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Unformatted text preview: Lecture Note 6
Dr. Jeff ChakFu WONG
Department of Mathematics
Chinese University of Hong Kong ÛÓÒ Ñ Ø º
Ù º Ùº MATH 2510
Linear Algebra and Its Applications
Winter, 2011
Produced by Jeff ChakFu WONG 1 D ETERMINANTS
1. Deﬁnition and Properties
2. Cofactor Expansion and Applications ETERMINANTS 2 Our aim is:
1. to know the notion of a determinant,
2. to study some of its properties, and
3. to calculate the determinant of the given matrix via the
reduction to triangular form. ETERMINANTS 3 D EFINITION AND P ROPER TIES
1. Permutation
2. Deﬁnition of Determinant
3. Properties of Determinants EFINITION AND P ROPERTIES 4 Deﬁnition 1
Let S = {1, 2, . . . , n} be the set of integers from 1 to n, arranged in ascending order.
A rearrangement j1 j2 . . . jn of the elements of S is called a permutation of S .
For example, let S = {1, 2, 3, 4}. Then 4123 is a permutation of S . It
corresponds to the function f : S → S deﬁned by
f (1) = 4
f (2) = 1
f (3) = 3
f (4) = 2. EFINITION AND P ROPERTIES 5 We can put
• any one of the n elements of S in the ﬁrst position,
• any one of the remaining n − 1 elements in the second position
• any one of the remaining n − 2 element in the third position, and so on,
• until the nth position can be ﬁlled by last remaining element.
Thus, there are
n(n − 1)(n − 2) · · · 2 · 1 (1) permutations of S ; we denote the set of all permutations of S by Sn .
The expression in Equation (1) is denoted
n! , EFINITION AND n factoral P ROPERTIES 6 We have
1! = 1 = 1
2! = 2 · 1 = 2
3! = 3 · 2 · 1 = 6
4! = 4 · 3 · 2 · 1 = 24
5! = 5 · 4 · 3 · 2 · 1 = 120
6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040
8! = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320
9! = 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 362, 880. EFINITION AND P ROPERTIES 7 Example 1 • S1 consists of only 1! = 1 permutation of the set {1}, namely, 1;
• S2 consists of 2! = 2 · 1 = 2 permutations of the set {1, 2}, namely, 12 and 21;
• S3 consists of 3! = 3 · 2 · 1 = 6 permutations of the set {1, 2, 3}, namely 123,
231, 312, 132,213 and 321.
A permutation j1 j2 · · · jn of S = {1, 2, · · · , n} is said to an inversion if a large
integer jr precedes a smaller js .
A permutation is called even or odd according to whether the total number
of inversions in it is even or odd.
Thus, the permutation 4132 of S = { 1, 2, 3, 4} has four inversions: 4 before 1, 4
before 3, 4 before 2, and 3 before 2. It is then an even permutation.
Remark: 0 inversion is even. EFINITION AND P ROPERTIES 8 Example 2 In S2 ,
• the permutation 12 is even, since it has no inversions;
• the permutation 21 is odd, since it has one inversion.
Example 3 • The even permutation in S3 are
– 123 (no inversions);
– 231(two inversions: 21 and 31); and
– 312 (two inversions: 31 and 32).
• The odd permutations in S3 are
– 132(one inversion: 32);
– 213 (one inversion: 21); and
– 321 (three inversions: 32,31,21). EFINITION AND P ROPERTIES 9 D EFINITION OF D ETERMINANT EFINITION OF D ETERMINANT 10 Deﬁnition 2
Let A = [aij ] be an n × n matrix. We deﬁne the determinant of A (written det(A) or A) by
det(A) = A = (±)a1j1 a2j2 . . . anjn , (2) where the summation ranges over all permutations j1 j2 . . . jn of the set S = {1, 2, . . . , n}.
The sign is taken as + or − according to whether the permutation j1 j2 . . . jn is even or odd.
Remarks:
• In each (±)a1j1 a2j2 . . . anjn of det(A), the row subscripts are in their natural
order, whereas the column subscripts are in the order j1 j2 . . . jn .
• Since the permutation j1 j2 . . . jn is merely a rearrangement of the numbers from
1 to n, it has no repeats.
• Thus each term in det(A) is a product of n elements of A each with its
appropriate sign, with exactly one element from each row and exactly one
element from each column.
• Since we sum over all the permutations of the set, S = {1, 2, · · · , n}, det(A) has
n! terms in the sum. Example 4 If A = [a11 ] is a 1 × 1 matrix, then S1 has only one permutation in it,
the identity permutation 1, which is even. Thus det(A) = a11 . EFINITION OF D ETERMINANT 12 Example 5 If A= a11 a12 a21 a22 is a 2 × 2 matrix, then to obtain det(A) we write down the terms
a1− a2− and a1− a2− , and ﬁll in the blanks with all possible elements of S2 ; the subscripts become 12 and
21.
• Since 12 is an even permutation, the term a11 a22 has a + sign associated with
it;
• since 21 is an odd permutation, the term a12 a21 has a − sign associated with it.
Hence
det(A) = a11 a22 − a12 a21 .
We can also obtain det(A) by forming the product of the entries on the line from left
to right in the following diagram and subtracting from this number the product of
the entries on the line from right to left (cf. Fig. 1). EFINITION OF D ETERMINANT 13 a a 11 a 21 12 a22
+ Figure 1:
Thus, if A= 2 −3 4 5 then det(A) = (2)(5) − (−3)(4) = 22. EFINITION OF D ETERMINANT 14 Example 6 If a11 A = a21 a31 a12 a13 a22 a23 a32 a33 then, to compute det(A), we write down the six terms
a1− a2− a3− , a1− a2− a3− , a1− a2− a3− , a1− a2− a3−
a1− a2− a3− and a1− a2− a3− , All the elements of S3 are used to ﬁll in the blanks, and if we preﬁx each term by +
or by  according to whether the permutation used is even or odd, we ﬁnd that
det(A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32
− a11 a23 a32 − a12 a21 a33 − a13 a22 a31 . (3) We can also obtain det(A) as follows. Repeat the ﬁrst and second columns of A as
shown below. From the sum of the products of the entries of the lines from left to
right, and subtract from this number the products of the entries on the lines from
right to left (verify) (cf. Figure 2): EFINITION OF D ETERMINANT 15 a 11 a12 a13 a11 a12 a 21 a22 a23 a21 a22 a a a a31 a  31
 32 33  + 32 + + Figure 2: EFINITION OF D ETERMINANT 16 Elementary Associated Even Signed Product Permutation or Elementary a1j1 a2j2 a3j3 (j1 , j2 , j3 ) Odd Product a11 a22 a33 (1,2,3) Even a11 a22 a33 a11 a23 a32 (1,3,2) Odd a11 a23 a32 a12 a21 a33 (2,1,3) Odd a12 a21 a33 a12 a23 a31 (2,3,1) Even a12 a23 a31 a13 a21 a32 (3,1,2) Even a13 a21 a32 a13 a22 a31 (3,2,1) Odd a13 a22 a31 Table 1: Using the matrices and the elementary products. Warning: It should be emphasized that the methods given for evaluating
det(A) in Examples 5 and 6 do not apply for n ≥ 4. EFINITION OF D ETERMINANT 17 Example 7 Let 1 A= 2 3 2
1
1 3 3 . 2 Evaluate det(A).
