Let's take an example of system of equations(actually they are all same)




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:




[-2, 1,0] & so on.


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?


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

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  • $\begingroup$ Then I have infinite linearly independent solutions. But as I said in textbook it has given that it has only 2 linearly independent solutions $\endgroup$ – Ajay vishwanath Oct 5 '18 at 13:33
  • $\begingroup$ The independent solutions are (-2,1,0) and (-4,0,1). The power of the fact they are linearly independent means their linear combinations form a solution space $\endgroup$ – NazimJ Oct 5 '18 at 13:39

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