The last set of problems and their solutions:
1) Recall t^A, found in Ex. 1 and let B =
(-1...1)
(1....0)
a) Find AB and thence: t^(AB)
Solution:
AB =
(2...1) (-1....1)
(3...1) (1.....0)
=
(-1.....2)
(-2.....3)
Then: t^(AB) =
(-1.....-2)
(2.......3)
b) Verify that: t^AB = t^B t^A
Solution:
From Ex. (1) in previous linear algebra blog we found t^A=
(2.....3)
(1.....1)
Given the matrix for B in part (a) then, t^B =
(-1...1)
(1.....0)
then: t^B t^A =
(-1....1)(2....3)
(1.....0) (1....1)
=
(-1.......-2)
(2.........3)
2) Find the trace of: R3(Θ) =
(cos(Θ)..........sin(Θ)..........0)
(-sin (Θ)......cos(Θ)...........0)
(0 ..................0..................1)
Solution:
Irrespective of dimension the trace Tr is the sum of the diagonal elements. Then, Tr(R3(Θ)) =
cos(Θ) + cos(Θ) + 1 = 2cos(Θ) + 1
3) Let A =
(cos Θ .....cos φ)
(cos φ .....sin Θ)
And let B = t^A
Find: AB
From this, t^A =
(cos Θ....cos φ)
(cos φ......sin Θ)
Then AB=
(cos Θ .....cos φ)(cos Θ....cos φ)
(cos φ .....sin Θ)(cos φ......sin Θ)
=
(cos^2 Θ + cos^2 φ...............cos Θcos φ + cos φ sin Θ)
(cos φ cos Θ + sin Θcos φ .............cos^2 φ + sin ^2Θ)
4) Find the traces of the following 3 x 3 matrices:
a) M1 =
(-i.....0.......0)
(0.......-7.....0)
(0.......0.......4)
Solution:
tr(M1) = -i - 7 + 4 = -3 - i
b) M2 =
(3.....-2.......4)
(1.......-4.....1)
(-7.......-3......-3i )
Solution:
tr (M2) = 3 + (-4) + (-3i) = -1 - 3i
c) A =
(1.....-1......1)
(2.......4.....1)
(3.......0.......1)
Solution:
tr(A) = 1 + 4 + 1 = 6
d) B =
(3.....1.......2)
(1.......1.....0)
(-1.......2.......1)
tr(B) = 3 + 1 + 1 = 5
5) For the last two parts (c and d) of (4), show: tr(AB) = tr(BA)
This requires multiplying two 3 x 3 matrices:
Then AB =
(1.....-1......1) (3.....1.......2)
(2.......4.....1)(1.......1.....0)
(3.......0.......1)((-1....2....1)
=
(1....2.......3)
(9.......8.....5)
(8.......5.......7)
tr(AB) = 1 + 8 + 7 = 16
BA =
(3.....1.......2)(1....-1...1)
(1.......1.....0)(2.....4....1)
(-1.......2.....1)(3....0...1)
=
(11....1......6)
(3.......3.....2)
(6.......9.......2)
tr(BA) = 11 + 3 + 2 = 16
Then: tr(AB) = tr(BA)
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