Homogeneous Equation with Examples

HOMOGENEOUS EQUATION: Each equation of the system of following linear equations

$a_{11}x_1+a_{12}x_2+a_{13}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$ .

$a_{21}x_1+a_{22}x_2+a_{23}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$ .

$a_{31}x_1+a_{32}x_2+a_{33}x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;(3)$ .

is always satisfied by $x_1= 0,\;\; x_2= 0\;\; and \;\;x_3 = 0$ , so such a system is always consistent. The solution $(0, 0, 0)$ of the above homogeneous equations $(i),\;\; (ii),\;\; and\;\; (iii)$ is called the trivial solution. Any other solution of equations $(i),\;\; (ii)\;\; and\;\; (iii)$ other than the trivial solution is called a non-trivial solution.

The above system can be written as

$AX = O\;\;$ where $\;\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

If $\left|A\right|\neq0$ , then $A$ is non-singular and $A^{-1}$ exists, that is

$A^{-1}(AX)=(A^{-1}A)X=OX=O$

$\;\;(A^{-1}A)X=O$

$⇒X=O$

$i.e,\;\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ .

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 $(i), (ii)$ and $(iii)$ by $A_{11},\;\; A_{21}\;\;$ and $A_{31}$ respectively and adding the resulting equations (where $A_{11}, A_{21}$ and $A_{31}$ are cofactors of the corresponding elements of $A$ ),
we have
$(a_{11}+A_{11} + a_{21} A_{21}+a_{31} A_{31})x_1+(a_{12} A_{11} + a_{22} A_{21}+a_{32} A_{31})x_2 +(a_{13} A_{11} + a_{23} A_{21}+a_{33} A_{31})x_3= 0$ .

that is,

$\left|A\right|x_1=0$ .

Similarly, we can get

$\left|A\right|x_2=0\;\;\; and\;\;\; \left|A\right|x_3=0$

For a non-trivial solution, at least one of $x_1 ,\; x_2\;$ and $x_3$ is different from zero

Let $x_1 ≠ 0$ , then from $\left|A\right|x_1=0$ ., we have $\left|A\right|=0$ .

For example, the system

$\left.\begin{array}{l}x_1+x_2+x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i)\\x_1-x_2+3x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)\\x_1+3x_2-x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iii)\\\end{array}\right\}$ .

has a non-trivial solution because

$\left|A\right|=\begin{vmatrix}1&1&1\\1&-1&3\\1&3&-1\end{vmatrix}$

$⇒\left|A\right|=\begin{vmatrix}1&0&0\\1&-2&2\\1&2&-2\end{vmatrix}$ .

by $\;\;C_2 -C_1, C_3-C_1$ .

$⇒\left|A\right|=\begin{vmatrix}-2&2\\2&-2\end{vmatrix}$

$⇒\left|A\right|=0$

Adding $(i) \;and\; (ii)$ , we get

$\;\;\;2x_1+4x_3=0$

$⇒x_1=-2x_3$

And subtracting $(ii) \;and \;(i)$ , we get

$\;\;\;2x_2 - 2x_3 = 0$

$⇒ x_2 = x_3$

Putting $x_1 = -2x_3$ and $x_2 = x_3$ in (III), we see that $(-2x_3) + 3(x_3) - x_3= 0$ , which shows that the equation $(i), (ii) \;and \;(iii)$ are satisfied by

$x_1 = -2t, x_2= t \;\;and \;\;x_3= t$ for any real value of t.
Thus the system consisting of $(i), (ii) \;and \;(iii)$ has infinitely many solutions.

But the system

$\left.\begin{array}{l}x_1+x_2+x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i)\\x_1-x_2+3x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)\\x_1+3x_2-2x_3=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iii)\\\end{array}\right\}$ .