Solution of a system of linear equations in a non-quadratic matrix

Question

I have a system of linear equations that makes up a matrix (i.e. Non-square) that I need to solve - or at least try to solve, to show that there is no solution for the system. (most likely there will be no solution) NxM

As I understand it, if my matrix is ​​not square (overdetermined or underdetermined), then the exact solution cannot be found . Am I considering it right? Is there a way to convert my matrix into a square matrix to calculate deterministic, apply the Gauss rule, Cramer rule, etc.

It may be worth mentioning that the coefficients of my unknowns can be zero, so in some rare cases it would be possible to have a zero column or a zero row.

Answer 1

Regardless of whether your matrix is ​​square, this is not what defines the solution space. This is the rank of the matrix compared to the number of columns that determines this (see rank oddness theorem ). In the general case, you can have zero, one, or an infinite number of solutions of a linear system of equations depending on the ratio of rank and residual.

, , , , , x0 Null (A) . x = x0 + xn, xn Null (A). , , , . , . 1, , x0, , .

Answer 2

, :

[ 1 0 0 ][x] = [1]
[ 0 0 1 ][y]   [1]
         [z] 

(, : x = z = 1). , , :

[ 1 0 ][x] = [1]
[ 0 1 ][y]   [1]
[ 1 1 ]      [2]

( = = 1). , , , "" ( ), .

Answer 3

- .

, (SVD), .

Answer 4

Ax = b, A m n . - , , , (m n). - , , .

, , , b, , A. ( ).

, A, b. ? , , , . , .

, . . , . aT b. AT b.

: AT Ax = AT b. - , AT A:

x = (AT A)^-1 * AT b