Does the existence of a free variable mean there are infinitely many solutions?

Question

lets say I have an augmented matrix

$$\begin{bmatrix} 2&0&3\\ 0&1&0\\ 2&1&3 \end{bmatrix}$$

The question is: Is the matrix consistent with a unique solution, inconsistent, or consistent with an infinite solution?

row reducing gives:

$$\begin{bmatrix} 1&0&\frac32\\ 0&1&0\\ 0&0&0 \end{bmatrix}$$

the answer given is: consistent, unique solution.

However, I thought if there is a free variable that indicates infinite solutions? Wouldn't the 3/2 be a free variable?

Answer 1

Since we are considering an augmented matrix, the related system of equations is

• $2x=3$
• $y=0$
• $2x+y=3$

then the answer is correct, the system is consistent and we have a unique solution.