Linear Algebra Rank of a matrix let $ n \ge 3$ and create an $n$ x $n$ matrix $A$ by defining $A_{ij} = \alpha i +\beta j + \gamma$, where $\alpha , \beta$ and $\gamma$ are three arbitrary positive numbers. What is the rank of $A$?
Can someone please help me I don't understand how to solve this one 
 A: The rank of a matrix is defined by how many linearly independent rows and columns are there.
Now, if you subtract the first row from the second one, you will get elements $\beta, \beta, \beta,\beta ... $. After all, the corresponding numbers in these rows differ only by their memeber $\beta j$. The first row has $j=1$ the second $j=2$. 
The same difference is between any two adjacent rows. So the rank is 2. You can get every other row from the linear combination of the first two. With $\gamma=0$ it would be just $1$ because all rows would just be $j$ times the first one.
A: Let
$$B=\left(\begin{array}{ccccc}
1&0&0&\cdots&0\\
-1&1&0&\cdots&0\\
\vdots&&\ddots&&\vdots\\
0&&\cdots&-1&1
\end{array}\right)$$
That is, $1$'s in the main diagonal, and $-1$'s just under them. It is clear that $B$ is a regular matrix.
Then
$$AB=\left(\begin{array}{ccccc}
\beta&\beta&\ldots&\beta&A_{n1}\\
\beta&\beta&\ldots&\beta&A_{n2}\\
&&\vdots&&\\
\beta&\beta&\ldots&\beta&A_{nn}\\
\end{array}\right)$$
The rank of $BA$ (and, hence, of $A$) is $2$ if the numbers $A_{in}$ are not equal (that is, if $\alpha\neq0$) and $\beta\neq0$. If $\alpha=\beta=\gamma$, the rank is $0$. Otherwise, the rank is $1$.
