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Chapter 1
Systems of Linear Equations and Matrices
Section 1.
Exercise Set 1.
1. (a), (c), and (f) are linear equations in x 1 , x 2 (^) ,and x 3. (b) is not linear because of the term x x 1 3. (d) is not linear because of the term x −^2. (e) is not linear because of the term x 1 3 5/^. 3. (a) and (d) are linear systems. (b) is not a linear system because the first and second equations are not linear. (c) is not a linear system because the second equation is not linear. 5. By inspection, (a) and (d) are both consistent; x 1 (^) = 3, x 2 (^) = 2 , x 3 (^) = − 2 , x 4 = 1 is a solution of (a) and x 1 (^) = 1 , x 2 (^) = 3 , x 3 (^) = 2 , x 4 = 2 is a solution of (d). Note that both systems have infinitely many solutions. 7. (a), (d), and (e) are solutions. (b) and (c) do not satisfy any of the equations. 9. (a) 7 5 3 5 3 7 7
x y x y
Let y = t. The solution is 5 3 7 7
x y y t
(b) (^) 1 2 3 4
1 2 3 4
x x x x
x x x x
Let x 2 (^) = r , x 3 = s ,and x 4 (^) = t .The solution is
1
x = r − s + t −
2 3 4
x r x s x t
11. (a)
corresponds to
1 1 2 2
x x x x
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
(b)
corresponds to
1 3 1 2 3 2 3
x x x x x x x
(c)^7 2 13 1 2 4 0 1
⎢⎣ ⎥⎦ corresponds to
1 2 3 4 1 2 3
x x x x x x x
(d)
corresponds to
1 2 3 4
x x x x
13. (a) The augmented matrix for (^1) 1 1
x x x
is
(b) The augmented matrix for
1 2 3 2 3
x x x x x
is 6 1 3 4 0 5 1 1
(c) The augmented matrix for 2 4 5 1 2 3 1 2 3 4 5
x x x x x x x x x x x
is
⎢⎣ −^ − ⎥⎦
(d) The augmented matrix for x 1 (^) − x 5 = 7 is [1 0 0 0 − 1 7].
15. If ( a , b , c ) is a solution of the system, then 2 ax 1 (^) + bx 1 (^) + c = y 1 , 2 ax 2 (^) + bx 2 (^) + c = y 2 ,and 2 ax 3 (^) + bx 3 (^) + c = y 3 which simply means that the points are on the curve. 17. The solutions of x 1 (^) + kx 2 = c are x 1 (^) = c − kt , x 2 = t where t is any real number. If these satisfy x 1 (^) + lx 2 (^) = d ,then c − kt + lt = d or c − d = ( k − l ) t for all real numbers t. In particular, if t = 0, then c = d , and if t = 1, then k = l.
True/False 1.
(a) True; x 1 (^) = x 2 (^) = " = xn = 0 will be a solution.
(b) False; only multiplication by nonzero constants is acceptable.
(c) True; if k = 6 the system has infinitely many solutions, while if k ≠ 6, the system has no solution.
(d) True; the equation can be solved for one variable in terms of the other(s), yielding parametric equations that give infinitely many solutions.
(e) False; the system 3 5 7 2 9 20 6 10 14
x y x y x y
has the
solution x = 1, y = 2.
(f) False; multiplying an equation by a nonzero constant c does not change the solutions of the system.
(g) True; subtracting one equation from another is the same as multiplying an equation by −1 and adding it to another.
(h) False; the second row corresponds to the equation 0 x 1 (^) + 0 x 2 = − 1 or 0 = −1 which is false.
Section 1.
Exercise Set 1.
1. (a) The matrix is in both row echelon and reduced row echelon form.
(b) The matrix is in both row echelon and reduced row echelon form.
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
Multiply the second row by 1 3
and then add − 1
times the new second row to the third row and add −3 times the new second row to the fourth row. 1 1 2 1 1 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0
Add the second row to the first row. 1 0 0 1 1 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0
The corresponding system of equations is 1 2 0
x w y z
− = or^
x w y z
Let z = s and w = t. The solution is x = t − 1, y = 2 s , z = s , w = t.
