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Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #3
(due Thursday, September 22)
All problems are from Leon’s book (8th edition). Section 2.1: 3b, 3c, 3d, 3e, 3f, 3g
Section 2.2: 3f, 4, 7c, 7d MATH 304
Linear Algebra
Lecture 7:
Evaluation of determinants.
The Vandermonde determinant.
Cramer’s rule. Determinants
Determinant is a scalar assigned to each square matrix.
Notation. The determinant of a matrix
A = (aij )1≤i ,j ≤n is denoted det A or
a11 a12
a21 a22
.
.
.
.
.
.
an1 an2 . . . a1n
. . . a2n
... . .
.
.
. . . ann Principal property: det A = 0 if and only if a
system of linear equations with the coeﬃcient
matrix A has a unique solution. Equivalently,
det A = 0 if and only if the matrix A is invertible. Explicit deﬁnition in low dimensions
Deﬁnition. det (a) = a, ab
= ad − bc ,
cd a11 a12 a13
a21 a22 a23 = a11a22 a33 + a12a23a31 + a13a21a32 −
a31 a32 a33
−a13a22a31 − a12 a21a33 − a11a23 a32. +: −: * ∗
∗ ∗ ∗
* ∗
*
∗
∗
*
∗ ∗
∗ ∗ , ∗
*
* ∗
* ∗ , *
∗
∗ *
∗
∗
*
∗
∗ ∗ * , ∗ ∗ ∗ , * ∗
*
∗ ∗
∗
* *
∗
∗ ∗
∗
* * ∗ .
∗ ∗ * .
∗ Properties of determinants Determinants and elementary row operations:
• if a row of a matrix is multiplied by a scalar r ,
the determinant is also multiplied by r ;
• if we add a row of a matrix multiplied by a scalar
to another row, the determinant remains the same;
• if we interchange two rows of a matrix, the
determinant changes its sign. Properties of determinants Tests for singularity:
• if a matrix A has a zero row then det A = 0;
• if a matrix A has two identical rows then
det A = 0;
• if a matrix has two proportional rows then
det A = 0. Properties of determinants Special matrices:
• det I = 1;
• the determinant of a diagonal matrix is equal to
the product of its diagonal entries;
• the determinant of an upper triangular matrix is
equal to the product of its diagonal entries. Properties of determinants
Determinant of the transpose:
• If A is a square matrix then det AT = det A.
Columns vs. rows:
• if one column of a matrix is multiplied by a
scalar, the determinant is multiplied by the same
scalar;
• adding a scalar multiple of one column to
another does not change the determinant;
• interchanging two columns of a matrix changes
the sign of its determinant;
• if a matrix A has a zero column or two
proportional columns then det A = 0. Row and column expansions
Given an n×n matrix A = (aij ), let Mij denote the
(n − 1)×(n − 1) submatrix obtained by deleting the
i th row and the j th column of A.
Theorem For any 1 ≤ k , m ≤ n we have that
n (−1)k +j akj det Mkj , det A =
j =1 (expansion by kth row )
n (−1)i +maim det Mim . det A =
i =1 (expansion by mth column) Signs for row/column expansions +
− + −
.
.
. −
+
−
+
.
.
. +
−
+
−
.
.
. −
+
−
+
.
.
. ···
· · · · · · · · ·
... 123
Example. A = 4 5 6.
789
Expansion by the 1st row: ∗2∗
∗∗ 3
1 ∗∗ ∗ 5 6 4 ∗ 6 4 5 ∗ ∗ 89
7∗9
78 ∗
det A = 1 56
46
45
−2
+3
89
79
78 = (5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0. 123
Example. A = 4 5 6.
789
Expansion by the 2nd 1∗
∗2∗
4 ∗ 6 ∗ 5
7∗9
7∗
det A = −2 column: 3
1∗3
∗ 4 ∗ 6 9
∗8∗ 46
13
13
+5
−8
79
79
46 = −2(4 · 9 − 6 · 7) + 5(1 · 9 − 3 · 7) − 8(1 · 6 − 3 · 4) = 0. 123
Example. A = 4 5 6.
789
Subtract the 1st row from the 2nd row and from
the 3rd row:
123
123
123
4 5 6 = 3 3 3 = 3 3 3 =0
789
789
666
since the last matrix has two proportional rows. 12 3
Another example. B = 4 5 6 .
7 8 13
Let’s do some row reduction.
