This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 92 Chapter 1 , {DiEEINifriON 13 Matrices and Systems of Linear Equations MATRIX INVERSES AND THEIR PROPERTIES In the preceding sections the matrix equation
AX = b ( 1) has been used extensively to represent a system of linear equations. Equation (1) looks,
symbolically, like the single linear equation ax = b, (2)
where a and b are real numbers. Since Eq. (2) has the unique solution
x = a“‘b
when a 9’: 0, it is natural to ask whether Eq. ( 1) can also be solved as
x=A4n In this section we investigate this question. We begin by deﬁning the inverse of a
matrix, showing how to calculate it, and then showing how the inverse can be used to
solve systems of the form Ax = b. The Matrix Inverse For a nonzero real number a, the inverse of a is the unique real number (1‘1 having the
property that 'a 2 an" : l. (3) a
In Eq. (3), the number I is the multiplicative identity for real number multiplication. In
an analogous fashion, let A be an (n x n) matrix. We now ask if we can ﬁnd a matrix A‘l with the property that
A"A = AA" = 1. (4) (In Eq. (4) 1 denotes the (n x n) identity matrix; see Section 1.6.) We formalize the idea suggested by Eq. (4) in the next deﬁnition. Note that the
commutativity condition A'l A = AA‘l means that A and A'I must be square and of
the same size; see Exercise 75. Let A befan,(n x We saythat'A,is§iiivertible we cantﬁnd an ‘ matrixéi‘l'such that? . iff‘”"f’ ‘ ’ ; ‘ r ‘ . . .. . ' ' ' ‘V A‘IA f=”AA‘.11=' 71.. 7 The is aninver‘se: 'A. ’ (Note: It is shown in Exercise 77 that if A is invertible, then A" is unique.
As an example of an invertible matrix, consider A: ~.
3 4 ‘. I ' EXAMPLE 1 Solution 1.9 Matrix Inverses and Their Properties 93 It is simple to show that A is invertible and that A“ is given by A_l_ —2 1
_ 3/2 —1/2 ' (To show that the preceding matrix is indeed the inverse of A, we need only form the
products A'1A and AA“1 and then verify that both products are equal to I .)
Not every square matrix is invertible, as the next example shows. An inverse for A must be a (2 x 2) matrix
a b
B 2
such that AB = BA = I . If such a matrix B exists, it must satisfy the following equation:
1 0 l 2 a b a + 2c b + 2d
01_36 cd_3a+6c3b+6d' The preceding equation requires that a + 2c = l and 3a + 6c = 0. This is clearly
impossible, so A has no inverse. ﬂ Let A be the (2 x 2) matrix Show that A has no inverse. Using Inverses to Solve Systems of Linear Equations One major use of inverses is to solve systems of linear equations. In particular, consider
the equation Ax = b, (5) where A is an (n x n) matrix and where A" exists. Then, to solve Ax = b, we might
think of multiplying both sides of the equation by A“: A" Ax = A‘lb or x '2 A‘lb. The preceding calculations suggest the following: To solve Ax = b we need only compute the vector x given by x = A"b. (6)
To verify that the vector x = A"b is indeed a solution, we need only insert it into the
equation: Ax = A(A"b)
= (AA")b (by associativity of multiplication)
= I b (by Deﬁnition 13)
= b. (because I is the identity matrix) 94 Chapter 1 I l I . LEMMA Proof Matrices and Systems of Linear Equations Existence of Inverses As we saw earlier in Example I, some matrices do not have an inverse. We now turn our
attention to determining exactly which matrices are invertible. In the process, we will
also develop a simple algorithm for calculating A". Let A be an (n x n) matrix. If A does have an inverse, then that inverse is an (n x n)
matrix B such that 143:]. (Of course, to be an inverse, the matrix B must also satisfy the condition BA = I . We
will put this additional requirement aside for the moment and concentrate solely on the
condition AB = I.) Expressing B and I in column form, the equation AB = I can be rewritten as A[b1,b2,... (7a) abn]=[elve29”ven] or
[Abh Abz.    , Abn] = [eh 62. “Henl If A has an inverse, therefore, it follows that we must be able to solve each of the
following n equations: (7b) AX = e.
