Thread: matrix proof 1

Can some 1 please help me with this as i really hate and sturggle doing matrix proof

Originally Posted by nerdo

Can some 1 please help me with this as i really hate and sturggle doing matrix proof

for b, A =
[1 0]
[0 1]
B =
[0 1]
[1 0]
Then AB =/= BA


so for c, no, by part b.

Once again thanks for the help

Originally Posted by GaloisTheory1

for b, A =
[1 0]
[0 1]
B =
[0 1]
[1 0]
Then AB =/= BA


so for c, no, by part b.

Sorry but i think u got this wrong as AB=BA if u calculate it.

Originally Posted by GaloisTheory1

for b, A =

[1 0]

[0 1]

B =

[0 1]

[1 0]

Then AB =/= BA

so for c, no, by part b.

Originally Posted by nerdo

Sorry but i think u got this wrong as AB=BA if u calculate it.

Correct. The example given does not answer the question. Also, since one of them is the identity matrix they obviously commute .....

Surely it's not too hard to find two matrices that satisfy the requirements of (b). In fact, I bet you could choose to 2x2 matrices at random that would work ..... Then the answer to c is obvious.

As for (a), two trivial matrices that satisfy the requirements are

[1, 1]
[1, 1]

[2, 2]
[2, 2]

It is not too hard to construct less trivial examples .....

Originally Posted by mr fantastic

Correct. The example given does not answer the question. Also, since one of them is the identity matrix they obviously commute .....

Surely it's not hard too find two matrices that satisfy the requirements of (b). In fact, I bet you could choose to 2x2 matrices at random that would work. Then the answer to c is obvious.

As for (a), two trivial matrices that satisfy the requirements are

[1, 1]
[1, 1]

[2, 2]
[2, 2]

It is not too hard to construct less trivial examples .....

Thanks for the help, i hav know actually managaed to solve this question