9. In Exercise 5, the following row echelon matrix occurred. 1 1 2 8 0 1 5 9 0 0 1 2
The corresponding system of equations is 1 2 3 2 3 3
x x x x x x
or
1 2 3 2 3 3
x x x x x x
Since x 3 (^) = 2 , x 2 = 5 2( ) − 9 = 1 ,and x 1 (^) = − − 1 2 2( ) + 8 = 3 .The solution is x 1 (^) =3, x 2 (^) = 1, x 3 (^) =2.
11. From Exercise 7, one row echelon form of the
augmented matrix is
The corresponding system is 2 1 2 0
x y z w y z
− = or^
x y z w y z
Let z = s and w = t. Then y = 2 s and x = 2 s − 2 s + t − 1 = t − 1. The solution is x = t − 1, y = 2 s , z = s , w = t.
13. Since the system has more unknowns (4) than equations (3), it has nontrivial solutions. 15. Since the system has more unknowns (3) than equations (2), it has nontrivial solutions. 17. The augmented matrix is
Interchange the first and second rows, then add −2 times the new first row to the second row. 1 2 0 0 0 3 3 0 0 1 1 0
Multiply the second row by
− ,then add − 1
times the new second row to the third row. 1 2 0 0 0 1 1 0 0 0 2 0
Multiply the third row by
, then add the new third row to the second row. 1 2 0 0 0 1 0 0 0 0 1 0
Add −2 times the second row to the first row. 1 0 0 0 0 1 0 0 0 0 1 0
The solution, which can be read from the matrix, is x 1 (^) = 0 , x 2 = 0 , x 3 (^) =0.
19. The augmented matrix is 3 1 1 1 0 5 1 1 1 0
Multiply the first row by 1 3
, then add −5 times
the new first row to the second row. 1 1 1 3 3 3 8 2 8 3 3 3
Multiply the second row by 3 8
1 1 1 3 3 3 1 4
SSM: Elementary Linear Algebra Section 1.
Add 1 3
− times the second row to the first row.
1 4 1 4
This corresponds to the system
1 3
2 3 4
x x
x x x
or
1 3
2 3 4
x x
x x x
Let x 3 (^) = 4 s and x 4 (^) = t .Then x 1 = − s and x 2 (^) = − s − t .The solution is x 1 (^) = − s , x 2 = − s − t , x 3 (^) = 4 , s x 4 = t.
21. The augmented matrix is
⎢⎣ −^ − ⎥⎦
Interchange the first and second rows. 1 0 1 3 0 0 2 2 4 0 2 3 1 1 0 2 1 3 2 0
⎢⎣^ − − ⎥⎦
Add −2 times the first row to the third row and 2 times the first row to the fourth row. 1 0 1 3 0 0 2 2 4 0 0 3 3 7 0 0 1 1 8 0
⎡ −^ − ⎤
Multiply the second row by 1 2
, then add − 3 times the new second row to the third row and −1 times the new second row to the fourth row. 1 0 1 3 0 0 1 1 2 0 0 0 0 1 0 0 0 0 10 0
⎡ −^ − ⎤
Add 10 times the third row to the fourth row, − 2 times the third row to the second row, and 3 times the third row to the first row. 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0
The corresponding system is
w y x y z
or 0
w y x y z
Let y = t. The solution is w = t , x = − t , y = t , z = 0.
