Help with a couple linear algebra (group theory) problems?

Question

Here they are: http://www.geocities.com/jsprplc2006/eq3.jpg

I don't really know where to start, especially with #5.

Answer 1

Good ol' Group theory.

Four is pretty easy. The individual terms {a,b,c, d} are contained within the set of integers.

Therefore, if a is inside the set, then -a is, too.

Now, recall the definition of the determinant for a 2 X 2 matrix. It is ad-bc. Note the definition of the set is only 2X2 matrices with integer entries with determinant of one. So, the prefix for inverting is always one. So the inverse of the matrix would be:

{ d -b
-c a}

Since a term and its negation are both in Integers, the group is closed under inversion. Still a group.

Ok, to be a group, the set must be closed under its operation.

So if a exists in the set, so does a^-1 A and a-1 under the operation will get to the "unit" - this is the operation does not change its value. 1 is the unit for traditional multiplication, zero for addition. A & a-1 combined get to the unit. Now, we take the inverse of a^-1.. By definition a (operation) a^-1 is the unit. There fore a = (a^-1)^-1

I'll let you work out 5b. It should follow from the same logic, since all of these are closed. It looks like your operation is multiplication.

Answer 2

Symmetric skill that the entries under the main diagonal are equivalent to the corresponding entries above the diagonal. an consumer-friendly thank you to define a symmetric matrix is: M = M^T A matrix is skew-symmetric if M = -M^T. With the right definition, area a might desire to be uncomplicated. purely discover B^T. B^T = (A + A^T)^T = A^T + (A^T)^T = A^T + A = B The info for C is comparable. For area b, for any matrix M, enable A =(a million/2)M. Then define B and C as partly a.