Perhaps explicitly thinking of them as systems of equations (or transformations on a vector) would be helpful.
As an example, suppose you are asked to multiply matrices A and B, where A is [1 2, 0 4, −1 2] (the commas represent the end of a row) and B is [2 1 0, 3 1 2]. Start out by taking the rightmost matrix (B in this case) and converting it into a series of equations, one for each row. So since the first row is 2 1 0, the relevant equation will be 2x + 1y + 0z. Assign each of these equations to some other variable. So we now have
X = 2x + y
Y = 3x + y + 2z
Now do the same thing with the matrix on the left, except this time use the new variables you’ve introduced (X and Y), so the three equations you end up with (one for each row) will be
X + 2Y
4Y
-X + 2Y
Now that you have these formulae, substitute in the values of X and Y based on your earlier equations. You get
(2x + y) + 2(3x + y + 2z)
4(3x + y + 2z)
-(2x + y) + 2(3x + y + 2z)
Simplifying, you get
8x + 3y + 4z
12x + 4y + 8z
4x + y + 4z
The coefficients of these equations are the result of the multiplication. So the product of the two matrices is [8 3 4, 12 4 8, 4 1 4].
I’ll admit this is not the quickest way to go about multiplying matrices, but it might be easier for you to remember since it doesn’t seem as arbitrary. And maybe once you get used to thinking about multiplication this way, the usual visual rule will start making more sense to you.
Perhaps explicitly thinking of them as systems of equations (or transformations on a vector) would be helpful.
As an example, suppose you are asked to multiply matrices A and B, where A is [1 2, 0 4, −1 2] (the commas represent the end of a row) and B is [2 1 0, 3 1 2]. Start out by taking the rightmost matrix (B in this case) and converting it into a series of equations, one for each row. So since the first row is 2 1 0, the relevant equation will be 2x + 1y + 0z. Assign each of these equations to some other variable. So we now have
X = 2x + y
Y = 3x + y + 2z
Now do the same thing with the matrix on the left, except this time use the new variables you’ve introduced (X and Y), so the three equations you end up with (one for each row) will be
X + 2Y
4Y
-X + 2Y
Now that you have these formulae, substitute in the values of X and Y based on your earlier equations. You get
(2x + y) + 2(3x + y + 2z)
4(3x + y + 2z)
-(2x + y) + 2(3x + y + 2z)
Simplifying, you get
8x + 3y + 4z
12x + 4y + 8z
4x + y + 4z
The coefficients of these equations are the result of the multiplication. So the product of the two matrices is [8 3 4, 12 4 8, 4 1 4].
I’ll admit this is not the quickest way to go about multiplying matrices, but it might be easier for you to remember since it doesn’t seem as arbitrary. And maybe once you get used to thinking about multiplication this way, the usual visual rule will start making more sense to you.