23. The augmented matrix is
Interchange the first and second rows. 1 0 2 7 11 2 1 3 4 9 3 3 1 5 8 2 1 4 4 10
Add −2 times the first row to the second and fourth rows, and add −3 times the first row to the third row. 1 0 2 7 11 0 1 7 10 13 0 3 7 16 25 0 1 8 10 12
Multiply the second row by −1, then add 3 times the new second row to the third row and − 1 times the new second row to the fourth row. 1 0 2 7 11 0 1 7 10 13 0 0 14 14 14 0 0 15 20 25
⎢⎣ −^ − ⎥⎦
Multiply the third row by 1 14
− ,then add − 15 times the new third row to the fourth row. 1 0 2 7 11 0 1 7 10 13 0 0 1 1 1 0 0 0 5 10
⎢⎣ −^ − ⎥⎦
Multiply the fourth row by 1 5
− ,then add the
new fourth row to the third row, add −10 times the new fourth row to the second row, and add −7 times the new fourth row to the first row. 1 0 2 0 3 0 1 7 0 7 0 0 1 0 1 0 0 0 1 2
⎡ −^ − ⎤
Add 7 times the third row to the second row and 2 times the third row to the first row.
SSM: Elementary Linear Algebra Section 1.
The system has only the trivial solution, which
corresponds to sin α = 0 , cos β = 0 , tan γ= 0.
For 0 ≤ α ≤ 2 π , sin α = 0 ⇒ α = 0 , π , 2 π.
For 0 2 0 3 2 2
≤ β ≤ π , cos β= ⇒ β=^ π^ , π.
For 0 ≤ γ ≤ 2 π , tan γ= 0 ⇒ γ = 0 , π , 2 π.
Thus, the original system has 3 ⋅ 2 ⋅ 3 = 18 solutions.
35. Let X = x^2^ , Y = y^2 ,and Z = z^2 , then the system is 6 2 2 2 3
X Y Z
X Y Z
X Y Z
The augmented matrix is
Add −1 times the first row to the second row and −2 times the first row to the third row. 1 1 1 6 0 2 1 4 0 1 3 9
Multiply the second row by 1 2
− ,then add the new second row to the third row.
1 2 7 2
⎢⎣ −^ − ⎥⎦
Multiply the third row by 2 7
− ,then add 1 2 times the new third row to the second row and −1 times the new third row to the first row. 1 1 0 4 0 1 0 3 0 0 1 2
Add −1 times the second row to the first row. 1 0 0 1 0 1 0 3 0 0 1 2
This corresponds to the system
X
Y
Z
X = 1 ⇒ x = ± 1 Y = 3 ⇒ y = ± 3 Z = 2 ⇒ z = ± 2 The solutions are x = ±1, y = ± 3, z = ± 2.
37. (0, 10): d = 10 (1, 7): a + b + c + d = 7 (3, −11): 27 a + 9 b + 3 c + d = − 11 (4, −14): 64 a + 16 b + 4 c + d = − 14 The system is 7 27 9 3 11 64 16 4 14 10
a b c d a b c d a b c d d
and the augmented
matrix is
Add −27 times the first row to the second row and −64 times the first row to the third row. 1 1 1 1 7 0 18 24 26 200 0 48 60 63 462 0 0 0 1 10
Multiply the second row by 1 18
− ,then add 48
times the new second row to the third row.
4 13 100 3 9 9 19 214 3 3
Multiply the third row by
4 13 100 3 9 9 19 107 12 6
Add
− times the fourth row to the third row,
13 9
− times the fourth row to the second row,
and −1 times the fourth row to the first row.
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
4 10 3 3
Add 4 3
− times the third row to the second row
and −1 times the third row to the first row. 1 1 0 0 5 0 1 0 0 6 0 0 1 0 2 0 0 0 1 10
Add −1 times the second row to the first row. 1 0 0 0 1 0 1 0 0 6 0 0 1 0 2 0 0 0 1 10
The coefficients are a = 1, b = −6, c = 2, d = 10.
39. Since the homogeneous system has only the trivial solution, using the same steps of Gauss- Jordan elimination will reduce the augmented matrix of the nonhomogeneous system to the
form
1 2 3
d d d
Thus the system has exactly one solution.
