Excellent. Your request for a geometric interpretation shows me that you are on the right track in learning linear algebra! (Well, at least visualizing the standard 1-3 dimensions)
Consider the Reduced Row Echelon Form (RREF) of a matrix A, it concisely describes some of the subspace information associated with A.
The RREF tell us:
- rank : number of basis vectors in the column space/range
- nullity : number of basis vectors in the null space/kernel
- Invertibility/Linear independence : Whether the null space is trivial or not
The null space being trivial (i.e, consisting of only an appropriate null vector) implies that the column space of A occupies the entirety of it's dimension(equal to the column count of A) and that there is no linear combination of any vectors in it's range that reduce to 0 vector.
The process of matrix inversion is supposed to find a subspace which when multiplied with A gets projected to the appropriate identity matrix.
If there is any linear combination of columns of A that reduces to 0, then it cannot be reversed to map onto it's original linear combination, which means that the vector is nullified. (Mapped to the 0 vector). This is exactly what the rank of a matrix succinctly describes with mathematical beauty.
So, such linear vector combinations of non-invertible matrices are consumed by it's null space/kernel!
The exact same can be witnessed and verified on the column space and null space of the non-invertible matrix A' (which are incidentally, NOT coincidentally, the row space and left-null space of A)
If your matrix $A$ is $n\times n$, the search for an inverse matrix is the same as solving the $n$ linear systems
$$
Ax=e_i\qquad (i=1,2,\dots,n)
$$
where $e_i$ is the $i$-th column of the identity matrix. If the matrix is not invertible, then at least one of those systems must have no solution, say it's the one for $e_i$.
Performing row reduction on the augmented matrix $[A\mid e_i]$ yields that the last column must be a pivot column. If the pivot is on row $j$, then the row reduction of $A$ has a zero $j$-th row.
Best Answer
Hint :
Let $A$ be a square matrix such that $i^{th}$ column is zero.
For any $B\in M_{n\times n}$ what would be the $i^{th}$ column of $BA$?