Solution: Substituting in Equation (3), we ﬁnd that
det(A) = (1)(1)(2) + (2)(3)(3) + (3)(2)(1)
−(1)(3)(1) − (2)(2)(2) − (3)(1)(3) = 6. We could obtain the same result by using the easy method given previously (verify). EFINITION OF D ETERMINANT 18 Example 8 • The number of inversions in the permutation 54132 is 8. • The number of inversions in the permutation 52134 is 5.
• The permutation 52134 was obtained from 54132 by interchanging 2 and 4.
• The number of inversions differ by 3, an odd number. EFINITION OF D ETERMINANT 19 P ROPER TIES OF D ETERMINANTS ROPERTIES OF D ETERMINANTS 20 Theorem 0.1 The determinants of a matrix and its transpose are equal, that is
det(AT ) = det(A). ROPERTIES OF D ETERMINANTS 21 Example 9 Let A be the matrix of Example 7. Then 1 AT = 2 3 2
1
3 3 1 . 2 Substituting in Equation (3) we ﬁnd that
det(AT ) = (1)(1)(2) + (2)(1)(3) + (3)(2)(3)
− (1)(1)(3) − (2)(2)(2) − (3)(1)(3)
= 6 = det(A).
Theorem 0.1 will allow us to replace “row" by “column" in many of the
addition properties of determinants; we see how to do this in the following
theorem. ROPERTIES OF D ETERMINANTS 22 Theorem 0.2 If matrix B results from matrix A by interchanging two rows
(columns) of A, then det(B ) = − det(A).
Proof:
• Suppose that B results from A by interchanging rows r and s of A, say
r < s.
• Then we have brj = asj , bsj = arj , and bij = aij for i = r, i = s.
• Now
det(B ) = (±)b1j1 b2j2 . . . brjr . . . bsjs . . . bnjn = (±)a1j1 a2j2 . . . asjr . . . arjs . . . anjn = (±)a1j1 a2j2 . . . arjs . . . asjr . . . anjn – The permutation j1 j2 . . . js . . . jr . . . jn results from the permutation
j1 j2 . . . jr . . . js . . . jn by an interchange of the two numbers; the number of
inversions in the former differs by an odd number from the number of inversions
in the latter.
– This means that the sign of each term in det(B ) is the negative of the sign of
the corresponding term in det(A). Hence det(B ) = − det(A). ROPERTIES OF D ETERMINANTS 23 • – Now suppose that B is obtained from A by interchanging two
columns of A.
– Then B T is obtained from AT by interchanging two rows of AT .
– So det(B T ) = − det(AT ), but det(B T ) = det(B ) and
det(AT ) = det(A).
– Hence det(B ) = − det(A). ROPERTIES OF D ETERMINANTS 24 Example 10 We have
2
3 ROPERTIES OF D ETERMINANTS −1
2 =7 and 3 2 2 −1 = −7. 25 Theorem 0.3 If two rows (columns) of A are equal, then det(A) = 0.
Proof:
• Suppose that rows r and s of A are equal.
• Interchange rows r and s of A to obtain a matrix B .
• Then det(B ) = − det(A).
• On the other hand, B = A, so det(B ) = det(A).
• Thus det(A) = − det(A), so det(A) = 0. ROPERTIES OF D ETERMINANTS 26 Example 11
Using Theorem 0.3, it follows that it follows that
1 0 7 1 D ETERMINANTS 3 −1 ROPERTIES OF 2
2 3 = 0. 27 Theorem 0.4 If a row (column) of A consist entirely of zeros, then det(A) = 0.
Proof:
• Let the rth row of A consist entirely of zeros.
• Since each term in the deﬁnition for the determinant of A contains
a factor from rth row, each term in det(A) is zero.
• Hence det(A) = 0. ROPERTIES OF D ETERMINANTS 28 Example 12
Using Theorem 0.4 if follows that
1 5 6 0 D ETERMINANTS 3 4 ROPERTIES OF 2
0 0 = 0. 29 Theorem 0.5 If B is obtained from A by multiplying a row (column) of A by a
real number c, then det(B ) = c det(A).
Proof:
• Suppose that the rth row of A = [aij ] is multiplied by c to obtain
B = [bij ].
• Then bij = aij if i = r and brj = carj .
• We obtain det(B ) from Equation (2) as
det(B ) = (±)b1j1 b2j2 . . . brjr . . . bnjn = (±)a1j1 a2j2 . . . (carjr ) . . . anjn = c (±)a1j1 a2j2 . . . arjr . . . anjn = c det(A). • We can now use Theorem 0.5 to simplify the computation of det(A) by
factoring out common factors from rows and columns of A. ROPERTIES OF D ETERMINANTS 30 Example 13 We have
2
1 ROPERTIES OF 6
12 D ETERMINANTS =2 1 3 1 12 = (2)(3) 1 1 1 4 = 6(4 − 1) = 18. 31 Example 14 Solve det(A), where 1 A= 1 2 ROPERTIES OF D ETERMINANTS 2
5
8 3 3 6 32 Solution: We have
1 2 1 5 3 2 8 1 3
6 =2 2 3 1 5 3 1 4 3 1
= (2)(3) 2 1 1 5 1 1 4 1 = (2)(3)(0) = 0. Here we ﬁrst factored out 2 from the third row, then 3 from the third column, and
then used Theorem 0.3, since the ﬁrst and third columns are equal. ROPERTIES OF D ETERMINANTS 33 Theorem 0.6 If B = [bij ] is obtained from A = [aij ] by adding to each
element of the rth row (column) of A a constant c times the corresponding
element of the sth row (column) r = s of A, then det(B ) = det(A).