41. (a) If a ≠ 0, the sequence of matrices is 1 1 0 1 1 0 0 1 0 1.
b b a a ad bc a b a
a b c d (^) c d −
⎡ ⎤ ⎡^ ⎤^ ⎡^ ⎤
⎢ ⎥ →^ ⎢^ ⎥→⎢^ ⎥
If a = 0, the sequence of matrices is 0 1 1 (^0 0 0 ) 1 0 0 1.
d d b c d c c c d b (^) b
⎡ ⎤ ⎡ ⎤^ ⎡^ ⎤^ ⎡^ ⎤
⎢ ⎥ →^ ⎢ ⎥→^ ⎢^ ⎥^ →⎢^ ⎥
→ ⎡^ ⎤
(b) Since ad − bc ≠ 0, the lines are neither identical nor parallel, so the augmented matrix a^ b^ k c d l
⎢⎣ ⎥⎦ will reduce to
1 1
k l
⎢⎣ ⎥⎦ and the system has exactly one solution.
43. (a) There are eight possibilities. p and q are any real numbers. 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0
p p q
⎡ ⎤ ⎡^ p^ q^ ⎤ ⎡ p ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 0 1 0 0 0 0 0 0
and
(b) There are 16 possibilities. p , q , r , and s are any real numbers. 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0
p q r
p p q
p q r s
p q p q r
p p q
⎡ ⎤ ⎡ p q r ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
⎡ p q ⎤ ⎡ p ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
and
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
(j)
tr 3 tr 1 0 1 3 3 6 3 2 4 12 3 9 17 2 7 tr 2 3 5 9 1 5 17 3 5 25
( D E )
− = ⎜^ ⎢^ − ⎥^ − ⎢^ − ⎥⎟
⎝ ⎣^ ⎦^ ⎣^ ⎦⎠
= ⎜^ ⎢^ − − ⎥⎟
⎝ ⎣ −^ −^ − ⎦⎠
(k) 4 tr 7 4 tr 28 7 0 14 4 28 14 4 42 168
B = ⎛⎜^ ⎡⎢^ − ⎤⎥⎞⎟
(l) tr( A ) is not defined because A is not a square matrix.
5. (a)
AB
= ⎢^ − ⎥
⎢ ⎥ ⎣⎢^ ⎥⎦
= ⎢^ − ⋅ + ⋅ ⋅ + ⋅ ⎥
= ⎢^ − ⎥
(b) BA is not defined since B is a 2 × 2 matrix and A is 3 × 2.
(c)
E D
= ⎢^ − ⎥ ⎢^ − ⎥
= ⎢^ − ⎥
(d)
AB C
= ⎢^ − ⎥⎡^ ⎤
⎢ ⎥ ⎣⎢^ ⎥⎦
⎡ ⋅^ −^ ⋅^ ⋅^ −^ ⋅^ ⋅^ −^ ⋅ ⎤
= ⎢^ − ⋅ + ⋅ − ⋅ + ⋅ − ⋅ + ⋅ ⎥
= ⎢^ − ⎥
SSM: Elementary Linear Algebra Section 1.
(e)
1 2 4 1^ 1 3^ 4 4^ 1 1^ 4 2^ 1 5
1 1 0 1^ 2 3^ 0 4^ 2 1^ 0 2^ 2 5
A BC
= ⎢^ − ⎥ ⎡^ ⎤ ⎡^ ⎤
⎢ ⎥ ⎜⎝^ ⎢⎣^ ⎥ ⎢⎦ ⎣^ ⎥⎦⎟⎠
= ⎢^ − ⎥
⎢ ⎥ ⎢⎣^ ⋅^ +^ ⋅^ ⋅^ +^ ⋅^ ⋅^ +^ ⋅ ⎥⎦
= ⎢^ − ⎥
⎢ ⎥ ⎣⎢^ ⎥⎦
= ⎢^ − ⎥
(f)
CC^ T
= ⎡^ ⎤ ⎢^ ⎥
= ⎡^ ⋅^ +^ ⋅^ +^ ⋅^ ⋅^ +^ ⋅^ +^ ⋅ ⎤
= ⎡^ ⎤
(g)
T T
T
DA
= ⎜^ ⎢^ − ⎥ ⎢^ − ⎥⎟
⎝ ⎣^ ⎦ ⎣^ ⎦⎠
= ⎜^ ⎢^ − ⋅ − ⋅ + ⋅ − ⋅ + ⋅ + ⋅ ⎥⎟
⎝ ⎣ ⋅^ −^ ⋅^ +^ ⋅^ ⋅^ +^ ⋅^ +^ ⋅ ⎦⎠
= ⎜^ ⎢^ − ⎥⎟
⎝ ⎣^ ⎦⎠
T
⎟⎟
= ⎡^ − ⎤
SSM: Elementary Linear Algebra Section 1.
