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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