Proof: We prove the theorem for rows. We have bij = aij for i = r, and
brj = arj + casj , r = s, say r < s. The
det(B ) = (±)b1j1 b2j2 . . . brjr . . . bnjn = (±)a1j1 a2j2 . . . (arjr + casjr ) . . . asjs . . . anjn = (±)a1j1 a2j2 . . . arjr . . . asjs . . . anjn
+ (±)a1j1 a2j2 . . . (casjr ) . . . asjs . . . anjn The ﬁrst sum in this last expression is det(A); the second sum can be written as
c R OPERTIES OF D ETERMINANTS (±)a1j1 a2j2 . . . asjr . . . asjs . . . anjn 34 Note that
a11 ... a1n a21
.
.
. a22
.
.
. ... a2n
.
.
. as1
.
.
. as2
.
.
. ... asn
.
.
. rth row as1
.
.
. as2
.
.
. ... asn
.
.
. sth row an 1 (±)a1j1 a2j2 . . . asjr . . . asjs . . . anjn = a12 an 2 ... ann =0 because there are two equal rows. Hence det(B ) = det(A) + 0 = det(A). ROPERTIES OF D ETERMINANTS 35 Example 15 We have
1 2 3 2 −1 3 1 0 1 5
= 0 9 2 −1 3 1 0 1 obtained by adding twice the second row to the ﬁrst row. By applying the deﬁnition
of determinant to the second determinant, we can see that both have the value 4
(Verify). ROPERTIES OF D ETERMINANTS 36 Theorem 0.7 If a matrix A = [aij ] is upper (lower) triangular, then
det(A) = a11 a22 . . . ann ;
that is, the determinant of a triangular matrix is the product of the elements
on the main diagonal.
Proof:
• Let A = [aij ] be upper triangular (that is, aij = 0 for i > j ).
• Then a term a1j1 a2j2 . . . anjn in the expression for det(A) can be nonzero
only for 1 ≤ j1 , 2 ≤ j2 , . . ., n ≤ jn .
• Now j1 j2 . . . jn must be a permutation, or rearrangement, of
{1, 2, . . . , n}.
• Hence we must have jn = n, jn−1 = n − 1, . . . , j2 = 2, j1 = 1.
• Thus the only term of det(A) that can be nonzero is the product of the
elements on the main diagonal of A.
• Since the permutation 12 . . . n has no inversion, the sign associated with
it is +. Therefore, det(A) = a11 a22 . . . ann .
ROPERTIES OF D ETERMINANTS 37 • We leave the proof of the lower triangular case to the student. ROPERTIES OF D ETERMINANTS 38 Corollary 0.1
The determinant of a diagonal matrix is the product of the entries on its main
diagonal. Proof: Exercise. ROPERTIES OF D ETERMINANTS 39 Example 16 Let 2
3 A = 0 −4 0
0 4 5 , 3 3 B= 2 6 0
5
−8 0 0 , −4 C= −5 0 0 4 0 0 0 0 −6 Computer det(A), det(B ), det(C ). ROPERTIES OF D ETERMINANTS 40 Let 2 A= 0 0 3
−4
0 4 5 , 3 3 B= 2 6 0
5
−8 0 0 , −4 C= −5 0 0 4 0 0 0 0 −6 Computer det(A), det(B ), det(C ).
Solution: By Theorem 0.7, det(A) = −24, det(B ) = −60. By Corollary 0.1,
det(C ) = 120. ROPERTIES OF D ETERMINANTS 41 We now introduce the following notation for elementary row and
elementary column operations on matrices and determinants.
• Interchange rows (columns) i and j :
ri ↔ rj ( ci ↔ cj ) • Replace row (column) i by a nonzero value k times row (column) i
k ri → ri ( k ci → ci ) • Replace row (column) j by a nonzero value k times row (column) i +
row (column) j : k ri + rj → rj ( k ci + cj → cj ) Using this notation, it is easy to keep track of the elementary row and column
operations performed on a matrix. For example, we indicate that we have
interchanged the ith and j th rows of A as Ari ↔rj . We proceed similarly for
column operations.
ROPERTIES OF D ETERMINANTS 42 We can now interpret Theorem 0.3, 0.5, 0.6 in terms of this notation as follows. det(Ari ↔rj ) = − det(A), i = j
det(Akri →ri ) = k det(A),
det(Akri +rj →rj ) = det(A), i = j.
It is convenient to rewrite these properties in terms of det(A): det(A) = − det(Ari ↔rj ), i = j
1
det(Akri →ri ), k = 0
k
det(A) = det(Akri +rj →rj ), i = j.
det(A) = ROPERTIES OF D ETERMINANTS 43 We proceed similarly for column operations.
• Theorems 0.2, 0.5, 0.6, 0.7 are very useful in the evaluation of det(A).
• What we do is transform A by means of our elementary row operations
to a triangular matrix.
