Rank of an $n×n$ matrix $A$ is $n-1$. Can we say anything about the system $Ax=b$?

Question

I have been given this question and the options are

1. System has $n-1$ parameter family of solutions
2. One parameter family of solutions
3. No solution
4. A unique solution

But if the rank of the augmented matrix is not known, how can we conclude anything? If the rank of the augmented matrix is equal to that of the coefficient matrix $A$, then I can say that the system has a one parameter family of solutions; and otherwise no solution, right?

Answer 1

• "Unique solution" is not possible, because $Ax=0$ has infinitely many solutions.

• "$(n-1)$-parameter family of solutions" is not possible, as it would require $A$ to have rank 1 (unless $n=2$, in which case $n-1=1$).

• "$1$-parameter family of solutions" is possible if the system is consistent.

• "No solution" is possible if the system is inconsistent.

Answer 2

They can both be true by the Rouchè-Capelli's theorem, if $\rho(A)=\rho(A\big|b)\rightarrow \dim(\text{Sol})=n-\rho(A)$ otherwise the sistem is unsolvable (Sol is the solution space). In our case, it means that a)the system has 1-parameter solutions (which might coincide with the $n-1$ parameter space, if $n=2$ b)the system has no solution. Let's have a look at an example:

$\begin{align} \begin{bmatrix} 1 & 1 & 2\\ 1 & 1 & 1\\ 2 & 2 & 2\\ \end{bmatrix} X=(3,1,2)^{t}\\ \begin{bmatrix} 1 & 1 & 2\\ 1 & 1 & 1\\ 2 & 2 & 2\\ \end{bmatrix} X=(3,1,5)^{t} \end{align}$

While the first one has a 1-dimensional space of solution, the second one has no solution