I have a system of linear equations with n variables

\begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = \frac{1}{2}x_1\\[4pt] a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = \frac{1}{2}x_2\\[4pt] \qquad\dots\\[4pt] a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = \frac{1}{2}x_n\\ \end{cases}

where $a_{ij} \in \mathbb{Z}$ ($i, j \in \mathbb{N}$)

I have to show that this system has exactly one solution. As I see it, I should show that the matrix of this system is non-singular. I tried using Gauss method but didn't get far.

  • $\begingroup$ I made the assumption that Z and N standards for the integers and natural numbers respectively in my edit. Please check to see if that is intended. $\endgroup$ – Willie Wong Apr 29 '13 at 12:31

Hint You should move the "right hand side" to the left and re-write the system as

$$ (\mathbf{A} - \frac12 \mathbf{I})\vec{x} = 0 $$

where $\mathbf{I}$ is the identity matrix. Now analyze the matrix $\mathbf{A} - \frac12 \mathbf{I}$ to see if it is singular.

Hint 2 We want to consider $\det(\mathbf{A} - \frac12 \mathbf{I})$ and show that this is non-zero. Let us be more general and consider the characteristic polynomial $\det(\mathbf{A} - \lambda \mathbf{I})$ as a polynomial in $\lambda$. We want to show that $\frac12$ cannot be a root of this polynomial. This follows from the fact that monic polynomials in one variable with integer coefficients cannot have rational roots away from the integers.

  • $\begingroup$ A - (1/2)I = 0 only if |A| = 1/2. But since every a ∈ Z => (from determinant definition) |A| ∈ Z => |A| ≠ 1/2 and therefore the system matrix is nonsingular meaning that the system of linear equations has one solution. This is correct, right? (sorry for the bad formating) $\endgroup$ – käyrätorvi Apr 29 '13 at 12:44
  • $\begingroup$ $\mathbf{A}$ and $\mathbf{I}$ are matrices. How do you tell if a matrix is singular? $\endgroup$ – Willie Wong Apr 29 '13 at 12:45
  • $\begingroup$ The matrix is singular if the determinant of a matrix equals 0. It seems that my previous thought was wrong, for some reason i thought that |A-B| = |A| - |B| $\endgroup$ – käyrätorvi Apr 29 '13 at 12:49
  • $\begingroup$ How can I show that A - (1/2)I is never singular? $\endgroup$ – käyrätorvi Apr 29 '13 at 13:15
  • $\begingroup$ I don't understand your question: you are given a system of equations. You know that you need to compute the determinant to check if the determinant is zero. So just compute! $\endgroup$ – Willie Wong Apr 29 '13 at 13:43

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