Linear Algebra Solutions

SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013

MATH 54 LINEAR ALGEBRA

AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA

All the Page numbers, Theorem numbers, etc. without stating the reference are from the

textbook [1]. Other references will be specified.

1.9 The Matrix of a Linear Transformation

24. (a) False. In fact, the linear transformation x 7→ Ax can never map R3 to R4.

The range of the this linear transformation, by definition, is the span of the set of column

vectors of A. Since there are at most three pivots in the matrix A, in the echelon form, the

last row cannot have a pivot. Thus this span cannot be the whole R4.

(b) True. Page 73, Theorem 10. For every linear transformation T from Rn to Rm, there

is a unique m × n matrix A such that T (x) = Ax. (Note that in the end of Page 67, it

is said that every matrix transformation is a linear transformation, yet there are examples

of linear transformations that are not matrix transformations. This refers to general linear

transformation on general vector space where there is no preferred basis.)

(c) True. Again, Page 73, Theorem 10. (Although technically, it should be stated that

the given linear transformation from Rn to Rm is T .)

(d) False. One-to-one, (or injectivity), means every vector in the codomain, has at most

one preimage. For T to be a mapping, it is automatically, each vector in Rn maps to only

one vector in Rm.

(e) False. See Table 3 on Page 76.

26. As we have already seen, the standard matrix of T is

A =

[1 −2 3

4 9 −8

].

By subtracting 4 times first row from the second row, we get the echelon form[1 −2 3

0 17 −20

].

1

2 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA

There are two pivots, two rows and three columns, thus the columns of A span R2 but are

not linear independent. By Theorem 12 on Page 79, T is onto but not one-to-one.

28. As we see in the figure, the column vectors of the standard matrix A, a1, a2 are not

multiple to each other. Thus {a1, a2} span R2 and is linear independent. (See Page 60 for

linear independence. For span R2, one can count the pivots in the 2 × 2 matrix A. Since

column vectors are linear independent, there is a pivot in each column, thus totally there

are two pivots. This implies each row must have a pivot.) Again, by Theorem 12 on Page

79, T is onto and one-to-one.

35. (1) If a linear transformation T : Rn → Rm is onto, then m 6 n. Assume A is the

standard matrix of T , then the columns of A span Rm, or equivalently, in the echelon form

of A, there is a pivot in each row. Since A has m rows and n columns, this means A has

exactly m pivots and (since there can be at most one pivot in each column) m 6 n.

(2) If a linear transformation T : Rn → Rm is one-to-one, then m > n. (The argument

is similar to above, just switch the roles of rows and columns.) Assume A is the standard

matrix of T , then the columns of A are linear independent, or equivalently, in the eche-

lon form of A, there is a pivot in each column. Since A has m rows and n columns, this

means A has exactly n pivots and (since there can be at most one pivot in each row) m > n.

36. The reason is that this question is equivalent to (by definition) “Is there a preimage

for every vector in the codomain?” or (in terms of matrix equation, A is the standard

matrix of T ) “Does Ax = b have a solution for every b?”. In both forms, it is clear that

this is an existence question.

2.1 Matrix Operations

2. Let

A =

[2 0 −1

4 −5 2

], B =

[7 −5 1

1 −4 −3

], C =

[1 2

−2 1

], D =

[3 5

−1 4

], E =

[−5

3

](1)

A+ 3B =

[23 −15 2

7 −17 −7

].

SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 3

(2) 2C − 3E is not defined since C and E have different size.

(3)

DB =

[26 −35 −12

−3 −11 −13

].

(4) EC is not defined, since E has only one column and C has two rows.

4.

A− 5I3 =

0 −1 3

−4 −2 −6

−3 1 −3

, (5I3)A =

25 −5 15

−20 15 −30

−15 5 10

.

6.

A =

4 −3

−3 5

0 1

, B =

[1 4

3 −2

].

So

b1 =

[1

3

],b2 =

[4

−2

].

and we have

Ab1 =

−5

12

3

, Ab2 =

22

−22

−2

.From either of the two ways of calculation, (which do not have much difference) we get

AB = [Ab1 Ab2] =

−5 22

12 −22

3 −2

.8. B has the same number of rows as BC, which is 5.

10. It is easy to calculate that

AB = AC =

[−21 −21

7 7

].

