For two matrices to be multiplied, the length of the rows of the first matrix must match the length of the columns of the second (in other words, the number of columns in the first matrix must match the number of rows in the second).
The resulting matrix formed will have the same number of rows as the first matrix and the same number of columns as the second. That is, the product of a matrix A_(lm)Alm with ll rows and mm columns and a matrix B_(mn)Bmn with mm rows and nn columns will be a matrix C_(ln)Cln with ll rows and nn columns.
A_(color(blue)(l) color(red)(m))*B_(color(red)(m)color(green)(n)) = C_(color(blue)(l)color(green)(n))Alm⋅Bmn=Cln
To perform the actual multiplication, the entry in the i^(th)ith row and j^(th)jth column of CC is the sum of the products of the i^(th)ith row of AA and the j^(th)jth column of BB.
c_(ij) = sum_(k=1)^ma_(ik)b_(kj)cij=m∑k=1aikbkj
Let's look at an example. We will multiply the following matrices:
A = ((2,1,4),(3,-2,5)) and B = ((3,2,4,1),(2,2,6,3),(-8,0,2,-1))
Note that A is 2 by 3 and B is 3 by 4. Thus the resulting matrix C will be 2 by 4.
((2,1,4),(3,-2,5))((3,2,4,1),(2,2,6,3),(-8,0,2,-1))=((c_11,c_12,c_13,c_14),(c_21,c_22,c_23,c_24))
Again, to find c_(ij) we multiply the i^(th) row of A by the j^(th) row of B. So, to find c_11, for example,
((color(red)(2),color(red)(1),color(red)(4)),(3,-2,5))((color(red)(3),2,4,1),(color(red)(2),2,6,3),(color(red)(-8),0,2,-1))=((color(red)(c_11),c_12,c_13,c_14),(c_21,c_22,c_23,c_24))
c_11 = 2*3 + 1*2 + 4*(-8) = -24
Similarly, for c_23 we have
((2,1,4),(color(red)(3),color(red)(-2),color(red)(5)))((3,2,color(red)(4),1),(2,2,color(red)(6),3),(-8,0,color(red)(2),-1))=((c_11,c_12,c_13,c_14),(c_21,c_22,color(red)(c_23),c_24))
c_23 = 3*4+(-2)*6+5*2 = 10
After performing all of the operations, the result becomes
((2,1,4),(3,-2,5))((3,2,4,1),(2,2,6,3),(-8,0,2,-1))=((-24,6,22,1),(-35,2,10,-8))
(For practice, try seeing if you get the same result)
