<s>[INST] If I subtract 1 from the n,n entry of a Pascal Matrix, why does the determinant become zero? In problem 19.2, how did the author reach this statement: "Since the n, n entry multiplies its cofactor positively, the overall determinant drops by 1 to become 0."
I was able to solve it a different way, but I feel that my approach was less efficient. I'd like to understand the solution's approach.
Here is what I did:
$det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&19\end{bmatrix}) = det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&20\end{bmatrix}) - det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix})$.
and
$ det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix}) = (-1)^3 * det(\begin{bmatrix}0&0&0&1\\1&1&1&1\\1&2&3&4\\1&3&6&10\end{bmatrix}) = (-1)^3 * (-1) * det(\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix})$.
therefore
$det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&19\end{bmatrix}) = 1 - 1 = 0 $.
It feels like the solution skipped doing this step, I don't get it.
 [/INST] Note: I don't have enough reputation to comment so I will try to make my point with your equations.
Your approach seems to be more efficient than the answers here to be honest. As I understand you only need to show the determinant below is equal to 1.
$
det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix})=1
$
Since all pascal matrices has a determinant of 1 this includes:
$
det(\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix})=1
$
Thus the first equation can be shown easily. I think this is as intuitive and efficient as it gets.
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