If I subtract 1 from the n,n entry of a Pascal Matrix, why does the determinant become zero?

Question

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.

Answer 1

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 19 \\ \end{array} \right) $$

Compare

$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \\ \end{array} \right) $$

five by five

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 \\ 1 & 4 & 6 & 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 & 6 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 & 10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 69 \\ \end{array} \right) $$

$$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 1 & 3 & 3 & 1 & 0 \\ 1 & 4 & 6 & 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 & 6 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 & 10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \\ \end{array} \right) $$

Answer 2

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. $\blacksquare$

Answer 3

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.