The equation $Ax=b$ is non-homogeneous, meaning that $b \neq 0$
I also know that the linear system has no free variables. What is the number of possible solutions that the equation can have?
I think the answer is $1$, but I am not sure, can someone fact-check me?
Here is my reasoning
No free variables, means that there are no variables not in pivot columns (e.g. all columns are pivot columns) - so the augmented row-echelon form of $Ax=b$ would be something like
$$\begin{bmatrix} A_{11} & A_{12} & A_{13} & b_1\\ 0 & A_{22} & A_{23} & b_2\\ 0 & 0 & A_{33} & b_3 \end{bmatrix}$$
Since we know there are 3 pivot columns, it becomes this
$$\begin{bmatrix} A_{11} & 0 & 0 & b_1\\ 0 & A_{22} & 0 & b_2\\ 0 & 0 & A_{33} & b_3 \end{bmatrix}$$
Since $A_{11} A_{22} A_{33}$ span $R^3$, there can only be one solution whatever $b$ is
Am I right?
no free variables, but I don't see offhand how that excludes the case of $0$ solutions. $\endgroup$Pivot columns are an artifact of a method of solving linear systems, not an intrinsic property thereof$\endgroup$