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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)?
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)?
