Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – 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.

share|cite|improve this answer
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) – Faz3r Apr 29 '13 at 12:44
$\mathbf{A}$ and $\mathbf{I}$ are matrices. How do you tell if a matrix is singular? – Willie Wong Apr 29 '13 at 12:45
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| – Faz3r Apr 29 '13 at 12:49
How can I show that A - (1/2)I is never singular? – Faz3r Apr 29 '13 at 13:15
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! – Willie Wong Apr 29 '13 at 13:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.