Given that A and B are matrices and r is a real number.

Let us consider the matrix A is of order

\(\displaystyle{m}\times{n}\)

and let us consider the matrix B is of order

\(\displaystyle{n}\times{m}\)

\(\displaystyle{A}{\left({r}{B}\right)}={A}_{{{m}\times{n}}}{\left({r}{B}\right)}_{{{n}\times{m}}}\)

Hence the result for A(rB) will be of order

\(\displaystyle{m}\times{m}\)

\(\displaystyle{\left({r}{A}\right)}{B}={\left({r}{A}\right)}_{{{m}\times{n}}}{B}_{{{n}\times{m}}}\)

Hence the result for (rA)B will be of order

\(\displaystyle{m}\times{m}\)

Since the results for the above three is of order

\(\displaystyle{m}\times{m}\)

Therefore A(rB)= r(AB)= (rA)B

Let us consider the matrix A is of order

\(\displaystyle{m}\times{n}\)

and let us consider the matrix B is of order

\(\displaystyle{n}\times{m}\)

\(\displaystyle{A}{\left({r}{B}\right)}={A}_{{{m}\times{n}}}{\left({r}{B}\right)}_{{{n}\times{m}}}\)

Hence the result for A(rB) will be of order

\(\displaystyle{m}\times{m}\)

\(\displaystyle{\left({r}{A}\right)}{B}={\left({r}{A}\right)}_{{{m}\times{n}}}{B}_{{{n}\times{m}}}\)

Hence the result for (rA)B will be of order

\(\displaystyle{m}\times{m}\)

Since the results for the above three is of order

\(\displaystyle{m}\times{m}\)

Therefore A(rB)= r(AB)= (rA)B