A First Course in Linear Algebra, 2017a

570 Selected Exercise Answers
6.3.7 Solution is:
(1 − i) √ 2,−(1 + i) √ 2,−(1 − i) √ 2,(1 + i) √ 2.
These are just the fourth roots of −16. Then to factor, you get
( (
x − (1 − i) √ ))( (
2 x − −(1 + i) √ ))
2 ·
( (
x − −(1 − i) √ ))( (
2 x − (1 + i) √ ))
2
(
6.3.8 x 4 + 16 = x 2 − 2 √ )(
2x + 4 x 2 + 2 √ )
2x + 4 . You can use the information in the preceding problem.
Note that (x − z)(x − z) has real coefficients.
6.3.9 Yes, this is true.
(cosθ − isinθ) n = (cos(−θ)+isin(−θ)) n
= cos(−nθ)+isin(−nθ)
= cos(nθ) − isin(nθ)
6.3.10 p(x)=(x − z 1 )q(x)+r (x) where r (x) is a nonzero constant or equal to 0. However, r (z 1 )=0and
so r (x) =0. Now do to q(x) what was done to p(x) and continue until the degree of the resulting q(x)
equals 0. Then you have the above factorization.
6.4.1
(x − (1 + i))(x − (2 + i)) = x 2 − (3 + 2i)x + 1 + 3i
6.4.2 (a) Solution is: 1 + i,1− i
(b) Solution is: 1 6 i√ 35 − 1 6 ,− 1 6 i√ 35 − 1 6
(c) Solution is: 3 + 2i,3− 2i
(d) Solution is: i √ 5 − 2,−i √ 5 − 2
(e) Solution is: − 1 2 + i,− 1 2 − i
6.4.3 (a) Solution is : x = −1 + 1 2√
2 −
1
2
i √ 2, x = −1 − 1 2√
2 +
1
2
i √ 2
(b) Solution is : x = 1 − 1 2 i, x = −1 − 1 2 i
(c) Solution is : x = − 1 2 , x = − 1 2 − i
(d) Solution is : x = −1 + 2i, x = 1 + 2i
(e) Solution is : x = − 1 6 + 1 6√
19 +
( 16
− 1 6√
19
)
i, x = −
1
6
− 1 6√
19 +
( 16
+ 1 6√
19
)
i
7.1.1 A m X = λ m X for any integer. In the case of −1,A −1 λX = AA −1 X = X so A −1 X = λ −1 X. Thus the
eigenvalues of A −1 are just λ −1 where λ is an eigenvalue of A.