AX = 82 (7C)
Ax = e,,. In particular, if A is invertible, then the kth column of A" can be found by solving
Ax=ek,k=1,2,...,n. We know (recall Theorem I3) that all the equations listed in (7c) can be solved if A
is nonsingular. We suspect, therefore, that a nonsingular matrix always has an inverse.
In fact, as is shown in Theorem 15, A has an inverse if and only if A is nonsingular. Before stating Theorem 15, we give a lemma. (Although we do not need it here,
the converse of the lemma is also valid; see Exercise 70.) Let P, Q, and R be (n x n) matrices such that PQ = R. If either P or Q is singular,
then so is R. Suppose ﬁrst that Q is singular. Then there is a nonzero vector x1 such that Qxl = 0.
Therefore, using associativity of matrix multiplication, RX] = (PQ)X1
= P(QX1)
= P9
= 0. So. Q singular implies R is singular.
Now, suppose Q is nonsingular but the other factor, P, is singular. Then there is a
nonzero vector x; such that le = 0. Also, Q nonsingular means we can ﬁnd a vector 1.9 Matrix Inverses and Their Properties 95 X2 such that sz = x,. (In addition, note that X; must be nonzero because x, is nonzero.) Therefore,
RXz = (PQ)X2
= P(QX2)
= PX;
= 0.
Thus, if either P or Q is singular, then the product PQ is also singular. ﬂ We are now ready to characterize invertible matrices.  l I THEOREM 15 Let A be an (n x n) matrix. Then A has an inverse if and only if A is nonsingular.
 l Proof Suppose ﬁrst that A has an inverse. That is, as in equation (7a), there is a matrix B
such that AB = I . Now, as Exercise 74 proves, I is nonsingular. Therefore, by the
lemma, neither A nor B can be singular. This argument shows that invertibility implies
nonsingularity. For the converse, suppose A is nonsingular. Since A is nonsingular, we see from
equations (7a)—(7c) that there is a unique matrix B such that AB = I. This matrix B
will be the inverse of A if we can show that A and B commute; that is, if we can also
show that BA = I. We will use a common algebraic trick to prove BA = I . First of all, note that the
matrix B must also be nonsingular since AB = I . Therefore, just as with equations
(7a)—(7c), there is a matrix C such that BC = I . Then, combining the expressions
AB = I and BC = I, we obtain A = A1 = A(BC) = (AB)C = IC= C. Since A = C, we also have BA = BC 2 I . Therefore, BA = I, and this shows that B is
the inverse of A. Hence, A nonsingular implies that A is invertible. q Calculating the Inverse In this subsection we give a simple algorithm for ﬁnding the inverse of a matrix A,
provided that A has an inverse. The algorithm is based on the system of equations (7c): Ax=el,Ax=e2,...,Ax=e,,. We ﬁrst observe that there is a very efﬁcient way to organize the solution of these
I: systems; we simply row reduce the associated augmented matrix [A le. , e2, . . . , en].
The procedure is illustrated in the next example. ‘ I EXAMPLE 2 Find the inverse of the (3 x 3) matrix 96 Chapter 1 Matrices and Systems of Linear Equations Solution The augmented matrix [A  e1, e2, e3] is given by (Note that the augmented matrix has the form [A l I 1.)
