# Prove a 3x3 system of linear equations with arithmetic progression coefficients has infinitely many solutions

How can I prove that a 3x3 system of linear equations of the form:

$\begin{pmatrix} a&a+b&a+2b\\ c&c+d&c+2d\\ e&e+f&e+2f \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix} =\begin{pmatrix} a+3b\\ c+3d\\ e+3f \end{pmatrix}$

for $a,b,c,d,e,f \in \mathbb Z$ will always have infinite solutions and will intersect along the line $r= \begin{pmatrix} -2\\3\\0 \end{pmatrix} +\lambda \begin{pmatrix} 1\\-2\\1 \end{pmatrix}$

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First, consider the homogeneous system $$\left(\begin{array}{ccc} a & a+b & a+2b\\\ c & c+d & c+2d\\\ e & e+f & e+2f \end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right).$$ If $(a,c,e)$ and $(b,d,f)$ are not scalar multiples of each other, then the coefficient matrix has rank $2$, so the solution space has dimension $1$. The vector $(1,-2,1)^T$ is clearly a solution, so the solutions are all multiples of $(1,-2,1)^T$. That is, the solutions to the homogeneous system are $\lambda(1,-2,1)^T$ for arbitrary $\lambda$.

Therefore, the solutions to the inhomogeneous system are all of the form $\mathbf{x}_0 + \lambda(1,-2,1)^T$, where $\mathbf{x}_0$ is a particular solution to this system. Since $(-2,3,0)$ is a particular solution always, then all solutions have the described form.

If one of $(a,c,e)$ and $(b,d,f)$ is a multiple of the other, though, then there are other solutions: the matrix has rank $1$, so the nullspace has dimension $2$. Say $(a,c,e) = k(b,d,f)$ with $k\neq 0$, then there is another solution: $(-1-\frac{1}{k},1,0)$ would also be a solution to the system, so that the solutions to the inhomogeneous system would be of the form $$r = \left(\begin{array}{r}-2\\3\\0\end{array}\right) + \lambda\left(\begin{array}{r}1\\-2\\1\end{array}\right) + \mu\left(\begin{array}{r}-1-\frac{1}{k}\\1\\0\end{array}\right).$$ This includes the solutions you have above, but also others. (If $k=0$, then you can use $(0,-2,1)$ instead of $(-1-\frac{1}{k},1,0)$)

If $(b,d,f)=(0,0,0)\neq (a,c,e)$, then $(1,0,-1)$ can be used instead of $(-1-\frac{1}{k},1,0)$ to generate all solutions.

And of course, if $(a,b,c)=(b,d,f)=(0,0,0)$, then every vector is a solution.

In all cases, you have an infinite number of solutions that includes all the solutions you give (but there may be solutions that are not in that line).

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+1 for special cases. I suppose a simpler solution would be to split the system into two 2x2 systems, and then prove each one of these for instance through matrix algebra ($\mathbf{AX=B, X=A^{-1}B}$). Would that be a valid proof? –  Milosz Wielondek Mar 16 '11 at 8:56
@Milosz: First, you can't invert every matrix, so $AX=B$ does not necessarily imply that $X=A^{-1}B$ makes sense. Second: it depends on what systems you break it into. –  Arturo Magidin Mar 16 '11 at 15:50

This follows immediately from a basic theorem of linear algebra. Namely, the solution of a nonhomogeneous linear system can be expressed as the sum of any particular solution plus the general solution of the homogeneous system. Since your exhibited solution has this form it is necessarily a solution, i.e. $\rm\ \ L(x_0) = a_0,\ L(x) = 0\ \Rightarrow\ L(x_0+\lambda\ x)\ =\ L(x_0)+\lambda\ L(x)\ =\ a_0\:.\ \$ This sort of "shifted" vector space structure is known as an affine space.

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