Set of linear equations with coefficients - solution using matrices

Question

I have a set of linear equations: \begin{matrix} ax_{1}& {}+bx_{2}& {}+x_{3}& & =0\\ cx_{1}& {}+dx_{2}& &{}-x_{4} & =0\\ & {}-ex_{2}& {}+cx_{3}& {}+ax_{4}& =0\\ & {}+ex_{2}& {}+dx_{3}& {}+bx_4& =0 \end{matrix} I should find the conditions for the coeficients $(a,b,\ldots)$, so that this set of linear equations has a non-zero solution. I should use a matrix or determinant or any solution based on matrices, because it is what we are learning at the school now.

Sorry if I made any mistakes. I am not so good in English.

Thank You for Your help!!!!

UPDATE:

Regarding @Alex Silva answer, I've constucted the matrix A made of the coefficients in front of each x:

$ A=\begin{pmatrix} a &b &1 &0 \\ c &d &0 &-1 \\ 0 &-e &c &a \\ 0 &e &d &b \end{pmatrix} $

Then I transposed the matrix so that I could check the linear dependence of its columns: $ A^{T} = \begin{pmatrix} a &c &0 &0 \\ b &d &-e &e \\ 1 &0 &c &d \\ 0 &-1 &a &b \end{pmatrix} \sim \begin{pmatrix} 1 &0 &c &d \\ 0 &-1 &a &b \\ b &d &-e &e \\ a &c &0 &0 \end{pmatrix} $

Having the Gaussian elimination in mind, I thought: under the main diagonal I should have only zero. So b, d, a, c = 0

Then: $ A^{T}= \begin{pmatrix} 1 &0 &0 &0 \\ 0 &-1 &0 &0 \\ 0 &0 &-e &e \\ 0 &0 &0 &0 \end{pmatrix} $

I transpose the $A^{T}$ back and I get:

$ A=\begin{pmatrix} 1 &0 &0 &0 \\ 0 &-1 &0 &0 \\ 0 &0 &-e &0 \\ 0 &0 &e &0 \end{pmatrix} $

The last two lines are linearly dependent and I always get an non-zero solution. e can be any number.

Am I right?

Answer 1

Hint: Construct the matrix $\mathbf{A}$ such that $\mathbf{A}\mathbf{x} = \mathbf{0}$, and assure the linear dependence of its columns.

Answer 2

The condition on the coefficients for your system to have a non-zero solution is to ensure that the rank of your matrix is less than 4 (the number of variables). This is the same as saying that your equations must be linearly dependent.

Also, you don't need to transpose your matrix or distinguish in any way between rows and columns, because for any matrix the row-rank and column-rank (and determinant-rank if the matrix is square) are all the same.