This was a sample test question and was not the answer to the question. I'm just curious about what the column of 0 does.

For reference if needed, the question was "Which of the following is the coefficient matrix for a homogeneous system Ax = 0 with only the trivial solution"

Here's the Matrix:


I personally would assume that there would be only 1 solution as despite there being no pivot in the 4th column, the values in the 4th column are all 0.

(I'm definitely doing this part wrong) Writing this in parametric vector form, I would get

x1 = 0

x2 = 0

x3 = 0

x4 = free

and now I'm kinda lost about what to do..

Taking a guess, would the result be:

\begin{bmatrix}0\\0\\0\\1\end{bmatrix} * x4?

To be honest, I'm not exactly sure how to examine this matrix..

Simply put, how does the 0 column affect this matrix? Is the initial matrix equivalent to a 3x3 identity matrix? How many solutions are there? Infinitely many or just 1?

  • $\begingroup$ A zero column in reduced row echelon form means that the corresponding variable is a free variable. So in your case $x_4$ can be set to anything $\endgroup$ – Riley Oct 9 '17 at 22:13

If $A$ is the matrix of your system, then you have three equations in four variables, $$ x_1=0,\\ \ \\ x_2=0,\\ \ \\ x_3=0. $$ And $x_4$ does not appear in the equations, so you are free to choose it, and there are infinitely many solutions.

Another possible interpretation, is that the matrix you were given is an augmented matrix $\begin{bmatrix} A& 0\end{bmatrix}.$ In that case, the system would have three variables and the unique solution $x_1=x_2=x_3=0$.


If $A$ has a column of zeros, then there is a solution with a free variable to the homogeneous equation $A \vec x = \vec 0$. To see this, let's take a small case, e.g. $2 \times 2$. If we have

$$\left[\begin{array}{cc} a & 0 \\ b & 0\end{array}\right] \left[\begin{array}{cc} x \\ y\end{array}\right] = \left[\begin{array}{cc} 0 \\ 0\end{array}\right]$$

then this corresponds to a system of equations

$$ax + 0y = 0 \quad \text{ and } \quad bx + 0y = 0$$

Neither equation actually depends on $y$, so this is a free variable; thus, any vector of the form $[0, y]^T$ is a solution.


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