We now perform appropriate row operations to transform [A  I ] to reduced echelon
form. R2 — 2R1, R3—R1: R1 — 2R2, R3 + 3R2: R1 — 7R3, R2 + 2R3: l 0 0 54 —23 —7
0 l 0 — l6 7 2
0 0 l —7 3 1 Having the reduced echelon form above, we easily ﬁnd the solutions of the three systems
Ax = e], Ax = e2, Ax = e3. In particular, Ax = e1 has solution: Ax = e2 has solution: Ax = c; has solution: 54 —23 —7
x1 = —16 xz = 7 x3 = 2
—7 3 1
Therefore, A" = [x1, x2, x3] or
54 —23 —7
A" = —16 7 2 . d
7 3 1 This procedure illustrated in Example 2 can be summarized by the following algo
 rithm for calculating A". 1.9 Matrix Inverses and Their Properties 97 Computation of A‘1 To calculate the inverse of a nonsingular (n x n) matrix, we can proceed as
follows: Step 1. Form the (n x 2n) matrix [A  I].
Step 2. Use elementary row operations to transform [A l I ] to the form [I  B]. Step 3. Reading from this ﬁnal form, A"I = B. (Note: Step 2 of the algorithm above assumes that [A  I] can always be row reduced
to the form [I  B] when A is nonsingular. This is indeed the case, and we ask you to
prove it in Exercise 76 by showing that the reduced echelon form for any nonsingular
matrix A is l . In fact, Exercise 76 actually establishes the stronger result listed next in Theorem 16.) I I THEOREM 16 Let A be an (n x n) matrix. Then A is nonsingular if and only if A is row equivalent
to I . d The next example illustrates the algorithm for calculating A'l and also illustrates
how to compute the solution to Ax = b by forming x = A"b. ’ I I EXAMPLE 3 Consider the system of equations
xl + 2132 = —1 2x; + 5252 = 10. (a) Use the algorithm to ﬁnd the inverse of the coefﬁcient matrix A.
(b) Use the inverse to calculate the solution of the system. Solution (a) We begin by forming the (2 x 4) matrix [A  I], Al_12i0
[H— 2501' We next row reduce [A  I ] to [I  B] as follows: R2 ~2R]: R1 —2Rz: 98 Chapter 1 I l I EXAMPLE4 Matrices and Systems of Linear Equations [3?] (b) The solution to the system is x 2: A"'b where b=[:15] Now, A“b = [15, —8]T, so the solution is x1 = 15, x2 = —8. ﬂ Thus, A“l is the matrix Inverses for (2 x 2) Matrices There is a simple formula for the inverse of a (2 x 2) matrix, which we give in the remark
that follows. Remark Let A be a (2 x 2) matrix,
A = ,
c d (a) If A = 0, then A does not have an inverse.
(b) If A ;é 0, then A has an inverse given by A—‘—1 (11) (8)
_A —c a o ‘ Part (a) of the remark is Exercise 69. To verify the formula given in (b), suppose
that A 5A 0, and deﬁne B to be the matrix 1[ d—b]
B=— .
A —c a d—b BA_1 ab lad—be O 10
_A—ca cd_A 0 adbc_01. Similarly, AB = I, so B = A".
The reader familiar with determinants will recognize the number A in the remark as
the determinant of the matrix A. We make use of the remark in the following example. and set A = ad — bc. Then Let A and B be given by [6 3] [17]
A: and B: .
34 35 For each matrix, determine whether an inverse exists and calculate the inverse if it does
exist. Solution ’ l ’ EXAMPLE 5
I Solution THEOREM 17
l  Proof 1.9 Matrix Inverses and Their Properties 99
For the matrix A, the number A is
A = 6(4) — 8(3) = 0, so, by the remark, A cannot have an inverse. For the matrix B, the number A is A = 1(5) —7(3) = —16. 1 5—7
___—
B _ 16L 1]. .. A 2
A: .
[2 A—3: For what values of A is the matrix A nonsingular? Find A" if A is nonsingular. According to formula (8) Consider the matrix A The number A is given by
A =A()t—3)—4=)tZ—3}t—4=(A4)(A+l). Thus, A is singular if and only if A = 4 or A = — i. For values other than these two, A"1
is given by
l A — 3 —2 I”
A2 — 3A — 4 _—2 A ' Properties of Matrix Inverses A" = The following theorem lists some of the properties of matrix inverses. Let A and B be (n x n) matrices, each of which has an inverse. Then: 1. A‘l has an inversefand (.4")'l = A. 2 AB has an inverse, and (A3)“ = B"A". 3. If k is a nonzero scalar, then kA has an inverse, and (kA)‘l = (1/k)A'l.