(k)
tr 2 tr 4 1 2 2 2 2 5 0 2 1 6 4 6 1 3 3 0 1 1 3 2 1 1 3 1 12 2 8 tr 4 3 1 0 4 1 1 2 4 1 1 1 2 2 2 2 3 5 0 2 1 5 2 2 1 5 1 6 4 6
C T^ AT ET
+ = ⎜^ ⎢^ ⎥^ ⎡^ − ⎤+⎢^ ⎥⎟
⎜⎜ ⎢ ⎥ ⎢⎣^ ⎥⎦ ⎢ ⎥⎟⎟
⎝ ⎣^ ⎦^ ⎣^ ⎦⎠
= ⎢^ ⋅ + ⋅ − ⋅ + ⋅ ⋅ + ⋅ ⎥^ +⎢
tr 12 2 5 2 2 2 6 8 7 6 4 6 15 3 12 tr 14 0 7 12 12 13 15 0 13 28
= ⎜^ ⎢^ − ⎥^ +⎢^ ⎥⎟
⎝ ⎣^ ⎦^ ⎣^ ⎦⎠
= ⎜^ ⎢^ ⎥⎟
⎝ ⎣^ ⎦⎠
(l)
tr tr 1 1 2 4 1 1 2 4 1 3 2 5 1 1
6 1 1 4 3 2 6 3 1 1 3 5 tr 1 1 1 4 2 2 1 3 1 1 2 5 4 1 1 4 3 2 4 3 1 1 3 5
T EC T^ TA
⎜ ⎜^ ⎢^ ⎥ ⎢^ ⎥^ ⎟ ⎢^ ⎥⎟
⎝ ⎝^ ⎣^ ⎦ ⎣^ ⎦^ ⎠ ⎣^ ⎦⎠
= ⎜ ⎢^ − ⋅ + ⋅ + ⋅ − ⋅ + ⋅ + ⋅ ⎥
tr 7 8 1 2 14 28 1 1
3 0 tr 16 7 14 1 2 (^34 8 28 1 )
tr 16 3^ 7 1^ 14 1^16 34 34 0
T
T
⎜ ⎜^ ⎟ ⎢^ ⎥⎟
⎝ ⎠ ⎣^ ⎦⎠
⎜ ⎜ ⎢^ ⎥^ ⎟⎢^ ⎥⎟
⎝ ⎝^ ⎣^ ⎦^ ⎠⎣^ ⎦⎠
= ⎜^ ⎡^ ⎤⎢^ − ⎥⎟
⎜⎜ ⎢⎣^ ⎥⎦⎢ ⎥⎟⎟
⎝ ⎣^ ⎦⎠
= ⋅^ −^ ⋅^ +^ ⋅^ ⋅ 0) + (7 ⋅ 2) + (14 ⋅1)
tr 55 28 122 44 55 44 99
= ⎛^ ⎡^ ⎤⎞
7. (a) The first row of AB is the first row of A times B.
1
[ ] [ ]
[ ]
[ ]
B
= − ⎢^ ⎥
a
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
(b) The third row of AB is the third row of A times B. 3 6 2 4 0 4 9 0 1 3 7 7 5 0 0 63 0 4 63 0 12 45 63 67 57
[ ]
[ ]
[ ]
[ ]
B
= ⎢^ ⎥
a
(c) The second column of AB is A times the second column of B.
2
[ A ]
⎡ −^ ⎤ ⎡− ⎤
= ⎢^ ⎥ ⎢^ ⎥
= ⎢^ − + + ⎥
= ⎢^ ⎥
b
(d) The first column of BA is B times the first column of A.
