How to prove linear system consistency?

[untold socket]

The answer for the question below is "only one solution", but how to prove that the system is consistent?

[gusty forge]

Have you taken linear algebra already?

[untold socket]

The idea is that I started learning linear algebra from scratch (I've finished the course at uni already, but I was very bad), and the only things that I've learned in this book are row echelon forms (including reduced ones), elimination methods and, of course, linear systems.
I know more things, but I wanted to have a conclusion based on what I've learned this far from the book.

[gusty forge]

Okay nice. Have you looked at matrices and linear dependency already?

[untold socket]

It depends on what particularities you mean

I know determinants, inverses, transpose matrix, etc.
For linear dependency, I know the general idea, not so much though.

[gusty forge]

Well in this problem for example, consider the matrix form of the first equation, with some matrix A holding all the coefficients.

Then you get

A * (x,y,z) = 0

Now there are only two ways to get the zero vector in matrix-vector multiplication. Do you know what they are?

[untold socket]

You mean O matrix? I only know how to obtain the identity one

[gusty forge]

(x,y,z) is supposed to be a column vector

[untold socket]

Yes

[gusty forge]

The matrix is composed of the coefficients of the linear equation system.

[untold socket]

Yes, I know this.

[gusty forge]

Right, so under which circumstances can this matrix transform a vector to the zero vector?

[untold socket]

If the 3 lines intersect into a single point?

I know that the 0 dimensionality is a point

or planes in this scenario

[gusty forge]

But we are talking about matrices right now, not lines.

Matrices transform vectors when multiplied with them.

[untold socket]

yes, I know that dimensionality can change when multiplying

[gusty forge]

Only if the matrix has a certain property. Which would be...?

Talking about square matrices of course.

[untold socket]

You said square matrices, so my thought goes to det(M) neq 0? (Most likely not)

but it doesn't make sense

[gusty forge]

So you're saying only a matrix with determinant not equal to 0 can demote vectors to a lower dimension or 0, that is, project for example all 3D vectors onto a plane?

Not quite but close.

[untold socket]

Ah I remember something similar with what you're saying. I've seen something on youtube that said that only when the det is 0, then the dimensionality goes from a certain number to a lower one.

Or am I wrong.

[gusty forge]

No, that is exactly right.

You can think of the determinant of a matrix as a measure of change of space. If it is 0, then that means that the matrix projecting all vectors into a lower dimensional space.

[untold socket]

So some lines become orthogonal this way?

[gusty forge]

No. Forget about lines for a second, they're not relevant right now. Think about this exclusively in terms of vectors and how they may be transformed.

So in the initial task, what does it say about the matrix of coefficients, if there is only one vector for which the transformation results in the zero vector?

[untold socket]

Sorry for taking so much, but I don't really understand what we're trying to reach.
When there exist only one vector that transforms the result into the zero vector, I think about 3 leading 1s, no free variables, so only unique solution when trying to solve the system. (because the system either has 0, 1 or an infinite amount of solutions).
I don't really know what can be said about the matrix, besides the fact that it is **irreversible **. We cannot go back from the lower dimensionality to a higher one without losing information

[gusty forge]

So this is the equation system rewritten as a matrix-vector product. We want to know how many rows of the matrix are linearly independent. We know that, if any row is not linearly independent from the others, the determinant of the matrix will be zero and thus we could choose infinitely many (x,y,z)'s for which this equation would hold true. But that is not the case. There is only one vector for which it holds, and it is "trivial".

[untold socket]

Yes, the trivial solution for homogenous systems, I've heard about it.

[gusty forge]

So there are two scenarios here:
Either the determinant of the matrix is zero, in which case the solution vector might take many forms and still satisfy the equation.
Or the determinant is non-zero, which would mean that there is only one vector that can satisfy this equation: The zero vector.

Since it is stated in the task that there is only one, "trivial" solution (the zero vector), then that must mean that the determinant of the matrix is non-zero.

What implications does this have on the linear (in)dependence of the row and columns of the matrix?