# Linear Algebra Solution Types Proven with Rank

Is there a way to determine the solution type of an augmented matrix through row reduction and the rank of the matrix?

To be more specific, I'm not having trouble with understanding solution types or how to solve an augmented matrix. My question is that is it possible to determine the TYPE of solution (unique, consistent, no solution, etc) through determining the rank of the augmented matrix and comparing it to the rank of the matrix on its own.

If the matrix $A$ has columns $\vec a_1, \dots, \vec a_n$, then a system of equations $A \vec x = \vec b$ is equivalent to $$x_1 \vec a_1 + x_2 \vec a_2 + \dots + x_n \vec a_n = \vec b$$ that is, to trying to write $\vec b$ as a linear combination of the columns of $A$. In other words, we are checking if $\vec b$ is in the column space of $A$.

Looking at the rank of $A$ will tell us if the solution, whenever it exists, will be unique.

• If the rank of $A$ is equal to $n$, that means the dimension of the column space is equal to $n$: the columns are linearly independent. In this case, for every $\vec b$, there is either a single solution (if it is in the column space of $A$) or no solution (if it's not).

• On the other hand, if the rank of $A$ is less than $n$, that means the dimension of the column space is less than $n$: the columns are linearly dependent. In this case, for every $\vec b$, there is either an infinite set of solutions (if it is in the column space of $A$) or no solution (if it's not).

In either case, to distinguish these two cases, we compare the rank of $A$ to the rank of the augmented matrix $A\mid \vec b$.

• If $\operatorname{rank}(A) = \operatorname{rank}(A \mid \vec b)$, that means the column space doesn't change when we add a column $\vec b$. This can only happen if $\vec b$ was already in the column space, so there must be a solution.

• If $\operatorname{rank}(A) < \operatorname{rank}(A \mid \vec b)$, this means that the column space becomes bigger when we add $\vec b$. This can only happen if $\vec b$ is linearly independent from $\vec a_1, \dots, \vec a_n$, so there is no solution.