Can Someone prove this too?
I have a question, I have uploaded a picture of the proof, and I understood everything, however there is something that isn't clear, (it's in Hebrew but I think you can understand what he did.) So in the last part he defined a new permutation sigma from 1 to k1, 2 to k2,..... n to kn, and then he changed the the expression and said that it's equal to detA. However the sign(sigma) (of permutation) is missing? Is there a mistake there? becauseif there is no sign maybe it will be equal to -det(A) at the end? Appreciate it if you can help me out. Thank you.
Couldn’t resist opportunity to promote Einstein summation convention to answer your question:
Let A,B,C be matrices, and lAl be detA, etc.
If AB=C then lABl=lCl=lAllBl
1) lAl=eijka1ia2ja3k, definition of determinant
2) eijkariasjatk=erstlAl, Permute rst starting with 123. Any transposition corresponds to a transposition of rows, which changes sign.
=lBlersta1ra2sa3t, from 1) and 2)
One caveat, think of eijk… for larger matrices.
note: repeated index is a dummy index: aibi =ajbj=a1b1+a2b2+a3b3+…
also note bonus for determinant of matrix product.
Better than wading through all the summation signs, isn't it?