Why does a row of zeroes in a matrix of linear systems mean infinite solutions?
I'm just starting to learn linear algebra and something has been stumping me to no end, why does a row of zeroes, particularly in a $4\times 3$ matrix of linear systems, mean there's an infinite amount of solutions? I've read people saying it's because of the free variable which we can make whatever we want, so if we have $x=5t-4$ and $y=3t-5$ and $0=0$ (this being the third row of zeroes) then there's infinite solutions because $t$ can be whatever we want.
What I don't get is how if we looked at $t=2$ then $x$ and $y$ aren't equal. This may be my understanding in the geometry as I'm thinking of two lines on a graph, and that each intersection is a solution, which obviously these lines are different so they'd have one intersection. So basically what I'm asking is, what exactly is the free variable? Is the value of $x$ and $y$ important to the number of solutions? Should I stop thinking of these as lines intersecting on a graph? Lastly what should I be thinking about geometrically when I see this kind of matrix?
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$\begingroup$In a $4\times3$ matrix, every row corresponds to a plane, like
$$3x-y+2z=4.$$
If you choose two coordinates, the third is implied.
Three non-parallel planes intersect in a single point.
But the plane of equation
$$0x+0y+0z=0$$ is very special in that you can plug any coordinates and the equation is fulfilled. In fact, it doesn't define a plane but the whole space.
Then the intersection of two planes and the whole space is a straight line, with infinitely many points.
Similarly, with two zero rows, you get the intersection of a plane and the whole space, hence the same plane.
$\endgroup$ $\begingroup$Short answer: it affects the rank of the coefficient matrix. What is important for whether there are infinitely many solutions is the relation between that rank and the number of columns. The rank is at most the number of nonzero rows.
Consider an $m \times n$ system $Ax = b$ ($m$ equations in $n$ unknowns). If the augmented matrix $[A \mid b]$ has a row of $0$'s, that row conveys no useful information: it just corresponds to the equation $0 = 0$. But it means the rank of $A$ is at most $m-1$. If the rank is less than $n$, and the system is consistent (so there is at least one solution), then there are infinitely many solutions.
$\endgroup$ $\begingroup$Each line in your augmented matrix represents a plane in space.
"Almost surely" 3 planes intersect at a single point.
"Almost surely" is math speak to say that if you are using a random number generator to create your equations, it will fail to happen with 0 probability. But if you are engineering the equations that is a different story.
If they intersect at a single point, we say that the equations are linearly independent.
The degenerate problems:
3 planes may intersect in a single line. Any pair of planes intersects forming a line, but the 3 lines are parallel. 2 planes (or all 3) are parallel to one another.
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