Linear system with constants in arithmetic progression I have stumbled upon a mathematical problem, that could be translated into English like this:

There's a system of n linear equations with n unknowns, and with constants that are in arithmetic progression like: 6x + 9y = 12 and 15x + 18y = 21. This system has a unique solution. Find this solution.

I tried to solve this problem for a few different values of n, and it seems that to find the solution, one has to assume n = 2. But how to prove mathematicaly that that such a system of linear equations has a unique solution if and only if n = 2?
Thank you in advance for your help.
 A: It's $$2x+3y=4$$ and $$5x+6y=7.$$ 
The first gives $$4x+6y=8$$ and from here $$x=-1$$ and $$y=2.$$
In the general we have $$ax+(a+d)y=a+2d$$ and
$$(a+3d)x+(a+4d)y=a+5d.$$
We have:
$$\Delta=a(a+4d)-(a+3d)(a+d)=-3d^2,$$
$$\Delta_x=(a+2d)(a+4d)-(a+5d)(a+d)=3d^2$$ and
$$\Delta_y=a(a+5d)-(a+3d)(a+2d)=-6d^2.$$
Thus, for $d\neq0$ we obtain:
$$x=\frac{\Delta_x}{\Delta}=-1$$ and $$y=\frac{\Delta_y}{\Delta}=2.$$
We can use also the following way.
Since $$(a+3d)x+(a+4d)y-(ax+(a+d)y)=a+5d-(a+2d),$$ we obtain $$d(x+y)=d,$$ which for $d\neq0$ gives $x+y=1$ or $y=1-x$.
Thus, $$ax+(a+d)(1-x)=a+2d$$ or
$$dx=-d$$ or $$x=-1$$ and $$y=2.$$
Actually, for $d=0$ your system has infinitely many solutions.
A: You are right, this is true if and only if $n = 2$. To prove this (the only if part), observe that if $a_1, a_2, a_3$ are in arithmetic progression, then $a_2 = \dfrac{a_1 + a_3} 2$.
Now, let the coefficient matrix be
$$A = \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}$$
where in each row, the entries are in arithmetic progression in the given order.
If $a$, $b$, and $c$ denote the three column vectors of $A$, then by our earlier observation, $b = \dfrac 1 2 (a + c)$. Thus, the three rows are linearly dependent, and the matrix is not invertible. Therefore, the system either has no solution, or has infinitely many solutions (since $\operatorname{rank} A$ is not equal to the number of unknowns). In fact, it has infinitely many solutions, if the right hand side of the system also follows the same arithmetic progression in each row.

$\DeclareMathOperator{\rank}{rank}$
A linear system $Ax = b$, where $A$ is an $m \times n$ matrix has a solution if $\rank A = \rank [A\ b]$, where $[A\ b]$ is the augmented matrix obtained by appending the column vector $b$ to the matrix $A$. If, further, this rank equals $n$, the number of unknowns (length of $x$), then the solution is unique.
The rank can be defined in different ways, but is equal to the total number of linearly independent rows as well the total number of linearly independent columns of the matrix.
Now, suppose $A$ whose columns are $A_1, A_2, \ldots, A_n$, and suppose that in each row of the augmented matrix $[A\ b]$, the entries are in arithmetic progression (as we move from the first column $A_1$ to the last, $b$). Then:


*

*$A_2 - A_1 = d$, where $d$ is the column vector of the respective common differences of the arithmetic progression — i.e., $a_{i, j + 1} - a_{ij} = d_i$, $i = 1, \ldots, m$, $j = 1, \ldots, n - 1$; and $b_i - a_{in} = d_i$ as well.

*Therefore, the $j$th column of $A$ can be written as $A_1 + (j - 1)d = (2 - j)A_1 + (j - 1)A_2$, which is a linear combination of $A_1$ and $A_2$. Similarly, $b = (1 - n)A_1 + n A_2$ as well.

*Thus, $\rank A = \rank [A\ b] = 2$ (unless all the common differences are zero, i.e., $d = 0$, in which the rank is $1$).


Therefore, the system always has a solution. The system is unique iff $n = 2$ (and $d \ne 0$).
A: For $n>2$, the columns vectors of the matrix are linearly dependent and the matrix not full rank.
Now for $n=2$,
$$\begin{cases}ax+(a+r)y=a+2r,\\bx+(b+s)y=b+2s.\end{cases}$$
For this system to have a unique solution, the condition
$$\Delta:=as-br\ne0$$ must be fulfilled.
Then for any choice of $a,b,r,s$,
$$\begin{cases}x=-1,\\y=2.\end{cases}$$
