Can a linear system of equations be solved if it has all zeros in the column matrix on the right hand side of the equation?

Question

I have an awkward linear system of equations that I am trying solve. But it has a column matrix containing only $0$s on the right hand side. I want to know if this type of system can even be solved. I tried constructing a simpler set of equations to see if it is possible (I assumed $x=1, y=2, z=3$ and made equations with these values in mind). The simpler system of equations:

$-2x + y = 0,$

$3y -2z = 0,$

$3x - z = 0 $

But I cannot think of way that can be used to solve even this simple system. Conventional way of inversing the co-efficient matrix and then multiplying it with the column matrix on the right hand side yields another column matrix containing only 0 s. Please tell me if these systems can even be solved, and if so, how.

Answer 1

The comment by @Hagen von Eitzen is a perfect answer.

In your case, the system is a system of homogeneous equations.

Such systems always have the zero vector as a solution. In addition it may have other solutions. This can be found by either inspection by calculating the determinant.

Calculating the determinant: $$\det \begin{pmatrix}-2&1&0\\ \:0&3&-2\\ \:3&0&-1\end{pmatrix} = 0$$

The determinant "determines" whether the system has a unique solution.This occurs precisely if the determinant is non-zero.

So in your case, the system has a trivial solution $x=0$,$y=0$,$z=0$ in addition to infinitely number of solutions. To find the general form of those infinite solutions, you can let $z=Constant$, so let $$z=k$$.

Now use this value in equation (2) to get:

$$y=\frac{2k}{3}$$

Now use this value for $y$ in equation (1) to determine the value of $x$:

$$-2x+ \frac{2k}{3}=0$$

This leads to:

$$x=\frac{k}{3}$$

Answer 2

We can represent the system in matrix form $Ax=0$ as

$$\begin{bmatrix}-2& 1 &0 \\0&3&-2\\3&0&-1\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$$

Recall that we can view the above matrix multiplication as a combination of the columns of matrix $A$ (i.e. $x_1\cdot \vec{a_1} + \dots + x_3\cdot \vec{a_3}$). A useful high-level interpretation of the question of whether we are going to be able to find a non-trivial solution: is there some linear combination of the columns of $A$ that give the zero vector?

The answer to the above is not immediately obvious in this case. So along with many other methods for determining the answer to this (such as computing the determinant of $A$), you could perform Gaussian Elimination on the matrix to shed light on the nullspace (the subspace containing all solutions $x$ to our $Ax=0$). For more on this, see Strang: Introduction to Linear Algebra; Section 3.2)