Homogeneous Equation with Examples

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




Homogeneous Equation

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





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


has a non-trivial solution because



by [katex] \;\;C_2 -C_1, C_3-C_1[/katex].



Adding [katex](i) \;and\; (ii)[/katex] , we get



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


has only the trivial solution.

because in this case



by [katex] \;\;C_2 -C_1, C_3-C_1[/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.

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