I forgot to mention.. the attached image is the matrix given in the question.
I have a maths question which reads:
Let M be the set of all real 2 * 2-matrices of the form:
Show that this set of matrices is closed under matrix multiplication, and that matrix multiplication is commutative on this set.
Should I use real numbers to solve this problem or use what the question gave??
If so the answer is no.
That M is closed under matrix multiplication is equivalent asking you
to prove that for any real numbers a, b, c, d, there exist two real numbers
e,f such that:
and that multiplication is comutative on M equivalent asking youCode:[a 0] [c 0] = [e 0] [0,b] [0 d] [0 f]
to prove that for any real numbers a, b, c, d that:
RonLCode:[a 0] [c 0] = [c 0] [a 0] [0,b] [0 d] [0 d] [0 b]
Take a subset R of S such that,
a*b in R for all a,b in R
Then we say R is closed under the binary operation *
Informally, closed mean whenever you take two numbers and you operate them you get the same number in the set.
If your matrix ended up being,
For a and b nonzero then your binary operation on this set would not have been closed because you have a diffrenet number (in this case a different matrix).
This is my 25th Post
Here's how I'm going to completely answer the question:
Code:Let a,b,c,d,e,f be real numbers: [a 0] [c 0] = [e 0] [0 b] [0 d] = [0 f] .: M is closed under matrix multiplication because the product of two M members is also in the M set. Let A = [a 0] [0 a] B = [b 0] [0 b] AB = [ab 0] [0 ab] BA = [ab 0] [0 ab] AB = BA .: matrix multiplication is commutative on this set.
I think you meant
Anyways that is not the biggest problem, you need tos show that,
[0 ba] is an element of these matricies.
Because a,b are real numbers and ab is closed under multiplication. That is the important step you omitted.
Because you need to show that,
is in the set which is true since ab is an element of the set.