Vtyjkyuk

Today I learnt non square matrices generally cannot have determinants because the matrix cannot fulfill all properties of determinants as seen in square matrices. So I want to know if given a system of linear equations whose matrix is non square can be solved if it can't have determinants.

Totally. Search Gauss elimination, and you would find something related. Also for overdetermined systems, we can find the best approximation in some sense. Check Linear Algebra textbooks of you like.

They don't use Gaussian elimination for non-square matrices. This is a series of charts from the matlab site for the backslash operator which is used to solve systems of equations i.e. $ Ax= b$

ml-divide full, for non-sparse matrices

ml-divide for sparse matrices

as you can see it when it isn't square it uses the $QR$ decomposition. That is

$$ Ax =b \ QRx = b \ Rx = Q^-1b $$
then it uses back substitution to solve this

It is also possible to use the SVD decomp

$$ Ax=b \ U Sigma V^T x = b \ x = V Sigma^-1U^Tb $$

A fruitful way of thinking about an $n times m$ matrix $A$ is as a function $A:mathbbR^m rightarrow mathbbR^n$. So $A$ acts on every vector $v in mathbbR^m$ by left multiplication, giving a new vector $Avin mathbbR^n$. If $n=m=2$ you can think of $A$ as a way of moving around the plane, keeping straight lines straight, and keeping the origin fixed. Because of this, $A$ sends every $1 times 1$ square in the plane to a parallelogram, and all of those parallelograms have the same area. The (absolute value of the) determinant of $A$ tells you that area.

If general for $m=n$, the matrix (though of as a function) $A:mathbbR^n rightarrow mathbbR^n$ sends any $underbrace1 times dots times 1_n textrm times$ hypercube to a parallelotope and all such parallelotopes have the same volume. $det(A)$ tells you the signed $n$ -dimensional volume of such a parallelotope.