4. AT has an inverse, and (ATVl = (A")T. 1. Since AA‘1 = A"A = I, the inverse of A'l is A; that is, (A")‘1 = A. 2. Note that (AB)(B"A") = A(BB")A" = A(1A")= AA“ = I. Simi
larly, (B"A")(AB) = 1, so, by Deﬁnition 13, B"A" is the inverse for AB.
Thus (A8)" = B“A". 3. The proof of property 3 is similar to the proofs given for properties 1 and 2 and
is left as an exercise. ‘ 100 Chapter 1 Matrices and Systems of Linear Equations 4. It follows from Theorem 10, property 2, of Section 1.6 that AT(A")T  (A"A)T = [T = 1. Similarly, (A"')TAT = 1. Therefore, A7 has inverse
m")? It Note that the familiar formula (ab)‘1 = (1“‘b'I for real numbers is valid only
because multiplication of real numbers is commutative. We have already noted that matrix multiplication is not commutative, so, as the following example demonstrates,
(AB)1 9e A"B“. I l l EXAMPLE 6 Let A and B be the (2 x 2) matrices l 3 3 —2
A = and B = .
2 4 l —1 1. Use formula (8) to calculate A‘l, B", and (A3)“. 2. Use Theorem 17, property 2, to calculate (A8)".
3. Show that (A3)“l ¢ A"B". Solution For A the number A is A = 1(4) —— 3(2) = —2, so by formula (8) A_]_ —2 3/2
_ 1—1/2 ' For B the number A is 3(—l)—1(—2)= —1, so
B_l _ l —2
_ l 3 .
6 —5
AB = ,
[ m 8 ]
—4 5/2
AB “I = .
( ) [ _5 3 ]
By Theorem 17, property 2, (AB)4=B_A_!= 1—2 —2 3/2 _ —4 5/2
1—3 1—1/2_—53'
Finally,
_ _ —2 3/2 1 —2 —1/2 1/2
A 'B '= = I
[ 1 —1/2][1—3] [ 1/2 —1/2]¢(AB) ' I“ The following theorem summarizes some of the important properties of nonsingular
matrices. The product AB is given by so by formula (8) I l ' THEOREM 18  I EXAMPLE 7
l Solution 1.9 Matrix Inverses and Their Properties 101 Let A be an (n x n) matrix. The following are equivalent: (a) A is nonsingular; that is, the only solution of Ax = 0 is x = 0. (b) The column vectors of A are linearly independent. (c) Ax = b always has a unique solution. (d) A has an inverse. (e) A is row equivalent to l . d IllConditioned Matrices In applications the equation Ax = b often serves as a mathematical model for a physical
problem. In these cases it is important to know whether solutions to Ax = b are sensitive
to small changes in the righthand side b. If small changes in b can lead to relatively
large changes in the solution x, then the matrix A is called illconditioned. The concept of an illconditioned matrix is related to the size of A ‘I. This connection is explained after the next example. The (n x n) Hilbert matrix is the matrix whose ijth entry is 1 / (i + j — I). For example,
the (3 x 3) Hilbert matrix is l 1/2 1/3
l/2 1/3 1/4
1/3 1/4 1/5
Let A denote the (6 x 6) Hilbert matrix, and consider the vectors b and b + Ah:
1 l
2 2
b: l , b Ah: 1
L414 L4142
l 1
2 2 Note that b and b + Ab differ slightly in their fourth components. Compare the solutions
ofo =bandAx=b+Ab. We used MATLAB to solve these two equations. If x] denotes the solution of Ax = b, and xz denotes the solution of Ax = b + Ab, the results are (rounded to the nearest
integer): 6538 ~6539
l85706 185747
—1256237 —1256519 x1 = and x2 =
3271363 3272089
, 3616326 —3617120
71427163 1427447 102 Chapter 1 Matrices and Systems of Linear Equations (Note: Despite the fact that b and b + Ab are nearly equal, x1 and X2 differ by almost 800 in their ﬁfth components.) 1. Example 7 illustrates that the solutions of Ax = b and Ax = b + Ab may be quite
different even though Ab is a small vector. In order to explain these differences, let X!
