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:
(a 0)
(0 b)
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??
Are you asking if you should replace a snd b with particular numbers?
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]
Given a set S with an algebraic binary operation *
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.
For example,
If your matrix ended up being,
[0 a]
[b 0]
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
[a 0]
[0 b]
Not,
[a 0]
[0 b]
Anyways that is not the biggest problem, you need tos show that,
[ab 0]
[0 ba] is an element of these matricies.
Why?
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,
[ab 0]
[0 ab]
is in the set which is true since ab is an element of the set.