12. Let B = [b1 b2], then since AB = 0, b1 and b2 are solutions to the matrix equation

Ax = 0. By the elementary row operation that adds 23

times of the first row to the second


row, we have

A ∼[

3 −6

0 0

].

Therefore the general solutions of Ax = 0 is

x = x2

[2

1

].

For example, we can choose

B =

[2 4

1 2

].

16. (a) True. This follows directly from the definition.

(b) False. In fact, AB = [Ab1 Ab2 Ab3].

(c) True. In (AB)T = BTAT , we take B = A, then we get (A2)T = (AT )2.

(d) False. In fact (ABC)T = CTBTAT not equal to CTATBT in general. A counter example

is

A =

[0 1

0 0

], B =

[0 0

1 0

], C =

[1 0

0 1

].

Then

(ABC)T =

[1 0

0 0

]6=[

0 0

0 1

]= CTATBT .

(e) True. Since (A+ B)T = AT + BT , by simple induction on n, we have the transpose of

a sum of n matrices equals the sum of their transposes.

18. The third column of AB is also all zeros. Since if B = [b1 b2 · · · bn], then

AB = [Ab1 Ab2 · · · Abn]. The third column is Ab3 = A0 = 0.

20. The first two columns of AB are also equal. Same as Problem 18. If b1 = b2, then

Ab1 = Ab2.

22. Again, assuming B = [b1 b2 · · · bn] and since the columns of B are linear dependent,

we can find c1, . . . , cn not all zero such that

c1b1 + · · ·+ cnbn = 0.

Therefore applying A to the left we have

c1Ab1 + · · ·+ cnAbn = A(c1b1 + · · ·+ cnbn) = A0 = 0.

This shows that the columns of AB, Ab1, . . . , Abn are linear dependent.


24. Since the columns of A span R3. We can find the solutions x = xj to Ax = ej,

j = 1, 2, 3 where e1, e2, e3 are standard unit vectors in R2. Then D = [x1 x2 x3] satisfies

AD = I3.

27.

uTv = vTu = −3a+ 2b− 5c

(think this as a 1× 1 matrix).

uvT =

−3a −3b −3c

2a 2b 2c

−5a −5b −5c

, vuT =

−3a 2a −5a

−3b 2b −5b

−3c 2c −5c

.28. By the formula (AB)T = BTAT and (AT )T = A, we have uTv = (vTu)T = vTu

since this is a 1×1 matrix, its transpose is the same as itself. Also we have uvT = (vuT )T .

32. For j = 1, . . . , n, the j-th column of AIn is Aej where e1, . . . , en are standard unit

vector in Rn. Assume A = [a1 · · · an], then

Aej =

(∑i 6=j

0ai

)+ 1aj = aj

Therefore AIn has exactly the same columns as A, or AIn = A.

34. Use the formula (AB)T = BTAT repeatedly, (ABx)T = (Bx)TAT = xTBTAT .

2.2 The Inverse of a Matrix

2. Use Theorem 4 on Page 105, since determinant of the matrix is 3 · 5− 2 · 8 = −1 6= 0,[3 2

8 5

]−1= (−1)−1

[5 −2

−8 3

]=

[−5 2

8 −3

].

4. Again, use Theorem 4 on Page 105, since determinant of the matrix is 2 · (−6) − 4 ·(−4) = 4 6= 0, [

2 −4

4 −6

]−1=

1

4

[−6 4

−4 2

]=

[−3

21

−1 12

].


6. Since [7 3

−6 −3

]−1=

1

7(−3)− 3(−6)

[−3 −3

6 7

]=

[1 1

−2 −73

],

The solution to the given system, which is equivalent to[7 3

−6 −3

] [x1x2

]=

[−9

4

]is [

x1x2

]=

[7 3

−6 −3

]−1 [ −9

4

]=

[1 1

−2 −73

] [−9

4

]=

[−5263

].

7. (a) Since detA = 2 6= 0, we have

A−1 =1

2

[12 −2

−5 1

]=

[6 −1

−52

12

].

Therefore the solutions to Ax = bj, j = 1, 2, 3, 4 are

x1 = A−1b1 =

[−9

4

],

x2 = A−1b2 =

[11

−5

],

x3 = A−1b3 =

[6

−2

],

x4 = A−1b4 =

[13

−5

].