denote the solution of Ax = b and X; the solution of Ax = b + Ab. Therefore, Ax; = b
and Ax; = b + Ab. To assess the difference, xz — x1, we proceed as follows: sz—Ax1=(b+Ab)—b=Ab. Therefore, A(xz — XI) = Ab, or
X; — XI = A"‘Ab. If A‘1 contains large entries, then we see from the equation above that x2 — x. can be large even though Ab is small. The Hilbert matrices described in Example 7 are wellknown examples of ill
conditioned matrices and have large inverses. For example, the inverse of the (6 x 6) Hilbert matrix is 36 —630 —630 14700 A4 _ 3360 ——88200
—7560 21 1680 7560 ~220500 —2772 83160 3360 —7560 7560 —2772 — 88200 21 1680 —220500 83160
564480  141 1200 1512000 —582120 ~ 141 1200 3628800 —3969000 1552320 4410000 — 1746360
698544 1512000 —3969000
—582120 1552320 l746360 Because of the large entries in A", we should not be surprised at the large difference
between x. and x2, the two solutions in Example 7. ‘ EXERCISES In Exercises 1—4, verify that B is the inverse of A by
showing that AB = BA = I . 1A 74 B 3—4
‘—53’ "—57 2.A=[310], B=[ 1_1]
210 —.2 .3
—l —2 11 013
3.A=l: 1 3—15 , B: 5 5 4
0—1 5 1 ll
100 l 0 0
4.A=[210 , B: 2 l 0
3 41 5—4 I In Exercises 5—8, use the appropriate inverse matrix
from Exercises 14 to solve the given system of linear
equations. 5.3x.+10x2=6 6.7x+4xz=5
2x1+10xz = 9 5x1+3x2 = 2 7. .\‘2 +313 =4 8. X] =2
5x; +5.17 +4x3 = 2 —2x. + 1‘2 = 3 x.+ x2+ x3=2 5xl4x2+x3=2 ln Exercises 9—12, verify that the given matrix A does
not have an inverse. (Hint: One of AB = l or BA = 1
leads to an easy contradiction] 000
9.A= 121
3 21 10.14: 000 4
l
3 \ONN 1
1
2 93—— 2 2 4 1
11.A= 117 12.A= 1
3 3 9 2 In Exercises 13—21, reduce [A  I] to ﬁnd A". In each
case, check your calculations by multiplying the given
matrix by the derived inverse. 13.11 14.23 [ml [67]
1s. 12 16. —1—2 11
[21] [13—15
0—1 5 17.100 18.135 210 [014 341 027
19.142 201—221
021 1—150
353 2—2112
0281 21.1231 l 0 2 l
2 l —3 0
l l 2 1 As in Example 5, determine whether the (2 x 2) matrices in Exercises 22—26 have an inverse. If A has an inverse,
ﬁnd A" and verify that A‘IA = I. , —3 2 2—2 22.21: 23.21: [ 11] [2 3] —1 3 2
24A=[ ] 25.A=[ 1] 21 4 2
26.A= 6—2 9—3 In Exercises 27—28 determine the value(s) of A for which
A has an inverse. 1—2 3
l4 ], 4_l 4 1A
2—32 1.9 Matrix Inverses and Their Properties 103 In Exercises 29—34, solve the given system by forming
x = A"'b, where A is the coefﬁcient matrix for the
system. 29.2x1+ x2=4 30. x1+x2=0 3x1+2x2=2 2x1+3x2=4
31. x1— x2=5 32.2x1+3x2=l 3x1—4x2=2 3x1+4x2=7
33. 3,131+ x2 =10 34. x1— X2 =10 —x1+3x2= 5 2X+3x2= 4 The following matrices are used in Exercises 3545. A_l_31 B_12 C_,_ —11
‘02’_21’_12' (9)
In Exercises 35—45, use Theorem 17 and the matrices in
(9) to form Q‘l, where Q is the given matrix. 35.Q=AC 36. Q=CA
37. Q = AT 38. Q = ATC
39. Q = CTAT 40. Q = 1314
41. Q = CB" 42. Q = B1
43. Q = 2.4 44. Q = 10C
45. Q = (4031 46. Let A be the matrix given in Exercise 13. Use the
inverse found in Exercise 13 to obtain matrices B and C such that AB = D and CA = E, where 2—1
12 3 D: and E: l l
10 2 0 3 47. Repeat Exercise 46 with A being the matrix given
in Exercise 16 and where 2—1
1 2 3
D: l l and E: .
