What is the Induction Process for Finding det(alpha*A)?

[DMOC]

Old topic: https://www.physicsforums.com/showthread.php?t=194725

I have a question that's the same as the one in the old topic/thread I linked to above. At the last post, I am trying to figure out how to carry out the induction process.

$\sum_{j=1}^n (-1)^{i+j} a_{ij} det(\alpha A_{ij})$

I understand that all $det(\alpha A_{ij})$ will be k x k matrices because they are basically (k+1)x(k+1) matrices but without the ij row/column so those matrices will fall under the range of induction, i.e. between 2 and k, inclusive.

But if I substitute $det(\alpha A_{ij})$ with $\alpha^n det(A_{ij})$ I will get $\sum_{j=1}^n (-1)^{i+j} a_{ij} \alpha^ndet(A_{ij})$

Where do I get the extra n so that it becomes alpha ^ n + 1. (Note: when I write out the actual proof, I'll use alpha^(k+1) instead of n + 1)?

 

[Dick]

a_ij is supposed to be the ijth element of the matrix alpha*A, not A. You are trying to find det(alpha*A). It should be alpha*a_ij.