Thread: Group formed by Matrix Multiplication

I'm not sure how to draw a matrix on this forum but this problem should still be readable.

Show that the set of all 2 x 2 matrices over R of the form

m b
0 1

with m=/=0 forms a group under matrix multiplication.

In this group, find all elements that commute with the element

2 0
0 1

Originally Posted by elninio

I'm not sure how to draw a matrix on this forum but this problem should still be readable.

Show that the set of all 2 x 2 matrices over R of the form

m b
0 1

with m=/=0 forms a group under matrix multiplication.

In this group, find all elements that commute with the element

2 0
0 1

For part 2 -
answer is
m 0
0 1

B =
2 0
0 1

To do this assume a matrix A in the general form

Now compute AB and BA
and equate the two matrices.

Part 1 is straight fwd, can you be specific where you are stuck. Take group axioms one by one as check if they apply to the set.

Originally Posted by elninio

I'm not sure how to draw a matrix on this forum but this problem should still be readable.

Show that the set of all 2 x 2 matrices over R of the form

m b
0 1

with m=/=0 forms a group under matrix multiplication.

In this group, find all elements that commute with the element

2 0
0 1

For the first part of your question you have both associativity and identity for free - matrix multiplication is associative, and the identity matrix is clearly of this form.

Thus, you need to show closure (multiply two such matrices together to show you have another matrix of this form) and that every element has an inverse. To show inverses exist, you need to show that every matrix will have non-zero determinant (I'm sure you remember how to compute the determinant of a 2x2 matrix!)

To find the elements, , of this form which commute with the given matrix, call it , simply calculate then . As we want these to be equal, we want the four coordinates to also be equal. Thus, find for which entries in these things are equal.