• Of course, we must keep track of how the determinant of the resulting
matrices changes as we perform the elementary row operations. ROPERTIES OF D ETERMINANTS 44 4 Example 17 Let A = 3 2 ROPERTIES OF D ETERMINANTS 3
−2
4 2 5 . Compute det(A). 6 45 Solution: We have
det(A) = = = = ROPERTIES OF 2 det(A 1 r →r ) Multiply row 3 by 1
2
3
23 4
32 3 −2 5 2 det 1
23 23 1 (−1)2 det 3 −2 5 Interchange rows 1 and 3 4
32
r1 ↔r3 1 −2 det 3 4 D ETERMINANTS 2 −2
3 3 5 −3r1 + r2 → r2
2
−4r1 + r3 → r3 Zero out below the (1,1) entry 46 = = = 1 −2 det 0 0 1 −2 det 0 0 1 −2 det 0 0 2
−8
−5
2
−8
−5
2
−8
0 3 −4 −10 3 −4 −10
5 − 8 r2 +r3 →r3 3 Zero out below the (2,2) entry. −4 . − 30
4 Next we compute the determinant of the upper triangular matrix
det(A) = −2(1)(−8) − ROPERTIES OF D ETERMINANTS 30
4 = −120 by Theorem 0.7. 47 Remark
The method used to compute a determinant in Example 17 will be referred
to as the computation via reduction to triangular form. ROPERTIES OF D ETERMINANTS 48 0 Example 18 Evaluate det(A) where A = 3 2 ROPERTIES OF D ETERMINANTS 1
−6
6 5 9 1 49 Solution: We have det(A) = = = = ROPERTIES OF 0
15 det 3 −6 9 2
61 3 −6 − det 0
1 2
6 1 −2 −3 det 0
1 2
6 1 −2 −3 det 0
1 0
10 D ETERMINANTS 9 5 1
3 r1 ↔r2 5 11 r →r1
31 3 Interchange rows 1 and 2 Multiply row 1 by 5 −5 −2r1 + r3 → r3 1
3 Zero out below the (1,1) entry 50 = = 1 −3 det 0 0 −2
1
0 1 (−3)(−55) det 0 0 3 5 −55
−2
1
0 −10r2 +r3 →r3 3 5 1 1
− 55 r1 →r1 Zero out below the (2,2) entry. 1
Multiply row 3 by − 55 Next we compute the determinant of the upper triangular matrix
det(A) = (−3)(−55)(1) = 165 ROPERTIES OF D ETERMINANTS by Theorem 0.7. 51 Theorem 0.8 The determinant of a product of two matrices is the product of
their determinants; that is
det(AB ) = det(A) det(B ). ROPERTIES OF D ETERMINANTS 52 Example 19 Let A= 1 2 3 4 B= and 2 −1 1 2 Then
 A  = −2
Also, and AB = B  = 5. 4 3 10 5 and
AB  = −10 = AB . ROPERTIES OF D ETERMINANTS 53 Remark
In Example 19 we also have (verify) BA = −1 0 7 10 , so AB = BA. However, BA = B A = −10 = AB 
As an immediate consequence of Theorem 0.8, we can readily compute
det(A−1 ) from det(A), as the following corollary shows. ROPERTIES OF D ETERMINANTS 54 Corollary 0.2: If A is nonsingular, then det(A) = 0 and
det(A−1 ) = 1
.
det(A) Proof: Exercise. ROPERTIES OF D ETERMINANTS 55 Example 20 Let A=
Then det(A) = −2 and A−1 = 1 2 3 4 . −2 1 3
2 1
−2 . Now,
det(A−1 ) = − ROPERTIES OF D ETERMINANTS 1
1
=
.
2
det(A) 56 C OFACTOR E XPANSION AND A PPLICATIONS
1. The Inverse of A Matrix
2. Cramer’s Rule OFACTOR E XPANSION AND A PPLICATIONS 57 Our main aim is:
• to develop a different method for evaluating the determinant of an
n × n matrix, which reduces the problem to the evaluation of
determinants of matrices of order n − 1.
We can repeat the process for these (n − 1) × (n − 1) matrices until we
get to 2 × 2 matrices. OFACTOR E XPANSION AND A PPLICATIONS 58 Deﬁnition 3 • Let A = [aij ] be an n × n matrix.
• Let Mij be the (n − 1) × (n − 1) submatrix of A obtained by deleting the ith
row and j th column of A.
• The determinant det(Mij ) is called the minor of aij .
• The cofactor Aij of aij is deﬁned as
Aij = (−1)i+j det(Mij ). OFACTOR E XPANSION AND A PPLICATIONS 59 Example 21 Let 3 A= 4 7 −1
5
1 2 6 . 2 1. Find the minor of aij : det(M12 ), det(M23 ), and det(M31 )
2. Find the cofactor Aij of aij : A12 , A23 , and A31 OFACTOR E XPANSION AND A PPLICATIONS 60 We ﬁrst evaluate the minor of aij :
3 −1 2 4 5 6 7 1 2 4 6 7 2 3 −1 7 1 det(M12 ) = det(M23 ) = 3 2 4 5 6 7 , −1
1 3 2 , −1 2 4 5 6 7 1 2 = 8 − 42 = −34; = 3 + 7 = 10; −1 2 5 6 A12 = (−1)1+2 det(M12 ) = (−1)(−34) = 34 A23 = (−1)2+3 det(M23 ) = (−1)(10) = −10 A31 = (−1)3+1 det(M31 ) = (1)(−16) = −16. det(M31 ) = = −6 − 10 = −16. If we consider the sign (−1)i+j as being located in position (i, j ) of an n× n
matrix, then the signs form a checkerboard pattern that has a + in the (1,1)
position. The patterns for n = 3 and n = 4 are as follows: +−+− +−+
− + − + − + − + − + − +−+
−+−+
The following theorem gives another method of evaluating determinants
that is not as computationally efﬁcient as reduction to triangular form. OFACTOR E XPANSION AND A PPLICATIONS 62 Theorem 0.9 Let A = [aij ] be an n × n matrix. Then for each 1 ≤ i ≤ n,
det(A) = ai1 Ai1 + ai2 Ai2 + . . . + ain Ain (4) (expansion of det(A) along the ith row).
and for each 1 ≤ j ≤ n
det(A) = a1j A1j + a2j A2j + . . . + anj Anj (5) (expansion of det(A) along the j th column).
Proof: The ﬁrst formula follows from the second by Theorem 0.1, that is, from
the fact that det(AT ) = det(A). We omit the general proof and consider the
3 × 3 matrix A = [aij ].
From Equation (3),
det(A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32
− a11 a23 a32 − a12 a21 a33 − a13 a22 a31 .
We can write the expression as
det(A) = a11 (a22 a33 − a23 a32 ) + a12 (a23 a31 − a21 a33 ) + a13 (a21 a32 − a22 a31 ).
