Why does $A$ have an inverse if $AX=B$, where X is a column matrix, $X=(x_1,...,x_n)^t$, has a unique solution for every $n\times1$-matrix $B$?

Question

My textbook in linear algebra states that the following statements are equivalent:

1. A has an inverse
2. A can be reduced to the indentity matrix, by applying row operations.
3. The equation system $AX=B$, where X is a column matrix, $X=(x_1,...,x_n)^t$, has a unique solution for every $n\times1$-matrix $B$.

I'm having a hard time visualizing/rationalizing 3) to myself. Why would this be the case and why would 3) be sufficient to prove that a matrix has an inverse?

Answer 1

For any vector $X$, $AX$ denotes a linear combination of the columns of $A$. Now if for every vector $B$, $AX=B$ has a solution, this means that every vector $B$ can be expressed as a linear combination of the columns of $A$. This implies that the column of $A$ form a basis for the vector space, and hence implying that they are linearly independent. Hence $A$ is full rank and will have an inverse.

Implication of uniqueness of solutions: Suppose there is a $B$ for which there are more than one solutions: say $AX_1=B=AX_2$ where $X_1\ne X_2$. Then we have $A(X_1-X_2)=0$ and $X_1-X_2\ne0$. Thus we have a non-trivial linear combination of the columns of $A$ that yields the $0$ vector. Hence the columns of $A$ are not linearly independent. Then the matrix will not have full rank, and hence will not be invertble