l 0 2
0 3
48. For what values of a is
1 1—1
A: 0 1 2
l l a
nonsingular?
49. Find (A8)", (324)", and (AT)1 given that
l 2 5 3 3 4
A": 316 andB": 513
2 8 l 7 6 —l 104 50. 51. 52. 53. 54. 55. 56. Find the (3 x 3) nonsingular matrix A if A2 =
AB + 2A, where
2 1 1
B = 0 3 2
l 4 1 Simplify (A‘lB)"(C"'A)"(B“C)" for (n xn)
invertible matrices A, B, and C. The equation x2 = 1 can be solved by setting
x2 — 1 = 0 and factoring the expression to obtain
(1: — 1)(x + l) = 0. This yields solutions x = 1 and
x = —1. a) Using the factorization technique given above,
what (2 x 2) matrix solutions do you obtain for
the matrix equation X 2 = I ? b)Showthat
[a l—az]
A:
l —a is a solution to X 2 = I for every real number a.
c) Let b = :H. Show that “ii—2’] is a solution to X 2 = I for every real number c. (1) Explain why the factorization technique used in
part (a) did not yield all the solutions to the
matrix equation X 2 = I. Suppose that A is a (2 x 2) matrix with columns 11
and v, so that A = [u, v], u and v in R2. Suppose
also that uTu = I, uTv = 0, and vTv = l. Prove
that ATA = I. [Himt' Express the matrix A as A: "101'": u] ’v "I
H2 02 142 U2 and form the product ATA.] Let u be a vector in R" such that uTu = 1. Let
A = I —uuT, where I is the (n x n) identity. Verify
that AA = A. [Hint Write the product uuTuuT as
uuTuuT = u(uTu)uT, and note that uTu is a scalar.] Suppose that A is an (n x n) matrix such that
AA = A, as in Exercise 54. Show that if A has
an inverse, then A = I . Let A = I — ava, where v is a nonzero vector in
R", I is the (n x n) identity, and a is the scalar given by a = 2/ (vrv). Show that A is symmetric and that
AA = I; that is, A“t = A. 57. 58. 59. 60. 61. Chapter 1 Matrices and Systems of Linear Equations Consider the (n x n) matrix A deﬁned in Exercise
56. For x in R", show that the product Ax has the
form Ax = x — Av, where A is a scalar. What is the value of A for a given x? Suppose that A is an (n x n) matrix such that
ATA = I (the matrix deﬁned in Exercise 56 is such
a matrix). Let x be any vector in R". Show that
Axl = llxll; that is, multiplication of x by A pro—
duces a vector Ax having the same length as x. bet u and v be vectors in R", and let I denote the
(n x n) identity. Let A = I + WT, and suppose
Wu 7’: 1. Establish the ShermanWoodbeny
formula: A‘l = I —auvT, a = l/(l + vTu). (10)
[Hint: Form AA", where A‘1 is given by formula (10)]
If A is a square matrix, we deﬁne the powers A2, A3,
and so on, as follows: A2 = AA, A3 = A(A2), and
so on. Suppose A is an (n x n) matrix such that
A3—2A2+3A—I=O. Show that AB = I, where B = A2 — 2A + 31.