OFACTOR E XPANSION AND A PPLICATIONS 63 Now,
A11 = (−1)1+1 a22 a23 a32 a33 A12 = (−1)1+2 a21 a23 a31 a33 A13 = (−1)1+3 a21 a22 a31 a32 = (a22 a33 − a23 a32 ) = (a23 a31 − a21 a33 ) = (a21 a32 − a22 a31 ) Hence
det(A) = a11 A11 + a12 A12 + a13 A13 ,
which is the expansion of det(A) along the ﬁrst row. OFACTOR E XPANSION AND A PPLICATIONS 64 The cofactors of A are We ﬁrst evaluate
a11 a12 a13 a21 a22 a23 a31 a32 a11 a33 , a12 a13 a21 a22 a23 a31 a32 a33 A11 = (−1)1+1 a22 a23 a32 a33 A12 = (−1)1+2 a21 a23 a31 a33 A13 = (−1)1+3 a21 a22 a31 a32 a11
, a12 a13 a21 a22 a23 a31 a32 a33 = (a22 a33 − a23 a32 ) = (a23 a31 − a21 a33 ) = (a21 a32 − a22 a31 ) Hence
det(A) = a11 A11 + a12 A12 + a13 A13 . OFACTOR E XPANSION AND A PPLICATIONS 65 If we now rewrite (3) as
det(A) = a13 (a21 a32 − a22 a31 ) + a23 (a12 a31 − a11 a32 ) + a33 (a11 a22 − a12 a21 ).
Similarly, we write
A13 = (−1)1+3 a21 a31 a22 a32 A23 = (−1)2+3 a11 a12 a31 a32 A33 = (−1)3+3 a11 a12 a21 a22 = (a21 a32 − a22 a31 ) = (a12 a31 − a11 a32 ) = (a11 a22 − a12 a21 ) we also derive that
det(A) = a13 A13 + a23 A23 + a33 A33 ,
which is the expansion of det(A) along the third column. OFACTOR E XPANSION AND A PPLICATIONS 66 Now, we evaluate
a11 a12 a21 a22 a23 a31 a32 a11 a13
a33 , a12 a13 a21 a22 a23 a31 a32 a33 a11
, a12 a13 a21 a22 a23 a31 a32 a33 Similarly, we write
A13 = (−1)1+3 a21 a31 a22 a32 A23 = (−1)2+3 a11 a12 a31 a32 A33 = (−1)3+3 a11 a12 a21 a22 = (a21 a32 − a22 a31 ) = (a12 a31 − a11 a32 ) = (a11 a22 − a12 a21 ) we also derive that
det(A) = a13 A13 + a23 A23 + a33 A33 , OFACTOR E XPANSION AND A PPLICATIONS 67 Example 22 To evaluate the determinant
1 2 −3 4 −4 2 1 3 3 0 0 −3 2 0 −2 3 , We note that it is best to expand along either the second column of the third row
because they each have two zeros. Obviously, the optimal course of action is to
expand along a row or column having the largest number of zeros because in that
case the cofactors Aij of those aij that are zero do not have to be evaluated, since
aij Aij = (0)Aij = 0. Thus, expanding along the third row, we have OFACTOR E XPANSION AND A PPLICATIONS 68 1 −3 4 −4 2 1 3 3 0 0 −3 2 0 −2 3 2 −3 4 2 1 3 0 = 2 −2 3 (−1)3+1 (3) (6) 1 1 OFACTOR −4 2 3 0 3 −4 1 3 −2 3 4 2 +(−1)3+3 (0) 2 4 2 + (−1)3+2 (0) −3 E XPANSION AND A PPLICATIONS 1
+ (−1)3+4 (−3) 2 −3 −4 2 1 2 0 −2 . 69 We now evaluate
2 −3 4 2 1 3 0 −2 3 by expanding along the ﬁrst column, obtaining
1 3 −2 3 −3 4 −2 3 1 2 −3 −4 2 1 2 (−1)1+1 (2) 0 −2 +(−1)2+1 (2) +0 = (1)(2)(9)+(−1)(2)(−1) = 20. Similarly, we evaluate by expanding along the third row, obtaining
(−1)3+1 (2) OFACTOR 2 −3 2 1 +0+(−1)3+3 (−2) E XPANSION AND A PPLICATIONS 1 2 −4 2 = (1)(2)(8)+(1)(−2)(10) = −4. 70 Substituting in Equation (6), we ﬁnd the value of the given determinant as
(1)(3)(20) + 0 + 0 + (−1)(−3)(−4) = 48.
On the other hand, evaluating the given determinant by expanding along the ﬁrst
column, we have
2 3 0 0 −3 0 (−1)1+1 (1) 1
−2 2 3 2
2 1 3 −2 3 + (−1)4+1 (2) 0 0 −3 −2 3 2 4 0 +(−1)3+1 (3) −3 4 0 + (−1)2+1 (−4) −3 −3 4 2 1 3 0 0 −3 = (1)(1)(−12) + (−1)(−4)(−12) + (1)(3)(20) + (−1)(2)(−24) = 48. OFACTOR E XPANSION AND A PPLICATIONS 71 T HE I NVERSE OF A M ATRIX HE I NVERSE OF A M ATRIX 72 It is interesting to ask what
ai1 Ak1 + ai2 Ak2 + . . . + ain Akn
is for i = k because, when we answer this question, we shall obtain another
method for ﬁnding the inverse of a nonsingular matrix. HE I NVERSE OF A M ATRIX 73 Theorem 0.10 If A = [aij ] is n × n matrix, then
ai1 Ak1 + ai2 Ak2 + . . . + ain Akn = 0 f or i = k; (7) a1j A1k + a2j A2k + . . . + anj Ank = 0 f or j = k. (8) Proof: We prove only the ﬁrst formula. The second follows from the ﬁrst one by
Theorem 0.1.
• Consider the matrix B obtained from A by replacing the kth row of A by
its ith row.
• Thus B is a matrix having two identical rows – the ith and kth rows.
• Then det(B ) = 0.
• Now expand det(B ) along the kth row.
• The elements of the kth row of B are ai1 , ai2 , . . . , ain .
• The cofactors of the kth row are Ak1 , Ak2 , . . . , Akn .
HE I NVERSE OF A M ATRIX 74 • Thus from Equation (1) we have
0 = det(B ) = ai1 Ak1 + ai2 Ak2 + . . . + ain Akn ,
which is what we want to show.
