# Usual Matrix Multiplication

In usual matrix muktiplication, we multiply rows of first matrix to the column of second matrix. My worries are

1. Why we can't multipy column of first matrix with row of second matrix in matrix multiplication

2. Why not we use natural multiplication of corresponding entries just like in addition. From where this idea comes ?

• 1. Multiplication is not commutative. If you multiply column of first with rows of second, that's just like doing the matrix product in the other order. 2. The idea is that matrices represent linear functions, and the matrix product is supposed to correspond to composition of functions.
– Nick
Mar 6, 2020 at 3:22
• @Nick Why not give this as an answer? Mar 6, 2020 at 4:42
• @Taj you might find this post helpful Mar 6, 2020 at 9:43
• I believe the question can be rephrased as follows: why matrix multiplication is defined the way it is and not any other way? The answer is that this specific multiplication corresponds to composition of linear operators. Mar 6, 2020 at 9:43
• If there are two matrice A and B and you are saying BA value =AB which is not correct because matrix multiplication is not commutative Mar 6, 2020 at 10:08

1. Why we can't multipy column of first matrix with row of second matrix in matrix multiplication

$$A\cdot B$$ multiplies rows of $$A$$ with columns of $$B$$. If you want to multiply columns of $$A$$ with rows of $$B$$ you can $$A^T\cdot B^T = (B\cdot A)^T$$

However, multiplying rows with columns is just a convention used everywhere. We could use linear algebra similarly if we used consistently multiplying columns with rows, however this would not gain anything, it would just cause confusion if we mixed up two conventions. Moreover, the current convention has the advantage that composition $$(A\circ B)x = A(B(x)) = (AB)x = A(Bx)$$ just looks the same like ordinary function composition.

1. Why not we use natural multiplication of corresponding entries just like in addition. From where this idea comes ?

You can represent linear functions from $$K^n$$ to $$K^m$$ by $$x\mapsto Ax$$ with $$A\in K^{m\times n}$$ and $$x\in K^n$$ a row vector. This only works when you define matriX multiplication as usual.

In addition, $$C=B \cdot A$$ represents the composition of two linear mappings $$K^n \stackrel A \to K^m \stackrel B \to K^\ell$$ only of you use the canonical matrix multiplication.

Using component-wise multiplication doesn't gain you much. This is similar to the situation when you construct the a multiplication in $$\Bbb R^2$$ (the Complex numbers) by a rule that is not so obvious. If you used $$(x_1,y_1)\cdot (x_2,y_2) := (x_1\cdot x_2,y_1\cdot y_2)$$ Then the result is not a field, i.e. you cannot find inverse for multiplication for non-zero elements like in $$(0,0)=(1,0)\cdot (0,1)$$ i.e. you cannot define division properly.

In the context of linear mappings this means that you can define inverse mappings for all matrices with non-zero determinant, and the determinant has nice multiplication properties:

$$\det(A\cdot B) = \det(A)\cdot \det(B)$$

$$\det(A^{-1}) = \det(A)^{-1}$$ if $$K$$ is a field. Even if $$K$$ is not a field and just, say a ring, using matrix multiplication is the way to go in 99.9% of cases.

• I think the key in this answer is the part about composition of two linear mappings. All the good arguments you give for the second question seem to contradict your statement "We could use linear algebra similarly if we used consistently multiplying columns with rows, however this would not gain anything, it would just cause confusion if we mixed up two conventions."
– Surb
Mar 6, 2020 at 9:47