Suppose that A is (n x n) and A2+blA +b01=o, (H) where b0 ,5 0. Show that AB = I , where B =
("l/bo)[A + bill. It can be shown that when A is a (2 x 2) matrix such that
A“1 exists, then there are constants b1 and be such that
Eq. (1 1) holds. Moreover, b0 # 0 in Eq. (11) unless A
is a multiple of I . In Exercises 62—65, ﬁnd the constants
b; and be in Eq. (1 1) for the given (2 x 2) matrix. Also,
verify that A'1 =(—1/bo)[A + bill. 62.
64.
66. 63. A in Exercise 15.
65. A in Exercise 22. a) If linear algebra software is available, solve the
systems Ax = b, and Ax = b2, where 0.932 0.443 0.417 A in Exercise 13. A in Exercise 14. A = 0.712 0.915 0.887 ,
0.632 0.514 0.493
1 1.01
bl = 1 . 112 = 1.0l
—l — 1.01 Note the large difference between the two 7
solutions. 67. 68. 69. 70. 71. b) Calculate A“ and use it to explain the results of
part (a). :1) Give examples of nonsingular (2 x 2) matrices
A and B such that A + B is singular. b) Give examples of singular (2 x 2) matrices A
and B such that A + B is nonsingular. Let A be an (n x n) nonsingular symmetric matrix.
Show that A”l is also symmetric. a) Suppose that AB = 0, where A is nonsing
ular. Prove that B = 0. b) Find a (2 x 2) matrix B such that AB = 0,
where B has nonzero entries and where A is the matrix
1 l
A = .
l 1 Why does this example not contradict part (a)? Let A, B, and C be matrices such that A is nonsing~
ular and AB = AC. Prove that B = C. Let A be the (2 x 2) matrix I: a b ] A = , c d and set A = ad — be. Prove that if A = 0, then A is singular. Conclude that A has no inverse. [Hint
Consider the vector 45;]. also treat the special case when d = c = 0.] 72. 73. 74. 75. 76. 77. 78.
79. Supplementary Exercises 105 Let A and B be (n x n) nonsingular matrices. Show
that AB is also nonsingular. What is wrong with the following argument that if
AB is nonsingular. then each of A and B is also non singular?
Since AB is nonsingular, (AB)‘l exists. But
by Theorem 17, property 2, (A8)”1 = B"A". Therefore, A“1 and B‘l exist, so A and B are
nonsingular. Let A and B be (n x n) matrices such that AB is
nonsingular. a) Prove that B is nonsingular. (Hint: Suppose v is
any vector such that Bv = 9, and write (AB)v
as A(Bv).] b) Prove that A is nonsingular. [Hint By part (a),
B‘I exists. Apply Exercise 72 to the matrices
AB and B“‘.] Let A be a singular (n x n) matrix. Argue that at
least one of the systems Ax = ck, k = 1,2, ...,n,
must be inconsistent, where e1, e2, ...,e,, are the ndimensional unit vectors. Show that the (n x n) identity matrix, I. is non
singular. Let A and B be matrices such that AB = BA. Show
that A and B must be square and of the same order.
[Hint Let A be (p x q) and let B be (r x s). Now
show that p = r and q = 5.] Use Theorem 3 to prove Theorem 16. Let A be (n x n) and invertible. Show that A" is
unique. f SUPPLEMENTARY EXERCISES 1. ' Consider the system of equations 1'] =1
2x1+(a2+a—2)x2=a2—a—4. For what values of a does the system have inﬁnitely
many solutions? No solutions? A unique solution
in which x2 = 0? 2. Let
1—1—1 x]
A: 2—1 1 ,x= x2 ,and
—3 l—3 x3
121
b: I);
193 ...
View
Full Document
 Fall '11
 OSU
 Linear Algebra, Matrices, Invertible matrix, Inverses

Click to edit the document details