Prove the following A(rB)= r(AB)= (rA)B where r and s are real numbers and A and B are matrices

Prove the following A(rB)= r(AB)= (rA)B where r and s are real numbers and A and B are matrices

Question
Matrices
asked 2021-02-14
Prove the following
A(rB)= r(AB)= (rA)B
where r and s are real numbers and A and B are matrices

Answers (1)

2021-02-15
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
0

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