A $2n\times 2n$ matrix $A=[a_{ij}]$ has its elements as
$a_{ij}=\{\begin{array}{ll}\beta & \text{if }(i+j)\text{ is odd,}\\ -\beta & \text{if }(i+j)\text{ is even,}\end{array}$
where $n$ is any integer greater than 2 and $\beta$ is any non-zero real number. The rank of $A$ is

**Understanding the Rank of a Special $2n\times 2n$ Matrix**

This problem asks us to find the rank of a specific type of $2n\times 2n$ matrix, denoted by $A$. The matrix elements $a_{ij}$ are defined based on the sum of their indices, $i+j$. Let's break down the definition and analyze the matrix structure to determine its rank.

Matrix Definition and Elements

The matrix $A=[a_{ij}]$ is a $2n\times 2n$ matrix where $n>2$ and $\beta \ne 0$. The elements are given by the rule:

- $a_{ij}=\beta$ if $(i+j)$ is odd.
- $a_{ij}=-\beta$ if $(i+j)$ is even.

We can express this rule more compactly using powers of -1. Notice that:

- If $(i+j)$ is even, $(-1{)}^{i+j}=1$. The element is $-\beta$.
- If $(i+j)$ is odd, $(-1{)}^{i+j}=-1$. The element is $\beta$.

This implies that the element $a_{ij}$ can be written as:

$a_{ij}=-\beta (-1{)}^{i+j}$

Let's write down the first few elements of the matrix to see the pattern:

- $a_{11}$: $(1+1)=2$ (even) $⟹a_{11}=-\beta$
- $a_{12}$: $(1+2)=3$ (odd) $⟹a_{12}=\beta$
- $a_{21}$: $(2+1)=3$ (odd) $⟹a_{21}=\beta$
- $a_{22}$: $(2+2)=4$ (even) $⟹a_{22}=-\beta$

For a $2n\times 2n$ matrix, the structure looks like this:

$A=\left(\begin{array}{ccccccc}-\beta & \beta & -\beta & \beta & \dots & -\beta & \beta \\ \beta & -\beta & \beta & -\beta & \dots & \beta & -\beta \\ -\beta & \beta & -\beta & \beta & \dots & -\beta & \beta \\ \beta & -\beta & \beta & -\beta & \dots & \beta & -\beta \\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ -\beta & \beta & -\beta & \beta & \dots & -\beta & \beta \\ \beta & -\beta & \beta & -\beta & \dots & \beta & -\beta \end{array}\right)$

Analyzing Row Relationships in Matrix A

To find the rank of a matrix, we can examine the linear dependence between its rows (or columns). Let $R_i$ represent the $i$ -th row of matrix $A$. We want to see how $R_{i+1}$ relates to $R_i$.

Consider the elements of row $i$ and row $i+1$ at column $j$:

- $a_{ij}=-\beta (-1{)}^{i+j}$
- $a_{i+1,j}=-\beta (-1{)}^{(i+1)+j}$

Let's compare $a_{i+1,j}$ to $a_{ij}$: $a_{i+1,j}=-\beta (-1{)}^{i+j+1}=-\beta (-1{)}^{i+j}(-1{)}^1=(-1)\times (-\beta (-1{)}^{i+j})=-a_{ij}$

This relationship, $a_{i+1,j}=-a_{ij}$, holds true for all columns $j$ (from $1$ to $2n$). This means that the $(i+1)$ -th row is exactly $-1$ times the $i$ -th row:

$R_{i+1}=-R_i$

Let's apply this relationship:

- $R_2=-R_1$
- $R_3=-R_2=-(-R_1)=R_1$
- $R_4=-R_3=-R_1$
- $R_5=-R_4=-(-R_1)=R_1$

Continuing this pattern, we see that every row in the matrix $A$ is either a scalar multiple of $R_1$ (specifically, $R_k=(-1{)}^{k-1}{R}_1$) or identical to $R_1$ or $-R_1$. This demonstrates that all rows are linearly dependent on the first row, $R_1$.

Deriving the Rank of Matrix A

The rank of a matrix is defined as the dimension of its row space (or column space), which is the maximum number of linearly independent rows (or columns).

From our analysis, the row space of matrix $A$ is spanned by the first row, $R_1$.

The first row is $R_1=(-\beta ,\beta ,-\beta ,\beta ,\dots ,-\beta ,\beta )$.

Since we are given that $\beta$ is a non-zero real number ($\beta \ne 0$), the first row $R_1$ is not the zero vector.

Because the row space is spanned by a single, non-zero vector ($R_1$), the dimension of the row space is 1.

Therefore, the rank of the matrix $A$ is 1.

Conclusion: All rows of the matrix are linearly dependent, being scalar multiples of the first row. Since the first row is non-zero, the rank of the matrix is 1.