(b) The augmented matrix is

[A b1 b2 b3 b4] =

[1 2 −1 1 2 3

5 12 3 −5 6 5

].

By subtracting 5 times the first row from the second row, we get[1 2 −1 1 2 3

0 2 8 −10 −4 −10

].

By subtracting the second row from the first row, we get[1 0 −9 11 6 13

0 2 8 −10 −4 −10

].


By multiplying 12

to the second row, we get the reduced echelon form[1 0 −9 11 6 13

0 1 4 −5 −2 −5

]which equals [I A−1b1 A

−1b2 A−1b3 A

−1b4]. The last four columns are exactly the solu-

tions to the four equations.

8. By multiplying P−1 to the left-hand side and P to the right-hand side of the equation

A = PBP−1, we have

P−1AP = P−1PBP−1P = IBI = B

since P−1P = I.

10. (a) False. The elementary row operations that reduce A to the identity In are

equivalent to multiply A−1 from the left. Therefore it will change A−1 to A−1A−1 which

does not necessarily equal to In.

(b) True. Since by definition, A−1A = AA−1 = I, this shows that A−1 is also invertible

and (A−1)−1 = A.

(c) False. If A,B are invertible n×n matrices, then by Theorem 6 part (b) on page 107,

we have (AB)−1 = B−1A−1 which does not necessarily equal to A−1B−1.

(d) True. Let xj be the solution of Ax = ej. Then for any b = (b1, · · · , bn) ∈ Rn, since

b = b1e1 + · · ·+ bnen, we get that x = b1x1 + · · ·+ bnxn satisfies the equation Ax = b. In

other words, the columns of A span Rn. Therefore in the reduced echelon form of A, there

is a pivot in every row. So there are exactly n pivots, i.e. the reduced echelon form is In.

In other words, A is row equivalent to In. Thus A is invertible. (Also see Theorem 8 on

page 114.)

(e) True. See Theorem 7 on page 109.

12. Since A is invertible, we have A−1A = I, therefore

D = ID = (A−1A)D = A−1(AD) = A−1I = A−1.

14. Since D is invertible, we can multiply D−1 to the right-hand side of the equation

(B − C)D = 0 to get

0 = (B − C)DD−1 = (B − C)I = B − C.


Therefore B = C.

16. Since B is invertible, we can write A = AI = A(BB−1) = (AB)B−1. Now both AB

and B−1 are invertible matrices, so the product A is also invertible.

18. Since B is invertible, as in Problem 16, A = (AB)B−1 = (BC)B−1 = BCB−1.

20. (a) Since (A − AX)−1 = X−1B, we have B = IB = (XX−1)B = X(X−1B) =

X(A − AX)−1. Now both X and (A − AX)−1 are invertible, we have that B is also

invertible.

(b) Since B = X(A−AX)−1, we have B(A−AX) = X, so BA−BAX = X and thus

BA = X +BAX = (I +BA)X.

We need to invert I +BA, but since X is invertible, we have I +BA = BAX−1 where all

of B,A,X−1 are invertible. Therefore I +BA is invertible and we can solve that

X = (I +BA)−1BA.

24. The same explanation as Problem 10 part d. The columns of A span Rn. Therefore

in the reduced echelon form of A, there is a pivot in every row. So there are exactly n

pivots, i.e. the reduced echelon form is In. In other words, A is row equivalent to In. Thus

A is invertible. (Also see Theorem 8 on page 114.)

25. We consider the following two separated cases:

Case 1: a = b = 0. Then the matrix A is row equivalent to[c d

0 0

]which can have at most one pivot. Therefore there is a row without a pivot, we know the

system Ax = 0 has infinite solutions.

Case 2: a and b are not both zero, then let x0 =

[−ba

], we can check

Ax0 =

[−ab+ ba

−cd+ da

]= 0.

Therefore Ax = 0 has more than one solution: at least 0 and x0.


30. [3 6 1 0

4 7 0 1

]∼[

1 2 13

0

4 7 0 1

]∼[

1 2 13

0

0 −1 −43

1

]∼[

1 2 13

0

0 1 43−1

]∼[

1 0 −73

2

0 1 43−1

].

Therefore [3 6

4 7

]−1=

[−7

32

43−1

].

