Question

If A= $open square brackets table row 1 cell negative 2 end cell 3 row cell negative 4 end cell 2 5 end table close square brackets space and space straight B space equals space open square brackets table row 1 3 row 4 5 row 2 1 end table close square brackets$ then find AB, BA.
Show that AB ≠ BA.

Solution

straight A equals open square brackets table row 1 cell negative 2 end cell 3 row cell negative 4 end cell 2 5 end table close square brackets comma space straight B equals open square brackets table row 2 3 row 4 5 row 2 1 end table close square brackets

A is of type 2 x 3 and B is of type 3 x 2.
$AB equals open square brackets table row 1 cell negative 2 end cell 3 row cell negative 4 end cell 2 5 end table close square brackets open square brackets table row 2 3 row 4 5 row 2 1 end table close square brackets space space space space equals space open square brackets table row cell 2 minus 8 plus 6 end cell cell 3 minus 10 plus 3 end cell row cell negative 8 plus 8 plus 10 end cell cell negative 12 plus 10 plus 5 end cell end table close square brackets equals open square brackets table row 0 cell negative 4 end cell row 10 3 end table close square brackets BA equals open square brackets table row 2 3 row 4 5 row 2 1 end table close square brackets open square brackets table row 1 cell negative 2 end cell 3 row cell negative 4 end cell 2 5 end table close square brackets space space space space space equals space open square brackets table row cell 2 minus 12 end cell cell negative 4 plus 6 end cell cell 6 plus 15 end cell row cell 4 minus 20 end cell cell negative 8 plus 10 end cell cell 12 plus 25 end cell row cell 2 minus 4 end cell cell negative 4 plus 2 end cell cell 6 plus 5 end cell end table close square brackets equals open square brackets table row cell negative 10 end cell 2 21 row cell negative 16 end cell 2 37 row cell negative 2 end cell cell negative 2 end cell 11 end table close square brackets therefore space space space space space AB space not equal to space BA.$
Since number of columns of A = number of rows of B =3
∴ AB is defined.
Again number of columns of B = number of rows of A = 2
∴ BA is defined.
Note 1. From above example, it is clear that though AB and BA are defined, yet AB ;≠ BA. Here A and B are of different types.
Note 2. Even if A and B are of same order, there are chances that AB≠ BA.
∴ we can say that matrix multiplication is not commutative.
Note 3. Even if AB ≠ BA in many cases, there are cases when AB = BA.

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Matrices

If A= $open square brackets table row 1 cell negative 2 end cell 3 row cell negative 4 end cell 2 5 end table close square brackets space and space straight B space equals space open square brackets table row 1 3 row 4 5 row 2 1 end table close square brackets$ then find AB, BA.
Show that AB ≠ BA.

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