# Show that G is an abelian under multiplication of matrices ?

Show that G is an abelian under multiplication of matrices ?
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Neelam Wainwright
Step 1
Let be two matrices. Now,
$AB=\left[\begin{array}{cc}{a}_{1}& {a}_{2}\\ {a}_{3}& {a}_{4}\end{array}\right]×\left[\begin{array}{cc}{b}_{1}& {b}_{2}\\ {b}_{3}& {b}_{4}\end{array}\right]$
$=\left[\begin{array}{cc}{a}_{1}{b}_{1}+{a}_{2}{b}_{3}& {a}_{1}{b}_{2}+{a}_{2}{b}_{4}\\ {a}_{3}{b}_{1}+{a}_{4}{b}_{3}& {a}_{3}{b}_{2}+{a}_{4}{b}_{4}\end{array}\right]$
and
$BA=\left[\begin{array}{cc}{b}_{1}& {b}_{2}\\ {b}_{3}& {b}_{4}\end{array}\right]×\left[\begin{array}{cc}{a}_{1}& {a}_{2}\\ {a}_{3}& {a}_{4}\end{array}\right]$
$=\left[\begin{array}{cc}{a}_{1}{b}_{1}+{a}_{3}{b}_{2}& {a}_{2}{b}_{1}+{a}_{4}{b}_{2}\\ {a}_{1}{b}_{3}+{a}_{3}{b}_{4}& {a}_{2}{b}_{3}+{a}_{4}{b}_{4}\end{array}\right]$
So, matrix multiplication does not follow commutative law.
Step 2
Hence, G is not an abelian under multiplication of matrices.
Jeffrey Jordon