Math 415 Spring 2011 Homework 2 Solutions
Sec 1.4, No 9: Let

0
 0
A=
 0
0
1
0
0
0
0
1
0
0

0
0 

1 
0


0
0


1 
0
, A3 = 
 0
0 
0
0
0
0
0
0
0
0
0
0


1
0


0 
0
, A4 = 
 0
0 
0
0
Show that An = O for n ≥ 4
Solution:

0

0
A2 = 
 0
0
0
0
0
0
1
0
0
0
0
0
0
0

0
0 

0 
0
0
0
0
0
Since A4 = O and O times any matrix is O again, the result follows.
Sec 1.4, No 16: Prove that if A is nonsingular, then AT is nonsingular and
(AT )−1 = (A−1 )T
[Hint: (AB)T = B T AT ]
Solution: We need to find a matrix B that satisfies the equation AT B = BAT = I. If we can do
that, then AT is nonsingular and its inverse is B. The question suggests that B = (A−1 )T will do
the job. Let’s just verify this and we will be done:
AT (A−1 )T = (A−1 A)T = I T = I
(A−1 )T AT = (AA−1 )T = I T = I
Sec 1.4, No 21: Given
R=
cos θ − sin θ
sin θ cos θ
show that R is nonsingular and that R−1 = RT .
Solution:Note that
T
R R=
cos θ sin θ
− sin θ cos θ
cos θ − sin θ
sin θ cos θ
cos2 θ + sin2 θ
− cos θ sin θ + sin θ cos θ
=
− sin θ cos θ + cos θ sin θ
sin2 θ + cos2 θ
1 0
=
=I
0 1
Similarly we can show directly that RRT = I. Therefore RT serves as an inverse of R, and so R is
nonsingular with R−1 = RT
Sec 1.4, No 23: Let u be a unit vector in Rn (i.e., uT u = 1) and let H = I − 2uuT . Show that
H is an involution, i.e. H 2 = I.
Solution: Since uT u = 1, we have
H 2 = (I − 2uuT )(I − 2uuT )
= II − 2IuuT − 2uuT I + 4uuT uuT
= I − 4uuT + 4u(uT u)uT
= I − 4uuT + 4uuT = I
Sec 1.5, No 8 b): Compute the LU factorization of
2 4
A=
−2 1
Solution: The simplification
2 4
−2 1
→
2 4
0 5
giving us U can be written in terms of elementary matrices as
1 0
2 4
EA = U where E =
,U =
1 1
0 5
This implies that
A=E
−1
U = LU where L =
1 0
−1 1
Sec 1.5, No 8 d): Compute the LU factorization of


−2 1
2
1 −2 
A= 4
−6 −3 4
Solution: Gaussian elimination

−2 1
 4
1
−6 −3

−2 1
 0
3
0 −6
gives us


2
−−−−−−→
−2  R2 + 2R1 
4


2
−−−−−−→
2  R3 + 2R2 
−2

−2 1 2
−−−−−−→
0
3 2  R3 − 3R1
−6 −3 4

−2 1 2
0 3 2 
0 0 2
In terms of elementary matrices this is


−2 1 2
E3 E2 E1 A = U =  0 3 2 
0 0 2






1 0 0
1 0 0
1 0 0
E1 =  2 1 0  , E2 =  0 1 0  , E3 =  0 1 0 
0 0 1
−3 0 1
0 2 1
Then A = E1−1 E2−1 E3−1 U = LU where



 

1 0 0
1 0 0
1 0 0
1
0 0
L = E1−1 E2−1 E3−1 =  −2 1 0   0 1 0   0 1 0  =  −2 1 0 
0 0 1
3 0 1
0 −2 1
3 −2 1
Extra Prob 1: Find the determinant of the

2
 1
1
matrix

0 1
−1 3 
0 7
Solution: Use a cofactor expansion along column 2 to get


2 0 1
2
1
det  1 −1 3  = − det
= −(14 − 1) = −13
1 7
1 0 7
Extra Prob 2: Find the determinant of the

1
 3
0
matrix

2 2
4 2 
0 3
Solution: Use a cofactor expansion along row 3 to get


1 2 2
1
2
= 3(4 − 6) = −6
det  3 4 2  = 3 det
3 4
0 0 3
Extra Prob 3: Find the determinant of the

1
 4
3
matrix

2 3
5 9 
−1 2
Solution: Subtract column 1 from column 3. Then you have a matrix with the second and third
columns the same. Thus the determinant is 0.
Extra Prob 4: If A is the 3 × 3 matrix with columns c1 , c2 , c3 , that is A = [c1 c2 c3 ], and
det A = 2, find the determinants of the matrices
a) [2c1 3c2 c3 ]
Solution: det [2c1 3c2 c3 ] = (2)(3) det [c1 c2 c3 ] = 6 det A = 12 (two type II operations)
a) [c1 + 2c2 c2 c3 ]
Solution: det [c1 + 2c2 c2 c3 ] = det [c1 c2 c3 ] = det A = 2 since this is a type III operation
a) [c1 + c2 c1 + c2 c3 ]
Solution: det [c1 + c2 c1 + c2 c3 ] = 0 since the first two columns are the same (subtract one
from the other to get a zero column)
a) [c2 c1 c3 ]
Solution: det [c2 c1 c3 ] = − det [c1 c2 c3 ] = − det A = −2 since this is a type I operation
Extra Prob 5: For what values of λ is the following matrix singular?
1−λ
2
3
2−λ
Solution:
det
1−λ
2
3
2−λ
= (1 − λ)(2 − λ) − 6 = λ2 − 3λ − 4 = (λ + 1)(λ − 4)
so the matrix is singular when (λ + 1)(λ − 4) = 0, i.e. λ = −1 and λ = 4