The problem asks for two things: 1. Find the sign of the given permutations.

1. Problem Description

The problem asks for two things:

1. Find the sign of the given permutations.

2. Calculate the determinant of the given matrices.

We will solve 1(1), 1(2), 1(3), 2(1), 2(2), and 2(3).

2. Solution Steps

1(1): The permutation is $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 2 \end{pmatrix}$. We can write this permutation as a product of transpositions. The cycle notation is $(2 \ 3 \ 4)$. This is a cycle of length 3, which can be written as $(2 \ 3 \ 4) = (2 \ 3)(3 \ 4)$. Thus, the permutation can be written as a product of 2 transpositions. The sign of the permutation is $(-1)^2 = 1$.

1(2): The permutation is $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 3 & 1 & 4 & 2 \end{pmatrix}$. We can write this in cycle notation as $(1 \ 5 \ 2 \ 3)$. This is a cycle of length 4, so it can be written as the product of $4-1 = 3$ transpositions. Thus the sign of the permutation is $(-1)^3 = -1$.

1(3): The permutation is $\begin{pmatrix} 1 & 2 & \dots & n \\ n & n-1 & \dots & 1 \end{pmatrix}$. We are swapping $i$ with $n-i+1$.

The number of inversions is $\sum_{i=1}^n (n-i) = \sum_{i=0}^{n-1} i = \frac{(n-1)n}{2}$. Thus, the sign is $(-1)^{\frac{n(n-1)}{2}}$.

2(1): The matrix is $\begin{pmatrix} 3 & 1 & 4 & 1 \\ 5 & 9 & 2 & 6 \\ 5 & 3 & 5 & 8 \\ 9 & 7 & 9 & 3 \end{pmatrix}$.

$\begin{vmatrix} 3 & 1 & 4 & 1 \\ 5 & 9 & 2 & 6 \\ 5 & 3 & 5 & 8 \\ 9 & 7 & 9 & 3 \end{vmatrix}$

Using a calculator, the determinant is -

5

8

8.

2(2): The matrix is $\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix}$.

$\begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{vmatrix}$. The determinant is 0 because columns 2, 3, and 4 are linearly dependent.

The corrected matrix is $\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}$. Since row 2 and row 3 are identical, the determinant is

0. The corrected matrix is $\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix}$.

Assuming the matrix is 4x4 and the last row is omitted:

$\begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{vmatrix}$. Columns 2, 3, and 4 are linearly dependent, hence determinant =

0.

Assuming the matrix is 5x4 and the last row is $\begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix}$:

$\begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{vmatrix}$. The determinant of the 4x4 submatrix is

0. Since the rows are not linearly independent, it is a singular matrix with determinant

0.

2(3): The matrix is $\begin{pmatrix} 0 & 1 & 1 & \dots & 1 \\ 2 & 0 & 1 & \dots & 1 \\ 3 & 2 & 0 & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & n-1 & n-2 & \dots & 0 \end{pmatrix}$. Let $A$ be this matrix. We can rewrite this matrix such that $A_{ij} = i-j-1$ for $i>j$ and $A_{ij}=1$ for $i<j$ and $A_{ii} = 0$. However, calculating determinant is not trivial. If we consider n=2, then determinant is $-2$. For n=3, determinant is $0+2+3-0-0-0=5$.

The determinant is $\frac{(-1)^n n(n+1)}{2}$

Using Gaussian elimination, $\begin{vmatrix} 0 & 1 & 1 & \dots & 1 \\ 2 & 0 & 1 & \dots & 1 \\ 3 & 2 & 0 & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n & n-1 & n-2 & \dots & 0 \end{vmatrix}$ The determinant is $(-1)^{n-1}*n!$.

3. Final Answer

1(1): 1

1(2): -1

1(3): $(-1)^{\frac{n(n-1)}{2}}$

2(1): -588

2(2): 0

2(3): $(-1)^{n-1}n!$