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CoormaBak9

Answered question

2021-01-10

Show that G is an abelian under multiplication of matrices ?

Neelam Wainwright

Skilled2021-01-11Added 102 answers

Step 1
Let

A = [\begin{matrix} a_{1} & a_{2} \\ a_{3} & a_{4} \end{matrix}] and B = [\begin{matrix} b_{1} & b_{2} \\ b_{3} & b_{4} \end{matrix}]

be two matrices. Now,

A B = [\begin{matrix} a_{1} & a_{2} \\ a_{3} & a_{4} \end{matrix}] \times [\begin{matrix} b_{1} & b_{2} \\ b_{3} & b_{4} \end{matrix}]

= [\begin{matrix} a_{1} b_{1} + a_{2} b_{3} & a_{1} b_{2} + a_{2} b_{4} \\ a_{3} b_{1} + a_{4} b_{3} & a_{3} b_{2} + a_{4} b_{4} \end{matrix}]

and

B A = [\begin{matrix} b_{1} & b_{2} \\ b_{3} & b_{4} \end{matrix}] \times [\begin{matrix} a_{1} & a_{2} \\ a_{3} & a_{4} \end{matrix}]

= [\begin{matrix} a_{1} b_{1} + a_{3} b_{2} & a_{2} b_{1} + a_{4} b_{2} \\ a_{1} b_{3} + a_{3} b_{4} & a_{2} b_{3} + a_{4} b_{4} \end{matrix}]

So, matrix multiplication does not follow commutative law.
Step 2
Hence, G is not an abelian under multiplication of matrices.

Jeffrey Jordon

Expert2022-01-24Added 2605 answers

Answer is given below (on video)

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