91Ó°ÊÓ

Let us take two matrices A and B of order 2 x 2 where n = p = m = 2 such that

$\mathrm{A}=\left(\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right)$and$\mathrm{B}=\left(\begin{array}{cc}{\mathrm{b}}_{11}& {\mathrm{b}}_{12}\\ {\mathrm{b}}_{21}& {\mathrm{b}}_{22}\end{array}\right)$

Then the matrix AB is of order 2 x 2 as well. Let us take any scalar k.

We have

$\mathrm{A}=\left(\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right)$and$\mathrm{B}=\left(\begin{array}{cc}{\mathrm{b}}_{11}& {\mathrm{b}}_{12}\\ {\mathrm{b}}_{21}& {\mathrm{b}}_{22}\end{array}\right)$

$\mathrm{kA}=\left(\begin{array}{cc}{\mathrm{ka}}_{11}& {\mathrm{ka}}_{12}\\ {\mathrm{ka}}_{21}& {\mathrm{ka}}_{22}\end{array}\right)$

$$\begin{gathered} \left(\mathrm{kA}\right)\mathrm{B}=\left(\begin{array}{cc}{\mathrm{ka}}_{11}& {\mathrm{ka}}_{12}\\ {\mathrm{ka}}_{21}& {\mathrm{ka}}_{22}\end{array}\right)\left(\begin{array}{cc}{\mathrm{b}}_{11}& {\mathrm{b}}_{12}\\ {\mathrm{b}}_{21}& {\mathrm{b}}_{22}\end{array}\right) \\ =\left(\begin{array}{cc}{\mathrm{ka}}_{11}{\mathrm{b}}_{11}+{\mathrm{ka}}_{12}{\mathrm{b}}_{21}& {\mathrm{ka}}_{11}{\mathrm{b}}_{12}+{\mathrm{ka}}_{12}{\mathrm{b}}_{22}\\ {\mathrm{ka}}_{21}{\mathrm{b}}_{11}+{\mathrm{ka}}_{22}{\mathrm{b}}_{21}& {\mathrm{ka}}_{21}{\mathrm{b}}_{12}+{\mathrm{ka}}_{22}{\mathrm{b}}_{22}\end{array}\right) \end{gathered}$$

$\mathrm{kB}=\left(\begin{array}{cc}{\mathrm{kb}}_{11}& {\mathrm{kb}}_{12}\\ {\mathrm{kb}}_{21}& {\mathrm{kb}}_{22}\end{array}\right)$

We have

$\mathrm{A}=\left(\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right)$and$\mathrm{B}=\left(\begin{array}{cc}{\mathrm{b}}_{11}& {\mathrm{b}}_{12}\\ {\mathrm{b}}_{21}& {\mathrm{b}}_{22}\end{array}\right)$

role="math" localid="1660193303568" $$\begin{gathered} \mathrm{A}=\left(\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right)\left(\begin{array}{cc}{\mathrm{b}}_{11}& {\mathrm{b}}_{12}\\ {\mathrm{b}}_{21}& {\mathrm{b}}_{22}\end{array}\right) \\ =\left(\begin{array}{cc}{a}_{11}{\mathrm{b}}_{11}+a_{12}{\mathrm{b}}_{21}& a_{11}{\mathrm{b}}_{12}+a_{12}{\mathrm{b}}_{22}\\ a_{21}{\mathrm{b}}_{11}+a_{22}{\mathrm{b}}_{21}& a_{21}{\mathrm{b}}_{12}+a_{22}{\mathrm{b}}_{22}\end{array}\right) \end{gathered}$$

$k\left(AB\right)=\left(\begin{array}{cc}{\mathrm{ka}}_{11}{\mathrm{b}}_{11}+{\mathrm{ka}}_{12}{\mathrm{b}}_{21}& {\mathrm{ka}}_{11}{\mathrm{b}}_{12}+{\mathrm{ka}}_{12}{\mathrm{b}}_{22}\\ {\mathrm{ka}}_{21}{\mathrm{b}}_{11}+{\mathrm{ka}}_{22}{\mathrm{b}}_{21}& {\mathrm{ka}}_{21}{\mathrm{b}}_{12}+{\mathrm{ka}}_{22}{\mathrm{b}}_{22}\end{array}\right)$

Consider a 2 x 2 matrix A, a 2 x 2 matrix B, and a scalar k. Then we have

(k A)B = A (k B) = k(AB).

Since this is true for n = p = m = 2, therefore it is true for all values of n, m, p.

Hence, if we have an n × p matrix A, a p × m matrix B, and a scalar k then

(k A)B = A (k B) = k(AB).