32. Since 1 2 −1

−4 −7 3

−2 −6 4

∼ 1 2 −1

0 1 −1

0 −2 2

∼ 1 2 −1

0 1 −1

0 0 0

,we know the original matrix is not row equivalent to the identity matrix. Therefore it is

not invertible.

33. Since 1 0 0 1 0 0

1 1 0 0 1 0

1 1 1 0 0 1

∼ 1 0 0 1 0 0

0 1 0 −1 1 0

0 1 1 −1 0 1

∼ 1 0 0 1 0 0

0 1 0 −1 1 0

0 0 1 0 −1 1

we have 1 0 0

1 1 0

1 1 1

−1 =

1 0 0

−1 1 0

0 −1 1

.Similarly,

1 0 0 0 1 0 0 0

1 1 0 0 0 1 0 0

1 1 1 0 0 0 1 0

1 1 1 1 0 0 0 1

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 1 0 0

0 1 1 0 −1 0 1 0

0 1 1 1 −1 0 0 1

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 1 0 0

0 0 1 0 0 −1 1 0

0 0 1 1 0 −1 0 1

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 1 0 0

0 0 1 0 0 −1 1 0

0 0 0 1 0 0 −1 1

.Therefore

1 0 0 0

1 1 0 0

1 1 1 0

1 1 1 1

−1

=

1 0 0 0

−1 1 0 0

0 −1 1 0

0 0 −1 1

.


In general, the corresponding n× n matrix is

A =

1 0 0 · · · 0 0

1 1 0 · · · 0 0

1 1 1 · · · 0 0...

......

. . ....

...

1 1 1 · · · 1 0

1 1 1 · · · 1 1

.

We can write A = [aij]16i,j6n where

aij =

{1 if i > j

0 if i < j

From the case of n = 3, 4 we can guess the inverse of A is

B =

1 0 0 · · · 0 0

−1 1 0 · · · 0 0

0 −1 1 · · · 0 0...

......

. . ....

...

0 0 0 · · · 1 0

0 0 0 · · · −1 1

.

We can also write B = [bij]16i,j6n where

bij =

1 if i = j

−1 if i = j + 1

0 if i < j or i > j + 1

We can check directly that the product C = AB = [cij]16i,j6n is actually the identity

matrix. Since

cij =n∑

k=1

aikbkj,

we first use the formula for aik to get

cij =i∑

k=1

bkj = b1j + · · ·+ bij.

Case 1: If i > j, then there are two nonzero terms in the above sum. cij = bjj + bj+1,j =

1− 1 = 0.

Case 2: If i = j, then there are only one nonzero term in the above sum. cij = bjj = 1.


Case 3: If i < j, then every term in the above sum is zero, so cij = 0.

This shows that

cij = δij =

{1 if i = j

0 if i 6= j

(Kronecker delta symbol, see [2]) or AB = I.

There is also an algebraic way to prove AB = I is as follows, let

J =

0 0 0 · · · 0 0

1 0 0 · · · 0 0

0 1 0 · · · 0 0...

......

. . ....

...

0 0 0 · · · 0 0

0 0 0 · · · 1 0

.

Then J is a nilpotent matrix: Jn = 0. (See [3]) In fact, Jk has entries 1 on the sub-diagonal:

i− j = k for k = 1, . . . , n− 1. Now we can write A = I + J + J2 + · · ·+ Jn−1, B = I − Jand we have

AB = (I + J + J2 + · · ·+ Jn−1)(I − J) = I − J + J − J2 + · · ·+ Jn−1 − Jn = I − Jn = I.

Remark: Actually this is a standard property for all nilpotent elements in a commutative

ring with a unit (See e.g. [4]).

The third way to prove this is by induction. We write An, Bn for the n × n matrices.

We need to prove AnBn = In for every n. It is obvious that this is true for n = 1 since

A1 = B1 = 1 (as either numbers or 1 × 1 matrices.) Now suppose An−1 = Bn−1 = In−1,

since

An =

[An−1 0

vTn−1 1

], Bn =

[Bn−1 0

wTn−1 1

],

where

vn−1 =

1...

1

1

, wn−1 =

0...