This theorem states that if we sum the products of the element of any row (or
column) times the corresponding cofactors of any other row (or column), then we
obtain zero. HE I NVERSE OF A M ATRIX 75 Example 23
Let 1 A = −2 4 2
3
5 3 1 . −2 Then
A21 = (−1)2+1 A23 = (−1)2+3 2 3 5 −2 1 2 4 5 = 19, A22 = (−1)2+2 1 3 4 −2 = −14, = 3. Now
a31 A21 + a32 A22 + a33 A23 = (4)(19) + (5)(−14) + (−2)(3) = 0
and
a11 A21 + a12 A22 + a13 A23 = (1)(19) + (2)(−14) + (3)(3) = 0. HE I NVERSE OF A M ATRIX 76 Now
a31 A21 + a32 A22 + a33 A23 = 0
and
a11 A21 + a12 A22 + a13 A23 = 0,
where
A21 a12 a13 a32 a33 = (−1)2+2 a11 a13 a31 a33 A23 A M ATRIX (−1)2+1 A22 H E I NVERSE OF = = (−1)2+3 a11 a12 a31 a32 77 Example 24
Let 1 A = −2 4 2
3
5 3 1 . −2 Then
A12 = (−1)2+1 A32 = (−1)2+3 −2 1 4 −2 1 3 −2 1 = 0, A22 = (−1)2+2 1 3 4 −2 = −14, = −7. Now
a13 A12 + a23 A22 + a33 A32 = (3)(0) + (1)(−14) + (−2)(−7) = 0
and
a11 A12 + a21 A22 + a31 A32 = (1)(0) + (−2)(−14) + (4)(−7) = 0. HE I NVERSE OF A M ATRIX 78 Now
a13 A12 + a23 A22 + a33 A32 = 0
and
a11 A12 + a21 A22 + a31 A32 = 0. A12 a21 a23 a31 a33 = (−1)2+2 a11 a13 a31 a33 A23 A M ATRIX (−1)1+2 A22 H E I NVERSE OF = = (−1)3+2 a11 a13 a21 a23 79 We may combine Equations (4) and (7) as
ai1 Ak1 + ai2 Ak2 + . . . + ain Akn = det(A) if i = k;
=0 (9) if i = k. Similarly, we may combine Equations (5) and (8) as
a1j A1k + a2j A2k + . . . + anj Ank = det(A)
=0 HE I NVERSE OF A M ATRIX if j = k; (10) if j = k. 80 Deﬁnition 4
Let A = [aij ] be an n × n matrix. The n × n matrix adjA, called the adjoint of A is
the matrix whose i, j th element is the cofactor Aji of aji . Thus A11 A12 adjA = .
.
. A1n A21 ... An1 A22
.
.
. ... An2
.
.
. A2n ... Ann . Remarks
1. The adjoint of A is formed by taking the transpose of the matrix of
cofactors of the elements of A.
2. It should be noted that the term adjoint has other meanings in linear
algebra in addition to its use in the above deﬁnition. HE I NVERSE OF A M ATRIX 81 Example 25 Let 3 5 1 −2
6
0 1 2 . −3 Compute adjA. HE I NVERSE OF A M ATRIX 82 Solution: The cofactors of A are We ﬁrst evaluate
3 −2 1 5 6 2 1 0 −3 A11 = (−1)1+1 A13 = (−1)1+3 HE I NVERSE OF A M ATRIX 3 2 0 −3 5 6 1 0 6 2 1 6 1 5 , −2
0 −3 3
, −2 1 5 6 2 1 0 −3 = −18; A12 = (−1)1+2 5 2 1 −3 = 17; = −6; 83 Then we evaluate
3 −2 1 5 6 2 1 0 −3 A21 = (−1)2+1 A23 = (−1)2+3 HE I NVERSE OF A M ATRIX 3 1 0 −3 3 −2 1 0 6 2 1 −2 1 5 , −2
0 −3 3
, −2 1 5 6 2 1 0 −3 3 1 1 −3 = −6; A22 = (−1)2+2 = −10; = −2. 84 Finally we evaluate
3 −2 1 5 6 2 1 0 −3 A31 = (−1)3+1 A33 = (−1)3+3 Then 3 1 6 2 3 6 A M ATRIX 0 −3 3 1 5 6 2 1 , −2
0 −3 A32 = (−1)3+2 3 1 5 2 = −1; = 28. adjA = HE I NVERSE OF 2 = −10; −2 5 6 1 −2 1 5 , −2 −18 −6 17 −10 −6 −2 −10 −1 . 28 85 Theorem 0.11 If A = [aij ] is an n × n matrix, then
A(adjA) = (adjA)A = det(A)In .
Proof: We have a11 a 21 .
.
.
A(adjA) = ai1 .
.
. an 1 a12 ... a22
.
.
. ... ai2
.
.
. ... an 2 ... a1n
a2n
.
.
.
ain
.
.
.
ann A11 A12 . . . A1n A21 ... Aj 1 ... An1 A22
.
.
. ... Aj 2
.
.
. ... An2
.
.
. A2n ... Ajn ... Ann . The i, j th element in the product matrix A(adjA) is, by Equation (9), HE I NVERSE OF A M ATRIX = det(A) if i = j = ai1 Aj 1 + ai2 Aj 2 + . . . + ain Ajn 0 if i = j. 86 This means that A(adjA) = det(A) 0 ... 0
.
.
. det(A)
.
.
. ... 0 0... 0 .. . 0
0
.
.
.
det(A) =
det(A) In . a real number The i, j th element in the product matrix (adjA)A is, by Equation (10)
= det(A) if i = j = A1i a1j + A2i a2j + . . . + Ani anj 0 if i = j. Thus (adjA)A = det(A)In . HE I NVERSE OF A M ATRIX 87 Example 26 Consider the matrix of Example 25. Then 3 5 1 −2
6
0 1 −18 2 −3 −6 17 −10 −1 = 28 −10 −6 −2 −94 0 0 −94 0 0 0 1 = −94 0 0 0 0 0 −94 0 1 1
0 and −18 −6 17 −10 −6 −2 3 −1 5 28
1 −10 −2
6
0 1 1 2 = −94 0 −3
0 0
1
0 0 0 1 We now have a new method for ﬁnding the inverse of a nonsingular matrix,
and we state this result as the following corollary. HE I NVERSE OF A M ATRIX 88 Corollary 0.3: If A is an n × n matrix and det(A) = 0, then A −1 1
=
(adjA) = det(A) A11
det(A)
A12
det(A) A21
det(A)
A22
det(A) .
.
. .
.
. A1n
det(A) A2n
det(A) ...
... An1
det(A)
An2
det(A) .
.
.
