Let's take an example of system of equations(actually they are all same)
x+2y+4z=0
2x+4y+8z=0
-3x-6y-12z=0
It is homogeneous system of linear equations. Therefore it has zero solution [0,0,0].
Rank of those coefficient matrix is found to be 1.
I read in textbook that it will have (n-r) linearly independent solutions where n= number of variables & r= rank of coefficient matrix.
So for above example it'll have (3-1)=2 linearly independent solutions.
I'll list some solutions out of infinitely many solutions:
[0,0,0]
[-6,1,1]
[-4,0,1]
[-2, 1,0] & so on.
Questions:
1)where are the 2 linearly independent solutions?
2)out of many solutions, 2 are found to be linearly independent. Is the rest of solutions Linearly dependent solutions?
3) what is actually meant by linearly dependence / independence?
Best Answer
If you reduce the coefficient matrix, you should find that the last 2 equations are a multiple of the first, so the system of equations reduces to $$ x + 2y + 4z=0 $$
As you noticed, if $y=z=0$, then $x=0$. So $(0,0,0)$ is a particular solution of the system.
Now we ask the question, what if $y$ and $z$ were not $0$? We can answer that question by making $y\neq0$, then $z\neq0$. You will notice that the vectors we end up with are NOT multiples of each other, so they are "linearly independent"
1) If $y=c\neq0\in\mathbb{R}$ and $z=0$, then $x=-2c$ and $(-2c,c,0)=c(-2,1,0)$ is a solution for any $c\in\mathbb{R}$.
2)If $z=d\neq0\in\mathbb{R}$ and $y=0$, then $x=-4d$ and $(-4d,0,d)=d(-4,0,1)$ is a solution for any $d\in\mathbb{R}$.
As mentioned earlier, there does not exist a $c$ and $d$ such that $c(-2,1,0)+d(-4,0,1)=0$. Therefore the vectors are not multiples of each other. This condition is the definition of linear independence