no. that is not what HallsofIvy said.

if A is invertible, we can solve the matrix equation AX = B for X:

X = A^{-1}B.

if A is invertible, we can also solve the equation XA = B for X:

X = BA^{-1}.

these are two totally different situations, and should NOT be confused. A^{-1}B and BA^{-1} are MOST OF THE TIME different matrices.

matrix multiplication is NOT like "ordinary multiplication", we cannot say that AB and BA are the same (even if BOTH are "the same size", nxn).

contrast this with a field (like the reals, or the rationals), it does not matter if we write:

x/y = x(1/y) or x/y = (1/y)x, because these are the same.

it's even worse, because we can have an mxn matrix A, with an nxm matrix B with AB = I (where I is mxm), but NO nxm matrix C with CA = I (where I is nxn).

for example, if:

\(\displaystyle A = \begin{bmatrix}2&3&0\\3&5&0 \end{bmatrix};\ B = \begin{bmatrix}5&-3\\-3&2\\0&0 \end{bmatrix}\)

then:

\(\displaystyle AB = \begin{bmatrix}1&0\\0&1 \end{bmatrix}\)

but for ANY matrix:

\(\displaystyle C = \begin{bmatrix}c_1&c_2\\c_3&c_4\\c_5&c_6 \end{bmatrix}\)

\(\displaystyle CA = \begin{bmatrix}2c_1+3c_2&3c_1+5c_2&0\\2c_3+3c_4&3c_3+5c_4&0\\2c_5+3c_6&3c_5+5c_6&0 \end{bmatrix}\)

which is NOT the 3x3 identity matrix no matter HOW we choose the c's, since the last column is all zeros.

so a matrix can have a "right-inverse" without having a "left-inverse". in such an unhappy state of affairs, division doesn't even make sense as a CONCEPT.

matrices are NOT numbers. they are an entirely different BREED of animal.