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}[/katex].
In this case the system of homogeneous equations possesses only the trivial solution.
Now we consider the case when the system has a non-trivial solution. Multiplying the equations [katex](i), (ii)[/katex] and [katex](iii)[/katex] by [katex]A_{11},\;\; A_{21}\;\;[/katex] and [katex]A_{31}[/katex] respectively and adding the resulting equations (where [katex]A_{11}, A_{21}[/katex] and [katex]A_{31}[/katex] are cofactors of the corresponding elements of [katex]A[/katex]),
we have
[katex](a_{11}+A_{11} + a_{21} A_{21}+a_{31} A_{31})x_1+(a_{12} A_{11} + a_{22} A_{21}+a_{32} A_{31})x_2 +(a_{13} A_{11} + a_{23} A_{21}+a_{33} A_{31})x_3= 0[/katex].
that is,
[katex]\left|A\right|x_1=0 [/katex].
Similarly, we can get
[katex]\left|A\right|x_2=0\;\;\; and\;\;\; \left|A\right|x_3=0[/katex]
For a non-trivial solution, at least one of [katex]x_1 ,\; x_2\;[/katex] and [katex]x_3[/katex] is different from zero
Let [katex]x_1 ≠ 0[/katex] , then from [katex]\left|A\right|x_1=0[/katex]., we have [katex]\left|A\right|=0[/katex].
For example, the system
[katex]\left.\begin{array}{l}x_1+x_2+x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i)\\x_1-x_2+3x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)\\x_1+3x_2-x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iii)\\\end{array}\right\}[/katex].
has a non-trivial solution because
[katex]\left|A\right|=\begin{vmatrix}1&1&1\\1&-1&3\\1&3&-1\end{vmatrix}[/katex]
[katex]⇒\left|A\right|=\begin{vmatrix}1&0&0\\1&-2&2\\1&2&-2\end{vmatrix}[/katex].
by [katex] \;\;C_2 -C_1, C_3-C_1[/katex].
[katex]⇒\left|A\right|=\begin{vmatrix}-2&2\\2&-2\end{vmatrix}[/katex]
[katex]⇒\left|A\right|=0[/katex]
Adding [katex](i) \;and\; (ii)[/katex] , we get
[katex]\;\;\;2x_1+4x_3=0[/katex]
[katex]⇒x_1=-2x_3[/katex]
And subtracting [katex](ii) \;and \;(i)[/katex] , we get
[katex]\;\;\;2x_2 – 2x_3 = 0[/katex]
[katex]⇒ x_2 = x_3[/katex]
Putting [katex]x_1 = -2x_3[/katex] and [katex]x_2 = x_3[/katex] in (III), we see that [katex](-2x_3) + 3(x_3) – x_3= 0[/katex], which shows that the equation [katex](i), (ii) \;and \;(iii)[/katex] are satisfied by
[katex]x_1 = -2t, x_2= t \;\;and \;\;x_3= t[/katex] for any real value of t.
Thus the system consisting of [katex](i), (ii) \;and \;(iii)[/katex] has infinitely many solutions.
But the system
[katex]\left.\begin{array}{l}x_1+x_2+x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i)\\x_1-x_2+3x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)\\x_1+3x_2-2x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iii)\\\end{array}\right\}[/katex].
has only the trivial solution.
because in this case
[katex]\left|A\right|=\begin{vmatrix}1&1&1\\1&-1&3\\1&3&-2\end{vmatrix}[/katex]
[katex]⇒\left|A\right|=\begin{vmatrix}1&0&0\\1&-2&2\\1&2&-3\end{vmatrix}[/katex].
by [katex] \;\;C_2 -C_1, C_3-C_1[/katex].
[katex]⇒\left|A\right|=\begin{vmatrix}-2&2\\2&-3\end{vmatrix}[/katex].
[katex]⇒\left|A\right|=6\neq0[/katex].
Solving the first two equations of the above system, we get
[katex]x_1= -2x-3[/katex] and [katex]x_2= x_3[/katex].
Putting [katex]x_1 = -2x_3[/katex] and [katex]x2= x_3[/katex] in the expression [katex]x_1+ 3x_2 -2x_3 [/katex]
we have,
[katex]- 2x_3 +3(x_3) – 2x_3= – x_3[/katex],
that is, the third equation is not satisfied by putting
[katex]x_1= -2x_3[/katex] and [katex] x_2= x_3[/katex]
But it is satisfied only if [katex]x_3=0[/katex].
Thus the above system has only the trivial solution.
you can also check homogenous function.