matrix proof 2

• Jan 24th 2009, 12:55 PM
nerdo
matrix proof 2

Attachment 9796
• Jan 24th 2009, 01:00 PM
GaloisTheory1
Quote:

Originally Posted by nerdo

Attachment 9796

for part a, multiply the matrices then take the determinate. that is det(ab). then take det(a), det(b) separately and then find det(a)det(b) then show they are equal.
• Jan 24th 2009, 01:02 PM
nerdo
Quote:

Originally Posted by GaloisTheory1
for part a, multiply the matrices then take the determinate. that is det(ab). then take det(a), det(b) separately and then find det(a)det(b) then show they are equal.

Oh i get it know, thanks for the help.
• Jan 24th 2009, 01:34 PM
SimonM
For the second part, take the determinant of both sides

\$\displaystyle \det(ABA^{-1}B^{-1}) = \det(A)\det(B)\det(A^{-1})\det(B^{-1}) = \det(AA^{-1})\det(BB^{-1}) = 1\$

\$\displaystyle \det(2I_2) = 2\$

Therefore no such matrices exist
• Jan 24th 2009, 02:46 PM
GaloisTheory1
Quote:

Originally Posted by SimonM
For the second part, take the determinant of both sides

\$\displaystyle \det(ABA^{-1}B^{-1}) = \det(A)\det(B)\det(A^{-1})\det(B^{-1}) = \det(AA^{-1})\det(BB^{-1}) = 1\$

\$\displaystyle \det(2I_2) = 2\$

Therefore no such matrices exist

thanks i thought it was false, i was trying to figure it out for a while.