Prove a(AB)=(aA)B=A(aB) ..

I suppose $A$ and $B$ are matrices.

You need to apply the definition of matrix product. Note that two matrices are equal, precisely when each of the corresponding entries are equal. Let me change the name of the scalar to $c$ .

Let $a_{ij}$ be the $(i,j)$ -entry of $A$ and let $b_{ij}$ be the $(i,j)$ -entry of $B$ . Suppose also that $A$ is an $m \times n$ matrix, and that $B$ is an $n \times p$ matrix.

Then the $(i,j)$ -entry of $AB$ is "the $i$ 'th row of $A$ times the $j$ 'th column of $B$ ". So this $(i,j)$ -entry of $AB$ is:

$\sum_{k=1}^n a_{ik}b_{kj}$ ,

and so the $(i,j)$ -entry of $c(AB)$ is

$c\sum_{k=1}^n a_{ik}b_{kj} = \sum_{k=1}^n ca_{ik}b_{kj},$

where the first expression is the entry that follows by applying the definition of the product directly, and where the equality follows from bringing the scalar $c$ into each term of the sum.

Consider then the product $(cA)B$ . We need to check that the $(i,j)$ -entry of this product is same expression as for $c(AB)$ . But $cA$ is the matrix, where the $(i,j)$ -entry is $ca_{ij}$ . The product $(cA)B$ has as its $(i,j)$ -entry the $i$ 'th row of this matrix multiplied by the $j$ 'th column of $B$ , so

$\sum_{k=1}^n ca_{ik}b_{kj}.$

But this is exactly the same expression as above.

Similarly you may show that A(cB) has the correct $(i,j)$ -entries.