0

−1

.Therefore by multiplication of partitioned matrices (see Section 2.4)

AnBn =

[An−1Bn−1 0

vTn−1Bn−1 + wTn−1 1

]=

[In−1 0

(BTn−1vn−1 + wn−1)

T 1

].


It remains to check that BTn−1vn−1 + wn−1 = 0 which is straightforward. In fact, BT

n−1vn−1is the sum of all the columns of BT

n−1, which is en−1 ∈ Rn−1, i.e. the negative wn−1.

34. Again we start with 3× 3 matrix 1 0 0 1 0 0

2 2 0 0 1 0

3 3 3 0 0 1

∼ 1 0 0 1 0 0

1 1 0 0 12

0

1 1 1 0 0 13

∼

1 0 0 1 0 0

0 1 0 −1 12

0

0 1 1 −1 0 13

∼ 1 0 0 1 0 0

0 1 0 −1 12

0

0 0 1 0 −12

13

Therefore 1 0 0

2 2 0

3 3 3

−1 =

1 0 0

−1 12

0

0 −12

13

.For 4× 4 matrix, we have

1 0 0 0 1 0 0 0

2 2 0 0 0 1 0 0

3 3 3 0 0 0 1 0

4 4 4 4 0 0 0 1

∼

1 0 0 0 1 0 0 0

1 1 0 0 0 12

0 0

1 1 1 0 0 0 13

0

1 1 1 1 0 0 0 14

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 12

0 0

0 1 1 0 −1 0 13

0

0 1 1 1 −1 0 0 14

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 12

0 0

0 0 1 0 0 −12

13

0

0 0 1 1 0 −12

0 14

∼

1 0 0 0 1 0 0 0

0 1 0 0 −1 12

0 0

0 0 1 0 0 −12

13

0

0 0 0 1 0 0 −13

14

.Therefore

1 0 0 0

2 2 0 0

3 3 3 0

4 4 4 4

−1

=

1 0 0 0

−1 12

0 0

0 −12

13

0

0 0 −13

14

.


In general, the corresponding n× n matrix is

A =

1 0 0 · · · 0 0

2 2 0 · · · 0 0

3 3 3 · · · 0 0...

......

. . ....

...

n− 1 n− 1 n− 1 · · · n− 1 0

n n n · · · n n

.

We can write A = [aij]16i,j6n where

aij =

{i if i > j

0 if i < j

From the case of n = 3, 4 we can guess the inverse of A is

B =

1 0 0 · · · 0 0

−1 12

0 · · · 0 0

0 −12

13· · · 0 0

......

.... . .

......

0 0 0 · · · 1n−1 0

0 0 0 · · · − 1n−1

1n

.

We can also write B = [bij]16i,j6n where

bij =

1j

if i = j

−1j

if i = j + 1

0 if i < j or i > j + 1

Again, we can check directly that the product C = AB = [cij]16i,j6n is actually the identity

matrix. Since

cij =n∑

k=1

aikbkj,

we first use the formula for aik to get

cij =i∑

k=1

ibkj = i(b1j + · · ·+ bij).

Case 1: If i > j, then there are two nonzero terms in the above sum. cij = i(bjj + bj+1,j) =

i(1j− 1

j) = 0.

Case 2: If i = j, then there are only one nonzero term in the above sum. cij = ibjj = ij

= 1.


Case 3: If i < j, then every term in the above sum is zero, so cij = 0.

This shows that

cij = δij =

{1 if i = j

0 if i 6= j

or AB = I.

The algebraic way to prove this is to write A = D(I + J + J2 + · · · + Jn−1) and B =

(I − J)D−1 where D is the diagonal matrix with diagonal entries 1, 2, . . . , n:

D =

1 0 0 · · · 0 0

0 2 0 · · · 0 0

0 0 3 · · · 0 0...

......

. . ....

...

0 0 0 · · · n− 1 0

0 0 0 · · · 0 n

.

To see A = D(I+J+J2+ · · ·+Jn−1), it is enough to use the expression of I+J+ · · ·+Jn−1

in the previous problem and the fact that multiply D on the left is equivalent to the row

operations that multiply the i-th row by i, i = 1, . . . , n. To see B = (I − J)D−1, it is

enough to see that multiply J on the left is equivalent to remove the last row, then move

every other row down for one level and finally put all zeroes in the first row. Now it is easy

to see AB = DD−1 = I using the previous problem.