... Ann
det(A) . Proof: By Theorem 0.11, A(adjA) = det(A)In , so if det(A) = 0, then
A 1
1
1
(adjA) =
[A(adjA)] =
(det(A)In ) = In .
det(A)
det(A)
det(A) Hence
A−1 = HE I NVERSE OF A M ATRIX 1
(adjA).
det(A) 89 Example 27 Again consider the matrix of Example 25. Then det(A) = −94, and A−1 = HE I NVERSE OF A M ATRIX 1
(adjA) = det(A) 18
94
17
− 94
6
94 6
94
10
94
2
94 10
94
1
94
− 28
94 . 90 Theorem 0.12 A matrix A is nonsingular if and only if det(A) = 0.
Proof:
• If det(A) = 0, then Corollary 0.2 gives an expression for A−1 , so A is
nonsingular.
• – We now prove the converse here.
– Suppose that A is nonsingular.
– Then
AA−1 = In .
– From Theorem 0.8 we have
det(AA−1 ) = det(A) det(A−1 ) = det(In ) = 1,
which implies that det(A) = 0. This completes the proof. HE I NVERSE OF A M ATRIX 91 Corollary 0.4: For an n × n matrix A, the homogeneous system Ax = 0 has a
nontrivial solution if and only if det(A) = 0.
Proof: Exercise. HE I NVERSE OF A M ATRIX 92 List of Nonsingular Equivalences The following statements are equivalent.
1. A is nonsingular.
2. x = 0 is the only solution to Ax = 0.
3. A is row equivalent to In .
4. The linear system Ax = b has a unique solution for every n × 1 matrix b.
5. det(A) = 0. HE I NVERSE OF A M ATRIX 93 C RAMER ’ S R ULE RAMER ’ S R ULE 94 Our aim is to use the result of Theorem 0.11 to obtain another method,
known as Cramer’s rule, for solving a linear system of n equations in n
unknowns with a nonsingular coefﬁcient matrix. RAMER ’ S R ULE 95 Let us consider the system of two equations in two unknowns a x +a x =b
11 1
12 2
1 a21 x1 + a22 x2 = b2 (11) possess the solution x1 x
2 b1 a22 − a12 b2
a11 a22 − a12 a21
a11 b2 − b1 a21
=
a11 a22 − a12 a21
= (12) proved that a11 a22 − a12 a21 = 0. The numerators and denominators in
Equation (11) has the unique solution b1 RAMER ’ S R ULE b1 b2 a22 a21 b2 a12 a11 a12 a21 = a11 a11 x1 a12 a22 a21 a22 , x2 = (13) 96 provided that the determinant of the matrix of coefﬁcients
a11
a21 RAMER ’ S R ULE a12
a22 =0 97 Theorem 0.13 (Cramer’s Rule) Let
a11 x1 + a12 x2 + ... + a1n xn = b1 a21 x1
.
.
. + a22 x2
.
.
. + ...
.
.
. + a2n xn
.
.
. = b2
.
.
. an1 x1 + an2 x2 + ... + ann xn = bn be a linear system of n equations in n unknowns and let A = [aij ] be the
coefﬁcient matrix so that we can write the given system as Ax = b, where b1 b2 b = . .
.
. bn If det(A) = 0, then the system has the unique solution
x1 = det(A1 )
,
det(A) x2 = det(A2 )
,...
det(A) xn = det(An )
det(A) where Ai is the matrix obtained from A by replacing the ith column of A by
b.
RAMER ’ S R ULE 98 Proof: If det(A) = 0, then, by Theorem 0.12, A is nonsingular. Hence
A A21
An1
11
. . . det(A)
det(A) det(A) A22
An2 A12
. . . det(A) det(A) det(A)
x1 b1 .
.
. x2 .
.
. b2 .
.
. −1
x= . =A b= A .
.
A2i
Ani
1i . . . det(A) .
. det(A) .
det(A) .
.
. .
.
.
xn
bn .
.
. A1n
A2n
Ann
. . . det(A)
det(A)
det(A) This means that
xi = A1i
A2i
Ani
b1 +
b2 + . . . +
bn
det(A)
det(A)
det(A) (1 ≤ i ≤ n). Now let a11 a21 Ai = .
.
. an 1
RAMER ’ S R ULE a12 . . . a1i−1 b1 a1i+1 ... a1n a22 . . .
.
.
. a2i−1 b2
.
.
. a2i+1
.
.
. ...
.
.
. a2n an 2 . . . ani−1 bn ani+1 ... ann . 99 • If we evaluate det(Ai ) by expanding along the ith column, we ﬁnd that
det(Ai ) = A1i b1 + A2i b2 + . . . + Ani bn .
Hence
xi = det(Ai )
det(A) for i = 1, 2 . . . , n.
• In this expression for xi , the determinant of Ai , det(Ai ) can be
calculated by any method.
• It was only in the derivation of the expression for xi that we had to
evaluate it by expanding along the ith column. Example 28 Consider the following linear system:
−2x1 + 3x2 − x3 = 1 x1 + 2x2 − x3 = 4 −2x1 − x2 + x3 = −3. The augmented matrix form is −2 3 −1 1 2 −1 −2 −1 1 1 4 . −3 Then
−2 RAMER ’ S R ULE −1 1 2 −1 −2 det(A) = 3
−1 1 = −2. 101 We ﬁrst evaluate
−2 3 −1 1 2 −1 −2 −1 1 A11 = (−1)1+1 A13 = (−1)1+3 −2 3 −1 1 2 −1 −2 −1 1 , 2 −1 −1 1 1 2 −2 −1 −2 3 −1 1 2 −1 −2 −1 1 1 −1 −2 1 , = 1; A12 = (−1)1+2 = 1; = 3; det(A) = a11 A11 + a12 A12 + a13 A13
= (−2)(1) + (3)(1) + (−1)(3)
= −2 RAMER ’ S R ULE 102 Hence
1 RAMER ’ S R ULE −1 4
x1 = 3
2 −1 −3 −1 1 det(A) = −4
= 2.
