Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 183
has solutions b 1 and b 2 , which are not both zero. Rewrite that as a linear system
(4 − x) · b 1 + −2 · b 2 = 0
1 · b 1 + (1 − x) · b 2 = 0
If x = 4 then the first equation gives that b 2 = 0, and then the second equation gives that b 1 = 0. The
case where both b’s are zero is disallowed so we can assume that x ≠ 4.
(−1/(4−x))ρ 1 +ρ 2 (4 − x) · b −→ 1 + −2 · b 2 = 0
((x 2 − 5x + 6)/(4 − x)) · b 2 = 0
Consider the bottom equation. If b 2 = 0 then the first equation gives b 1 = 0 or x = 4. The b 1 = b 2 = 0
case is disallowed. The other possibility for the bottom equation is that the numerator of the fraction
x 2 − 5x + 6 = (x − 2)(x − 3) is zero. The x = 2 case gives a first equation of 2b 1 − 2b 2 = 0, and so
associated with x = 2 we have vectors whose first and second components are equal:
( ( ) ( ( 1 4 −2 1 1 ⃗β 1 = (so
= 2 · , and λ
1)
1 1 1)
1)
1 = 2).
If x = 3 then the first equation is b 1 − 2b 2 = 0 and so the associated vectors are those whose first
component is twice their second:
( ( ) ( (
2 4 −2 2 2 ⃗β 2 = (so
= 3 · , and so λ
1)
1 1 1)
1)
2 = 3).
This picture
R 2 t
w.r.t. E 2
−−−−→ R 2 w.r.t. E T 2
⏐
⏐
id↓
id↓
shows how to get the diagonalization.
( )
2 0
=
0 3
R 2 w.r.t. B
(
1 2
1 1
t
−−−−→
D
R 2 w.r.t. B
) −1 (
4 −2
1 1
) (
1 2
1 1
Comment. This equation matches the T = P SP −1 definition under this renaming.
( ) ( ) −1 ( ) ( )
2 0 1 2
T = P =
p −1 1 2 4 −2
= S =
0 3 1 1
1 1 1 1
Five.II.2.7 (a) Setting up
( ) ( ) ( )
−2 1 b1 b1
(−2 − x) · b
= x · =⇒
1 + b 2 = 0
0 2 b 2 b 2 (2 − x) · b 2 = 0
gives the two possibilities that b 2 = 0 and x = 2. Following the b 2 = 0 possibility leads to the first
equation (−2 − x)b 1 = 0 with the two cases that b 1 = 0 and that x = −2. Thus, under this first
possibility, we find x = −2 and the associated vectors whose second component is zero, and whose
first component is free. ( ( ) ( ) (
−2 1 b1 b1 1
= −2 · ⃗β
0 2)
0
0 1 =
0)
Following the other possibility leads to a first equation of −4b 1 +b 2 = 0 and so the vectors associated
with this solution have a second component that is four times their first component.
( ) ( ) ( ) (
−2 1 b1 b1
1
= 2 ·
⃗β
0 2 4b 1 4b 2 =
1 4)
The diagonalization is this.
( ) −1 ( ) ( ) −1 ( )
1 1 −2 1 1 1 −2 0
0 4 0 2 0 4 0 2
(b) The calculations are like those in the prior part.
( ) ( ) ( )
5 4 b1 b1
(5 − x) · b
= x · =⇒
1 + 4 · b 2 = 0
0 1 b 2 b 2 (1 − x) · b 2 = 0
The bottom equation gives the two possibilities that b 2 = 0 and x = 1. Following the b 2 = 0
possibility, and discarding the case where both b 2 and b 1 are zero, gives that x = 5, associated with
vectors whose second component is zero and whose first component is free.
(
1 ⃗β 1 =
0)
)