We can also prove this by induction. We write An, Bn for the n× n matrices. We need

to prove AnBn = In for every n. It is obvious that this is true for n = 1 since A1 = B1 = 1

(as either numbers or 1× 1 matrices.) Now suppose An−1 = Bn−1 = In−1, since

An =

[An−1 0

vTn−1 n

], Bn =

[Bn−1 0

wTn−1

1n

],

where

vn−1 =

n...

n

n

, wn−1 =

0...

0

− 1n−1

.Therefore by multiplication of partitioned matrices (see Section 2.4)

AnBn =

[An−1Bn−1 0

vTn−1Bn−1 + nwTn−1 1

]=

[In−1 0

(BTn−1vn−1 + nwn−1)

T 1

].

It remains to check that BTn−1vn−1 + nwn−1 = 0 which is straightforward.


37. We assume

C =

[c11 c12 c13c21 c22 c23

].

Then the equation CA = I2 is equivalent to the system

c11 + c12 + c13 = 1

2c11 + 3c12 + 5c13 = 0

c21 + c22 + c23 = 0

2c21 + 3c22 + 5c23 = 1

.

We need to find solutions cij ∈ {−1, 0, 1}. It is easy to see that the solutions are

c11 = 1, c12 = 1, c13 = −1.

c21 = −1, c22 = 1, c23 = 0.

Therefore

C =

[1 1 −1

−1 1 0

].

Now we can compute AC:

AC =

−1 3 −1

−2 4 −1

−4 6 −1

6= I3.

(In fact, the product of a 3 × 2 matrix and a 2 × 3 matrix can have at most rank 2, but

the identity matrix has rank 3. So they can not equal to each other. See Section 2.7.)

38. We can choose

D =

1 0

1 1

1 1

0 1

.It is not possible that CA = I2 for some matrix C. Assume C = [c1 c2], then we have

CA = [c1 − c1 + c2 c1 − c2 c2].

Therefore the span of columns of CA: Span{c1,−c1+c2, c2−c2, c2} equals to Span{c1, c2},which cannot be R4. Thus CA 6= I4.


2.3. Characterizations of Invertible Matrices

2. Since [−4 2

6 −3

]∼[−4 2

0 0

]The matrix is not invertible.

4. Since there is a row of all zero, there are at most two pivots, so the matrix is not

invertible.

6. Since 1 −3 −6

0 4 3

−3 6 0

∼ 1 −3 −6

0 4 3

0 −3 −18

∼ 1 −3 −6

0 4 3

0 0 −634

,we know the matrix has three pivots, thus invertible.

8. Since this is an upper triangular matrix, which is already in the echelon form. There

are exactly four pivots, thus the matrix is invertible.

12. (a) True. Since AD = I where both A and D are n × n matrices, both A and D

are invertible (Invertible Matrix Theorem (a)⇔(j)(k)), also A = D−1. Therefore DA =

DD−1 = I.

(b) False. A can be any n × n matrix, for example, the zero matrix. Then the row

reduced echelon form of A is not I.

(c) True. Since the columns of A are linear independent, A is invertible, thus the columns

of A span Rn. (Invertible Matrix Theorem (e)⇒(a)⇒(h).)

(d) False. Since the equation Ax = b has at least one solution for each b, the columns

of A span Rn. Therefore A is invertible and x 7→ Ax is one-to-one. (Invertible Matrix

Theorem (h)⇒(a)⇒(f).)

(e) False. We can take b = 0 and A is not invertible, so the equation Ax = 0 has infinite

solutions.

14. A square lower triangular matrix is invertible if and only if all the diagonal entries

are nonzero. Let A be a lower triangular matrix, then A is invertible if and only if AT

is invertible. AT is a square upper triangular matrix. If all the diagonal entries of A are

non-zero, then AT is in echelon form and it has exact n pivots which are the diagonal


entries. So A is invertible. Conversely, if A is invertible, then the echelon form of AT must

have n pivots. This means that the diagonal entries must be the pivots, thus nonzero.

16. If A is invertible, then so is AT , thus the columns of AT must be linearly independent.

(Invertible Matrix Theorem (a)⇒(l) and (a)⇒(e).)

