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Solutions to Linear Algebra
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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
4 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 5
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 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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
].
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 7
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.
8 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 9
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
.
10 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 11
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
].
12 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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
.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 13
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.
14 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 15
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.
16 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 17
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.
18 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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
∣∣∣∣∣∣ ,
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 19
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.)
20 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.
SOLUTIONS TO HOMEWORK 3, DUE 02/13/2013 21
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.
22 MATH 54 LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS WITH PROF. STANKOVA
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.