1
[ B ]
= ⎢^ ⎥ ⎢ ⎥
⎡ −^ + ⎤
= ⎢^ + + ⎥
= ⎢^ ⎥
a
(e) The third row of AA is the third row of A times A. 3 3 2 7 0 4 9 6 5 4 0 4 9 0 24 0 0 20 36 0 16 81 24 56 97
[ ]
[ ]
[ ]
[ ]
A
= ⎢^ ⎥
a
(f) The third column of AA is A times the third column of A.
3
[ A ]
= ⎢^ ⎥ ⎢ ⎥
⎡ −^ + ⎤
= ⎢^ + + ⎥
= ⎢^ ⎥
a
9. (a)
AA
⎡^ − ⎤
= ⎢^ ⎥
⎡^ − ⎤ ⎡ ⎤ ⎡− ⎤
⎡ ⎤ ⎡ ⎤ ⎡^ − ⎤ ⎡ ⎤
⎡ ⎤ ⎡ ⎤ ⎡^ − ⎤ ⎡ ⎤
(b)
BB
= ⎢^ ⎥
11. (a)
A ,
= ⎢^ − ⎥
1 2 3
x x x
= ⎢^ ⎥
x
= ⎢^ −⎥
b
The equation is
1 2 3
x x x
⎢ − ⎥ ⎢^ ⎥= ⎢ −⎥
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
(b)^2 2 0 0 0 0
f ⎛⎜^ ⎞⎟^ = ⎛⎜^ + ⎞⎟^ =⎛⎜^ ⎞⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ y 1
1 2^ x
f ( x ) = x
(c)^4 4 3 3 3 3
f ⎛⎜^ ⎞⎟^ = ⎛⎜^ + ⎞⎟^ =⎛⎜^ ⎞⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 5^ y
x 4 8
x f ( x )
(d)^2 2 2 2 2 2
f ⎛⎜^ ⎞⎟^ = ⎛⎜^ − ⎞⎟^ =⎛⎜^ ⎞⎟ ⎝ −^ ⎠ ⎝ −^ ⎠ ⎝ − ⎠ y
x 2
–2 x
1
f ( x )
27. Let A =[ aij ].
11 12 13 11 12 13 21 22 23 21 22 23 31 32 33 31 32 33
a a a (^) x a x a y a z a a a y a x a y a z a a a z a x a y a z x y x y
= ⎢^ − ⎥
The matrix equation yields the equations 11 12 13 21 22 23 31 32 33 0
a x a y a z x y a x a y a z x y a x a y a z
For these equations to be true for all values of x , y , and z it must be that a 11 (^) = 1, a 12 (^) = 1, a 21 (^) =1, a 22 (^) = −1, and aij = 0 for all other i , j. There is
one such matrix,
A.
= ⎢^ − ⎥
29. (a) Both 1 1 1 1
⎢⎣ ⎥⎦ and^
⎡ −^ −⎤
⎢⎣ (^) − −⎥⎦ are square roots
of 2 2 2 2
(b) The four square roots of 5 0 0 9
⎢⎣ ⎥⎦ are
5 0 0 3
and
5 0 0 3
True/False 1.
(a) True; only square matrices have main diagonals.
(b) False; an m × n matrix has m row vectors and n column vectors.
(c) False; matrix multiplication is not commutative.
(d) False; the i th row vector of AB is found by multiplying the i th row vector of A by B.
(e) True
(f) False; for example, if A = ⎡⎢^14 −^31 ⎤⎥ ⎣ ⎦
and
1 0 B (^) 0 2 , = ⎡^ ⎤ ⎢⎣ ⎥⎦ then^
AB 4 2
= ⎡^ ⎤
⎢⎣ (^) −⎥⎦ and tr( AB ) = −1, while tr( A ) = 0 and tr( B ) = 3.
(g) False; for example, if 1 3 4 1
A = ⎡⎢^ ⎤⎥
and
1 0 0 2
B = ⎡⎢^ ⎤⎥,
then 1 4 6 2
( AB ) T = ⎡⎢^ ⎤⎥,
while
1 8 3 2
A BT^ T = ⎡⎢^ ⎤⎥.