18. No. By an elementary row operation that add the negative of one of the two rows to

another, we get a row with all zero. Therefore the matrix cannot have n pivot positions,

thus not invertible. (Assume the matrix is n× n.)

20. No. Because the equation Ax = b is consistent for every b in R5, the matrix A

is invertible. Therefore Ax = b has at most one solution for every b. (Invertible Matrix

Theorem (g)⇒(a)⇒(f), then use definition of one-to-one.)

22. Since EF = I, by Invertible Matrix Theorem (j)(k))⇒(a), we know E and F are

both invertible. Also E−1 = F . Therefore FE = E−1E = I = EF , i.e. E and F commute.

26. Since the columns of A are linear independent, A is invertible. (Invertible Matrix

Theorem (e)⇒(a).) So A2 is also invertible and as a consequence, the columns of A2 span

Rn. (Invertible Matrix Theorem (a)⇒(h).)

28. Since AB is invertible, let C = (AB)−1A, then CB = (AB)−1AB = I. Therefore B

is invertible. (Invertible Matrix Theorem (j)⇒(a)).

30. By definition, if x 7→ Ax is one-to-one, then each b can have at most one pre-

image. Ax = b cannot have more than one solutions for any b. Therefore If Ax = b has

more than one solutions for some b, then the transformation x 7→ Ax is not one-to-one. We

can also deduce that this transformation is not invertible. (By Invertible Matrix Theorem.)

32. Since the equation Ax = 0 has only the trivial solution, A has a pivot in every

column. Since A is an n × n matrix, A has exactly n pivots. Therefore A has a pivot in

every row. This shows that Ax = b has a solution for each b.


34. The standard matrix of T is

A =

[2 −8

−2 7

].

Since detA = −2 6= 0, its inverse is

A−1 = −1

2

[7 8

2 2

]=

[−7

2−4

−1 −1

].

Therefore T is invertible and T−1(x1, x2) = (−72x1 − 4x2,−x1 − x2).

3.1 Introduction to Determinants

2. Cofactor expansion across the first row:∣∣∣∣∣∣0 5 1

4 −3 0

2 4 1

∣∣∣∣∣∣ =(−1)1+10

∣∣∣∣ −3 0

4 1

∣∣∣∣+ (−1)1+25

∣∣∣∣ 4 0

2 1

∣∣∣∣+ (−1)1+31

∣∣∣∣ 4 −3

2 4

∣∣∣∣=0− 20 + 22 = 2.

Cofactor expansion down the second column:∣∣∣∣∣∣0 5 1

4 −3 0

2 4 1

∣∣∣∣∣∣ =(−1)1+25

∣∣∣∣ 4 0

2 1

∣∣∣∣+ (−1)2+2(−3)

∣∣∣∣ 0 1

2 1

∣∣∣∣+ (−1)3+24

∣∣∣∣ 0 1

4 0

∣∣∣∣=− 20 + 6 + 16 = 2.

6. Cofactor expansion across the first row:∣∣∣∣∣∣5 −2 4

0 3 −5

2 −4 7

∣∣∣∣∣∣ =(−1)1+15

∣∣∣∣ 3 −5

−4 7

∣∣∣∣+ (−1)1+2(−2)

∣∣∣∣ 0 −5

2 7

∣∣∣∣+ (−1)1+34

∣∣∣∣ 0 3

2 −4

∣∣∣∣=5 + 20− 24 = 1.

10. First we use the cofactor expansion across the second row:∣∣∣∣∣∣∣∣1 −2 5 2

0 0 3 0

2 −6 −7 5

5 0 4 4

∣∣∣∣∣∣∣∣ = (−1)2+33

∣∣∣∣∣∣1 −2 2

2 −6 5

5 0 4

∣∣∣∣∣∣ ,


then we use the cofactor expansion across the second column to get

∣∣∣∣∣∣1 −2 2

2 −6 5

5 0 4

∣∣∣∣∣∣ = (−1)1+2(−2)

∣∣∣∣ 2 5

5 4

∣∣∣∣+ (−1)2+2(−6)

∣∣∣∣ 1 2

5 4

∣∣∣∣ = 2(−17)− 6(−6) = 2.

Therefore ∣∣∣∣∣∣∣∣1 −2 5 2

0 0 3 0

2 −6 −7 5

5 0 4 4

∣∣∣∣∣∣∣∣ = −6.