−2 103 We ﬁrst evaluate
1 3 −1 4 2 −1 −3 −1 1 A11 = (−1)1+1 A31 = (−1)1+3 1 3 −1 4 2 −1 −3 −1 1 , 2 −1 −1 1 3 −1 2 −1 1 3 −1 4 2 −1 −3 −1 1 3 −1 −1 1 , = 1; A21 = (−1)1+2 = −2; = −1; det(A1 ) = a11 A11 + a21 A21 + a31 A31 = A11 b1 + A21 b2 + A31 b3
= (1)(1) + (4)(−2) + (−3)(−1)
= −4 RAMER ’ S R ULE 104 Hence
−2 RAMER ’ S R ULE −1 1
x2 = 1
4 −1 −2 −3 1 det(A) . 105 We evaluate
−2 1 −1 1 4 −1 −2 −3 1 A12 = (−1)1+2 A32 = (−1)3+2 −2 1 −1 1 4 −1 −2 −3 1 , 1 −1 −2 1 −2 −1 1 −1 −2 1 −1 1 4 −1 −2 −3 1 −2 −1 −2 1 , = 1; A22 = (−1)2+2 = −4; = −3; det(A2 ) = a12 A12 + a22 A22 + a32 A32 = A12 b1 + A22 b2 + A32 b3
= (1)(1) + (4)(−4) + (−3)(−3)
= −6 RAMER ’ S R ULE 106 Hence
1 −1 4
x1 = 3
2 −1 −3 −1 1 det(A)
−2 4
−3 1 det(A)
−2 RAMER ’ S R ULE 3 −6
= 3;
−2 = −8
= 4.
−2 1 1
x3 = = −1 −2 −4
= 2;
−2 −1 1
x2 = 1 = 2 4 −2 −1 −3 det(A) 107 Example 29 Consider the following linear system:
3x1 + 2x2 + 1x3 = 7 x1 − x2 + 3x3 = 3 5x1 + 4x2 − 2x3 = 1. The augmented matrix form is 3 1 5 2 1 −1 3 4 −2 7 3 . 1 Then
3 RAMER ’ S R ULE 1 1 −1 3 5 det(A) = 2
4 −2 = 13. 108 Hence
7 RAMER ’ S R ULE 1 3
x1 = 2
−1 3 1 4 −2 det(A) 109 We ﬁrst evaluate
7 2 1 3 −1 3 1 4 −2 A11 = (−1)1+1 A31 = (−1)1+3 7 3 4 −2 2 1 −1 3 −1 3 1 −1 1 3 , 2
4 −2 7
, 2 1 3 −1 3 1 4 −2 = −10; A21 = (−1)1+2 2 1 4 −2 = 8; = 7; det(A1 ) = (7)(−10) + (3)(8) + (1)(7)
= −39
Hence, x1 =
RAMER ’ S R ULE −39
13 = −3
110 Similarly,
3 1 1
x2 = 7
3 3 5 1 −2 det(A) = 6, and
3 RAMER ’ S R ULE 7 1
x3 = 2
−1 3 5 4 1 det(A) = 4. 111 Cramer’s rule for solving the linear system Ax = b, where A is an n × n, is as
follows:
Step 1. Compute det(A). If det(A) = 0, Cramer’s rule is not applicable. Use
GaussJordan reduction.
Step 2. If det(A) = 0, for each i
xi = det(Ai )
,
det(A) where Ai is the matrix obtained from A by replacing the ith column of A
by b. RAMER ’ S R ULE 112 Remark:
• Cramer’s rule is applicable only to the case where we have n equations
in n unknowns and where the coefﬁcient matrix A is nonsingular.
• If we have to solve a linear system of n equations in n unknowns whose
coefﬁcients matrix is singular, then we must use the GaussJordan
reduction method as discussed in earlier section.
• Cramer’s rule becomes computationally inefﬁcient for n > 4, and it is
better to use the GaussJordan reduction method. RAMER ’ S R ULE 113 P OLYNOMIAL INTERPOLATION OLYNOMIAL INTERPOLATION 114 Suppose we given the n distinct points (x1 , y1 ), (x2 , y2 ), . . . (xn , yn ).
Can we ﬁnd a polynomial of degree n − 1 or less that "interpolates"
the data, that is, passes through the n points? Thus, the polynomial
we seek has the form
y = an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0 .
The n given points can be used to obtain an n × n linear system
whose unknown are a0 , a1 , . . . , an−1 . It can be shown that this linear
system has a unique solution. Thus, there is unique interpolating
polynomial.
OLYNOMIAL INTERPOLATION 115 We consider the case where n = 3 in detail. Here we are given the
points (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), where x1 = x2 , x1 = x3 and x2 = x3
and seek the polynomial
y = a2 x2 + a1 x + a0 . (14) Substituting the given points in (14), we obtain the linear system
a2 x2 + a1 x1 + a0 = y1
1
a2 x2 + a1 x2 + a0 = y2
2 (15) a2 x2 + a1 x3 + a0 = y3 .
3
In what follows, we show that the linear system (15) has a unique
solution. Thus, there is a unique interpolating quadratic polynomial.
In general, there is a unique interpolating polynomial of degree
n − 1 passing through n given points. OLYNOMIAL INTERPOLATION 116 Example 30 Find the quadratic polynomial that interpolate the points
(1, 3), (2, 4), (3, 7).
Solution Setting up the linear system (15) we have
a2 + a1 + a0 = 3
4a2 + 2a1 + a0 = 4
9a2 + 3a1 + a0 = 7.
whose solution is (verify)
a2 = 1, a1 = −2, a0 = 4. Hence, the quadratic interpolating polynomial is
y = x2 − 2x + 4.
Its graph, shown in Figure 3, passes through the three given points. OLYNOMIAL INTERPOLATION 117 12 11 10 9 y 8 7 6 2 y = x −2x + 4 5 4 3
−1 −0.5 0 0.5 1 1.5
x 2 2.5 3 3.5 4 Figure 3: OLYNOMIAL INTERPOLATION 118 The coefﬁcient matrix of this linear system is x2
1 x2
2
x2
3 x1
x2
x3 1 1 1 whose determinant is the Vandermonde determinant, which has
the value
(x1 − x2 )(x1 − x3 )(x2 − x3 ).
• Since the three given points are distinct, the Vandermonde
determinant is not zero.
• Hence, the coefﬁcient matrix of the linear system in
Equation (15) is nonsingular, which implies that the linear system
has a unique solution.
• Therefore, there is a unique interpolating quadratic polynomial. OLYNOMIAL INTERPOLATION 119 ...
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 jeff
 Math, Linear Algebra, Algebra

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