(h) True; for a square matrix A the main diagonals of A and AT are the same.
(i) True; if A is a 6 × 4 matrix and B is an m × n matrix, then AT is a 4 × 6 matrix and BT is an n × m matrix. So BT AT is only defined if m = 4, and it can only be a 2 × 6 matrix if n = 2.
(j) True, (^11 ) 11 22
tr
tr
nn nn
cA ca ca ca c a a a c A
(k) True; if A − C = B − C , then aij − cij = bij − cij , so it follows that aij = bij.
SSM: Elementary Linear Algebra Section 1.
(l) False; for example, if 1 2 3 4
A = ⎡⎢^ ⎤⎥,
B = ⎡⎢^ ⎤⎥,
and 1 0 0 0
C = ⎡⎢^ ⎤⎥,
then 1 0 3 0
AC = BC = ⎡⎢^ ⎤⎥.
(m) True; if A is an m × n matrix, then for both AB and BA to be defined B must be an n × m matrix. Then AB will be an m × m matrix and BA will be an n × n matrix, so m = n in order for the sum to be defined.
(n) True; since the j th column vector of AB is A [ j th column vector of B ], a column of zeros in B will result in a column of zeros in AB.
(o) False; 1 0 1 5 1 5 0 0 3 7 0 0
Section 1.
Exercise Set 1.
5.^1
1 3 5 20 1 1 5 10
2 4 3 4^4
B −^ = ⎡⎢^ ⎤⎥
= ⎡^ ⎤
= ⎡^ ⎤
1 (^1 ) 1 3
D −^
⎡ ⎤ ⎡ ⎤ ⎡^ ⎤
9. Here, 1 2
a = d = ( e x^ + e − x )and
1 2
b = c = ( e x^ − e − x ),so
2 2
2 2 2 2
x x x x
x x x x
ad bc e e e e
e e e e
− −
− −
Thus,
1 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2
x x x x x x x x
x x x x x x x x
e e e e e e e e
e e e e e e e e
− −^ − − −
− − − −
= ⎢^ ⎥
11.^2
BT = ⎡⎢^ ⎤⎥;
1 1 1 5 5 1 3 1 20 10
( BT ) −^ ( B −) T
⎡ − ⎤ ⎡^ − ⎤
= ⎢ ⎥= ⎢^ ⎥=
13.^70
ABC = ⎡⎢^ ⎤⎥
70 79 122 45^122
( ABC ) −^ = ⎡⎢^ − − ⎤⎥
= ⎡^ − ⎤
6 ( 1 ) 4 ( 2 )^2 6 22
C −^ = ⎡⎢−^ −^ ⎤⎥^ = ⎡⎢^ −^ − ⎤⎥
B −^ = ⎡⎢^ ⎤⎥^ = ⎡⎢^ ⎤⎥
3 2 1 5^5 3 5
A −^ = ⎡⎢^ −^ ⎤⎥^ =⎡⎢^ −⎤⎥
C −^ B −^ A −^ = ⎛⎜^ ⎡⎢−^ −^ ⎤ ⎡⎥ ⎢^ ⎤ ⎡⎥ ⎢^ −⎤⎥⎞⎟
⎝ ⎣^ ⎦ ⎣−^ ⎦ ⎣−^ ⎦⎠
= ⎡^ − ⎤
3 2 1 7^1
A = A −^ −
= ⎡−^ − ⎤
− − − ⋅ ⎢⎣^ −^ −⎥⎦
= − ⎡^ −^ − ⎤
= ⎡^ ⎤
Thus
2 7 1 3 7 7
A.
= ⎢^ ⎥
17.^1
5 2 13 13 4 1 13 13
1 5 2 4^4
I + A = (( I + A ) −^ )−
= ⎡^ − ⎤
− ⋅ − ⋅ ⎢⎣^ − −⎥⎦
= − ⎡^ −⎤
SSM: Elementary Linear Algebra Section 1.