12. First, we use the cofactor expansion across the first row:

∣∣∣∣∣∣∣∣4 0 0 0

7 −1 0 0

2 6 3 0

5 −8 4 −3

∣∣∣∣∣∣∣∣ = (−1)1+14

∣∣∣∣∣∣−1 0 0

6 3 0

−8 4 −3

∣∣∣∣∣∣Then we use the cofactor expansion across the first row again:

∣∣∣∣∣∣−1 0 0

6 3 0

−8 4 −3

∣∣∣∣∣∣ = (−1)1+1(−1)

∣∣∣∣ 3 0

4 −3

∣∣∣∣ = (−1)3(−3).

Therefore ∣∣∣∣∣∣∣∣4 0 0 0

7 −1 0 0

2 6 3 0

5 −8 4 −3

∣∣∣∣∣∣∣∣ = 4(−1)3(−3) = 36.

(Actually, this is a lower triangular matrix. The determinant is the product of the diagonal

entries.)


14. First we use the cofactor expansion across the fourth row, then the cofactor expansion

down the last column, finally the cofactor expansion down the first column:∣∣∣∣∣∣∣∣∣∣

6 3 2 4 0

9 0 −4 1 0

8 −5 6 7 1

3 0 0 0 0

4 2 3 2 0

∣∣∣∣∣∣∣∣∣∣=(−1)4+13

∣∣∣∣∣∣∣∣3 2 4 0

0 −4 1 0

−5 6 7 1

2 3 2 0

∣∣∣∣∣∣∣∣ = (−1)4+13(−1)3+41

∣∣∣∣∣∣3 2 4

0 −4 1

2 3 2

∣∣∣∣∣∣=(−1)4+13(−1)3+41

[(−1)1+13

∣∣∣∣ −4 1

3 2

∣∣∣∣+ (−1)3+12

∣∣∣∣ 2 4

−4 1

∣∣∣∣]=3[−33 + 36] = 9.

18.∣∣∣∣∣∣1 3 5

2 1 1

3 4 2

∣∣∣∣∣∣ = (1 · 1 · 2) + (3 · 1 · 3) + (5 · 2 · 4)− (1 · 1 · 4)− (3 · 2 · 2)− (5 · 1 · 3) = 20.

20. The elementary row operation is to multiply k to the second row. The determinant

is also multiplied by k.

22. The elementary row operation is to add k times the second row to the first row. The

determinant remains the same.

24. The elementary row operation is to switch the first row and the second row. The

determinant becomes the negative of the original determinant.

26.

det

1 0 0

0 1 0

k 0 1

= 1.

28.

det

1 0 0

0 k 0

0 0 1

= k.

30.

det

0 0 1

0 1 0

1 0 0

= −1.


32. The determinant of an elementary scaling matrix with k on the diagonal is k.

36. Since

E =

[1 0

k 1

], A =

[a b

c d

],

we have

EA =

[a b

ka+ c kb+ d

].

We can calculate that det(E) = 1, det(A) = ad− bc and

det(EA) = a(kb+ d)− b(ka+ c) = ad− bc,

which shows that det(EA) = det(E) det(A).

38. Since

A =

[a b

c d

],

we have

kA =

[ka kb

kc kd

].

So we can calculate that

det(kA) = (ka)(kd)− (kb)(kc) = k2(ad− bc).

Therefore

det(kA) = k2 det(A).

40. (a) False. Cofactor expansions across rows or down columns are always the same.

(b) False. The determinant of a triangular matrix (either upper or lower) is the product

of the entries on the main diagonal.

43. In general, det(A+B) 6= detA+ detB. For example,

A =

[1 0

0 1

], B =

[1 1

1 1

], A+B =

[2 1

1 2

]then det(A+B) = 3, det(A) = 1, det(B) = 0.


References

[1] David C. Lay, R. Kent Nagle, Edward B. Saff, Arthur David Snider, Linear Algebra and Differential

Equations, Second Custom Edition for University of California, Berkeley.

[2] http://en.wikipedia.org/wiki/Kronecker delta

[3] http://en.wikipedia.org/wiki/Nilpotent

[4] Michael Atiyah, Ian G. Macdonald, Introduction to Commutative Algebra.

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