Theorem: Let A be a square matrix. there exists an invertible matrix B such that BA equals a triangular matrix.
case 1 :
if A = 0, then for every B : BA = 0, which is a triangular matrix.
if A is regular then take B = A inverse, and then BA=I ,a triangular matrix
cas3 : (problem here )
if A does not equal to the zero matrix and also not invertible:
BA=C is the matrix whose columns are : Ba1 Ba2 ... Ban.
let B, be B inverse. so multiply by B' from the left you get : B'c1 = a1; B'c2 = a2; ...; B'cn = an;
B' is regular therefore row equivalent to In and therefore there's exists a uniqe solution for any system of equations of the form above. we built T 1-1 and onto such that T(B') = B'c = a'
a' is a column of a matrix A such that BA = C.