(b) Note that the inverse of 1 3 3 1
⎡^ − ⎤
⎢⎣ (^) − − ⎥⎦ is
1 3 10 10 3 1 10 10
⎡− − ⎤ ⎡^ −^ − ⎤
⎢ ⎥=⎢^ ⎥
⎣ −^ ⎦ ⎢ − ⎥
Thus
1 3 1 1 3 10 10 3 1 10 10
A −.
1 1 1 3 3 3 3 1 3 (^1 3 1 3 ) 10 10 10 10 10 10 3 1 3 1 3 1 10 10 10 10 10 10 1 1 9 3 2 3 1 3 25 50 10 10 3 2 3 1 50 25 10 10 1 27
A −^ A −
= = ⎜^ ⎢ − − ⎥ ⎢ − − ⎥ ⎟⎢ − − ⎥
⎢⎣ −^ −^ ⎥ ⎢⎦ ⎣ − ⎥⎦
(c)^2
A A I
− + = ⎢^ − − ⎥^ − ⎢^ − ⎥^ +⎢^ ⎥
= ⎢^ − − ⎥
(d)
p A ( ) A I
= − = ⎢^ − ⎥^ −⎢^ ⎥
= ⎢^ − ⎥
(e) 2 2 18 0 0 3 0 0 1 0 0 0 16 12 0 1 3 0 1 0 0 12 16 0 3 1 0 0 1 16 0 0 0 14 15 0 15 14
p A ( ) = A − A + I ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ = ⎢^ − − ⎥^ − ⎢^ − ⎥^ +⎢^ ⎥ ⎢ (^) − ⎥ ⎢ (^) − −⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ = ⎢^ − − ⎥ ⎢ (^) − ⎥ ⎣ ⎦
(f)^3 2 27 0 0 6 0 0 4 0 0 0 26 18 0 2 6 0 4 0 0 18 26 0 6 2 0 0 4 25 0 0 0 32 24 0 24 32
p A ( ) = A − A + I ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ = ⎢^ − ⎥^ − ⎢^ − ⎥^ +⎢^ ⎥ ⎢ ⎥ ⎢ (^) − − ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ = ⎢^ − ⎥ ⎢ ⎥ ⎣ ⎦
Chapter 1: Systems of Linear Equations and Matrices SSM: Elementary Linear Algebra
27. Since a 11 (^) a 22 (^) " ann ≠0,none of the diagonal entries are zero.
Let
11
22
1 1
1
nn
a
a
a
B
AB = BA = I so A is invertible and B = A −^1.
29. (a) If A has a row of zeros, then by formula (9) of Section 1.3, AB will have a row of zeros, so A cannot be invertible.
(b) If B has a column of zeros, then by formula (8) of Section 1.3, AB will have a column of zeros, and B cannot be invertible.
31.^1 2 1 2 1 1 2 1 1 2 1 2 2 1 1 2 1 2 2 1 1 2 2 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 2
T T T T T T T T T T T T T
C B A BAC DA B C C
CA B A B C C B A BAC DA B C CA B A B C C
DA B C CA B A B
DA B C C B A CA B A B C B A
D CA B A BC B A
− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −
33.^1 1 1 1 1 1 1 1 1
1 1 1 1 1
AB AC D C D B A AC CD D
B A A C C DD
B
− − − − − − − − − − − − − − −
35. Let
11 12 13 21 22 23 31 32 33
b b b X b b b b b b
= ⎢^ ⎥
11 31 12 32 13 33 11 21 12 22 13 23 21 31 22 32 23 33
AX I
b b b b b b b b b b b b b b b b b b
⎢⎣ +^ +^ + ⎥⎦ ⎣ ⎦
Consider the equations from the first columns of the matrices. 11 31 11 21 21 31
b b b b b b
Thus b 11 (^) = b 31 (^) = − b 21 ,so 11 1 2
b = , 21 1 2
b = − , and 31 1 2
b =.
Working similarly with the other columns of both matrices yields
1 1 1 2 2 2 (^1 1 1 ) 2 2 2 1 1 1 2 2 2
A −.
