Why is the order of the matrices for multiplication not flipped?

In Week3, the Matrix Inverse Lecture at minute 1:13 - why is the order of not flipped?

In the earlier lecture, the professor introduced a concept where we flip the matrix order when multiplying two matrix transformations. But here the order stays the same i.e. Matrix 1 x Matrix 2.

Please see below

The slide shows:
Mat 1    x    Mat 2
|3  1|   x   |a  b|
|1  2|       |c  d|

when it should be
Mat 2    x    Mat 1
|a  b|   x   |3  1|
|c  d|       |1  2|

?

The lecture at 1:13 is discussing the product of two matrices that yield the identity matrix. That is a stand-alone fact. There is no reason to flip the order.

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In the matrix multiplication lecture, 1:45, we combine the two linear transformations by flipping the order during multiplication.


Isn’t the same thing happening in the lecture I’m referring?

So if the matrix with values a b c d represents the second transform, shouldn’t the order of multiplication matter this time as well? The result is the identity matrix.

The order of matrix multiplication always matters, right? Because as has been discussed, matrix multiplication is not commutative in general.

The point is that when you apply a linear transformation, you multiply by that matrix on the left. So if I have a matrix A that expresses one linear transformation and I have another matrix B that is a different transformation and I want to apply A first and then B to a vector v, the expression would be:

B \cdot (A \cdot v)

Seems pretty clear doesn’t it? Or am I just missing your point?

But there is one well known case in which matrix multiplication is commutative: when we deal with a matrix and its inverse. Of course not all matrices are invertible, but if A is invertible, then we have:

A \cdot A^{-1} = A^{-1} \cdot A = I

Of course another way to express that relationship would be:

(A^{-1})^{-1} = A

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Thanks for the reference to the other lecture.

Thanks @paulinpaloalto - the exception makes it clear now.

While I try to understand this better - it’s not wrong to stick to the order of matrix multiplication in the initial lecture. Since in the case of a matrix and its inverse, the operation is commutative anyways we can chose any order.

One other point that might be worth adding here w.r.t. my example about applying two linear transformations:

B \cdot (A \cdot v)

While matrix multiplication is not commutative, it is associative. So we have:

B \cdot (A \cdot v) = (B \cdot A) \cdot v

In other words, (B \cdot A) is the composition of those two transformations expressed as a single matrix.

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