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.
 A: Let's make this more general:

Proposition 1. Let $n$ be a positive integer. Let $A = \left(a_{i,j}\right)_{1\leq i\leq n, \  1\leq j\leq n}$ be an $n \times n$-matrix over a commutative ring $\mathbb{K}$. For each $u, v \in \left\{1,2,\ldots,n\right\}$, we let $A_{\sim u, \sim v}$ be the $\left(n-1\right)\times\left(n-1\right)$-matrix obtained from $A$ by removing the $u$-th row and the $v$-th column.
Let $p \in \mathbb{K}$. Let $B = \left(b_{i,j}\right)_{1\leq i\leq n, \  1\leq j\leq n}$ be the $n \times n$-matrix obtained from $A$ by subtracting $p$ from its $\left(n,n\right)$-th entry. (That is, each entry $b_{i,j}$ of $B$ equals the corresponding entry $a_{i,j}$ of $A$, except for the $\left(n,n\right)$-th entry $b_{n,n}$ of $B$, which is $a_{n,n} - p$ instead.)
Then, $\det B = \det A - p \det\left(A_{\sim n, \sim n}\right)$.

Proposition 1 tells you how the determinant of a matrix changes when you modify a single entry. This entry is the $\left(n,n\right)$-th entry in the case of Proposition 1, but you can similarly analyze any other entry (except that you need an extra $\left(-1\right)^{i+j}$ sign in the formula if you modify the $\left(i,j\right)$-th entry).
How would this help you with your question? Well, if $A$ is your original Pascal matrix of size $n$ (whose $\left(i,j\right)$-th entry is $\dbinom{i+j-2}{i-1}$ for all $i,j \in \left\{1,2,\ldots,n\right\}$), and if $p=1$, then the $B$ in Proposition 1 is precisely the modified Pascal matrix whose determinant you are looking for. Proposition 1 yields $\det B = \det A - \det\left(A_{\sim n, \sim n}\right)$ (since $p = 1$), but both $\det A$ and $\det\left(A_{\sim n, \sim n}\right)$ equal $1$ (because both $A$ and $A_{\sim n, \sim n}$ are Pascal matrices, and you know that those have determinant $1$), so this becomes $\det B = 1-1 = 0$ as you want to show.
It remains to prove Proposition 1:
Proof of Proposition 1. From $b_{n,n} = a_{n,n} - p$, we obtain $b_{n,n} - a_{n,n} = -p$.
Recall that the matrix $B$ differs from $A$ only in its $\left(n,n\right)$-th entry. Thus,
\begin{equation}
B_{\sim n, \sim j} = A_{\sim n, \sim j}
\qquad \text{ for each } j \in \left\{1,2,\ldots,n\right\}
\label{darij.1}
\tag{1}
\end{equation}
(since the matrices $B_{\sim n, \sim j}$ and $A_{\sim n, \sim j}$ are obtained from $B$ and $A$ in the same way, and this way does not depend on the $\left(n,n\right)$-th entries of $B$ and $A$).
Also,
\begin{equation}
b_{n,j} = a_{n,j}
\qquad \text{ for each } j \in \left\{1,2,\ldots,n-1\right\}
\label{darij.2}
\tag{2}
\end{equation}
(for the same reason).
Expanding the determinant of $B = \left(b_{i,j}\right)_{1\leq i\leq n, \  1\leq j\leq n}$ along the $n$-th row, we obtain
\begin{align}
\det B
&= \sum_{j=1}^n \left(-1\right)^{n+j} b_{n,j} \det\left(\underbrace{B_{\sim n, \sim j}}_{\substack{= A_{\sim n, \sim j} \\ \text{(by \eqref{darij.1})}}}\right) \\
&= \sum_{j=1}^n \left(-1\right)^{n+j} b_{n,j} \det\left(A_{\sim n, \sim j}\right) \\
&= \sum_{j=1}^{n-1} \left(-1\right)^{n+j} \underbrace{b_{n,j}}_{\substack{= a_{n,j} \\ \text{(by \eqref{darij.2})}}} \det\left(A_{\sim n, \sim j}\right) + \underbrace{\left(-1\right)^{n+n}}_{=1} b_{n,n} \det\left(A_{\sim n, \sim n}\right) \\
&= \sum_{j=1}^{n-1} \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) + b_{n,n} \det\left(A_{\sim n, \sim n}\right) .
\end{align}
On the other hand, expanding the determinant of $A = \left(a_{i,j}\right)_{1\leq i\leq n, \  1\leq j\leq n}$ along the $n$-th row, we obtain
\begin{align}
\det A
&= \sum_{j=1}^n \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) \\
&= \sum_{j=1}^{n-1} \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) + \underbrace{\left(-1\right)^{n+n}}_{=1} a_{n,n} \det\left(A_{\sim n, \sim n}\right) \\
&= \sum_{j=1}^{n-1} \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) + a_{n,n} \det\left(A_{\sim n, \sim n}\right) .
\end{align}
Subtracting the last two equalities from each other, we obtain
\begin{align}
& \det B - \det A  \\
& = \left( \sum_{j=1}^{n-1} \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) + b_{n,n} \det\left(A_{\sim n, \sim n}\right) \right) \\
& \qquad - \left( \sum_{j=1}^{n-1} \left(-1\right)^{n+j} a_{n,j} \det\left(A_{\sim n, \sim j}\right) + a_{n,n} \det\left(A_{\sim n, \sim n}\right) \right) \\
& = b_{n,n} \det\left(A_{\sim n, \sim n}\right) - a_{n,n} \det\left(A_{\sim n, \sim n}\right) \\
& = \underbrace{\left(b_{n,n} - a_{n,n}\right)}_{= -p} \det\left(A_{\sim n, \sim n}\right) \\
& = -p \det\left(A_{\sim n, \sim n}\right) .
\end{align}
Thus, $\det B = \det A - p \det\left(A_{\sim n, \sim n}\right)$. This proves Proposition 1. $\blacksquare$
A: 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.
