Matrix and scalar multiplication Say we have the following variables:


*

*A, a matrix that is nxn in size containing complex numbers

*B, a matrix that is also nxn in size containing complex numbers

*x, a scalar


If you multiply, does it matter whether you multiply in either of the following orders:


*

*(AB)x

*(Ax)B
if:


*

*x is a real number

*x is a complex number

 A: $x(AB)=(xA)B=A(xB)=A(Bx)=(AB)x=(Ax)B$
When matrices are being multiplied by a scalar element, the order in which multiplication takes place can be disregarded
A: It runs down to the axiom of commutativity of multiplication in fields (a scalar is element of a field) 
$$
\forall a, b \in F: a b = b a
$$
and the commutative property of scalar multiplication: 
$$
x A 
= x(a_1, \dotsc, a_n) 
= (x a_1, \dots, x a_n) 
= (a_1 x, \dots, a_n x)
= (a_1, \dotsc, a_n) x
= A x
$$
where $x \in F$ and $A$ is a matrix over $F$. For matrix multiplication we have
$$
x \sum_j a_{ij} b_{jk}
= \sum_j (x a_{ij}) b_{jk}
= \sum_j a_{ij} (x b_{jk})
= \left(\sum_j a_{ij} b_{jk} \right) x
$$
or
$$
x(AB) = (xA)B = A(xB) = (AB)x
$$
A: Scalar multiplication actually is short hand for multiplication by a diagonal matrix full of the scalar. In other words:
$$sA = \left[\begin{array}{ccc}s&0&0\\0&\ddots&0\\0&0&s\end{array}\right]A$$
You can convince yourself that this will be the same when multiplying from the left as from the right.