Add −4 times the 1st row to the 2nd row:
12 3
123
4 5 6 = 0 −3 −6
7 8 13
7 8 13
Add −7 times the 1st row to the 3rd row:
123
123
0 −3 −6 = 0 −3 −6
0 −6 −8
7 8 13 123
123
0 −3 −6 = 0 −3 −6
7 8 13
0 −6 −8
Expand the determinant by the 1st column:
123
−3 −6
0 −3 −6 = 1
−6 −8
0 −6 −8
Thus
det B =
= (−3)(−2) 12
−3 −6
= (−3)
−6 −8
−6 −8
12
= (−3)(−2)(−2) = −12
34 2 −2 0 3
−5 3 2 1 Example. C = 1 −1 0 −3, det C =?
2 0 0 −1
Expand the determinant by the 3rd column:
2 −2 0 3
2 −2 3
−5 3 2 1
= −2 1 −1 −3
1 −1 0 −3
2 0 −1
2 0 0 −1
Add −2 times the 2nd row to the 1st row:
009
2 −2 3
det C = −2 1 −1 −3 = −2 1 −1 −3
2 0 −1
2 0 −1 009
2 −2 3
det C = −2 1 −1 −3 = −2 1 −1 −3
2 0 −1
2 0 −1
Expand the determinant by the 1st row:
009
1 −1
det C = −2 1 −1 −3 = −2 · 9
20
2 0 −1
Thus
det C = −18 1 −1
= −18 · 2 = −36
20 Problem. For what values of a will the following
system have a unique solution? x + 2y + z = 1
−x + 4y + 2z = 2 2x − 2y + az = 3 The system has a unique solution if and only if the
coeﬃcient matrix is invertible. 1 21
A = −1 4 2, det A =?
2 −2 a 1 21
A = −1 4 2,
2 −2 a Add −2 times the
12
−1 4
2 −2 det A =? 3rd column to the 2nd column:
1
0
1
1
0
2
2 = −1
2 −2 − 2a a
a Expand the determinant by the 2nd column:
1
0
1
11
0
2 = −(−2 − 2a)
det A = −1
−1 2
2 −2 − 2a a
Hence det A = −(−2 − 2a) · 3 = 6(1 + a).
Thus A is invertible if and only if a = −1. More properties of determinants
Determinants and matrix multiplication:
• if A and B are n×n matrices then
det(AB ) = det A · det B ;
• if A and B are n×n matrices then
det(AB ) = det(BA);
• if A is an invertible matrix then
det(A−1) = (det A)−1.
Determinants and scalar multiplication:
• if A is an n×n matrix and r ∈ R then
det(rA) = r n det A. Examples −1 2 1
1 00
X = 0 2 −2, Y = −1 3 0.
0 0 −3
2 −2 1 det X = (−1) · 2 · (−3) = 6, det Y = det Y T = 3,
det(XY ) = 6 · 3 = 18, det(YX ) = 3 · 6 = 18,
det(Y −1) = 1/3, det(XY −1) = 6/3 = 2,
det(XYX −1) = det Y = 3, det(X −1 Y −1 XY ) = 1,
det(2X ) = 23 det X = 23 · 6 = 48,
det(−3X T XY −4) = (−3)3 · 6 · 6 · 3−4 = −12. The Vandermonde determinant
Deﬁnition. The Vandermonde determinant is
the determinant of the following matrix n
2
1 x1 x1 · · · x1 −1 n
2 1 x2 x2 · · · x2 −1 n
2
V = 1 x3 x3 · · · x3 −1 , .
... . . .
..
.
.
.
.
. .
2
n
1 xn xn · · · xn −1
where x1, x2, . . . , xn ∈ R. Equivalently,
V = (aij )1≤i ,j ≤n, where aij = xij −1. Examples.
• 1 x1
1 x2 • 2
1 x1 x1
2
1 x2 x2
2
1 x3 x3 = x2 − x1 .
1 x1
0
2
= 1 x2 x2 − x1 x2
2
1 x3 x3 − x1 x3 1
0
0
2
= 1 x2 − x1 x2 − x1 x2
2
1 x3 − x1 x3 − x1 x3
= (x2 − x1 ) = 1
x2
2
x3 − x1 x3 − x1 x3 = (x2 − x1 )(x3 − x1 )(x3 − x2 ). 2
x2 − x1 x2 − x1 x2
2
x3 − x1 x3 − x1 x3 = (x2 − x1 )(x3 − x1 ) 1 x2
1 x3 Theorem
1
1
1
.
.
.
1 x1
x2
x3
.
.
.
xn 2
x1
2
x2
2
x3
.
.
.
2
xn n
· · · x1 −1
n
· · · x2 −1
n
(xj − xi ).
· · · x3 −1 =
... .
.
1≤i <j ≤n
.
n
· · · xn −1 Corollary The Vandermonde determinant is not
equal to 0 if and only if the numbers x1 , x2, . . . , xn
are distinct. Let x1, x2, . . . , xn be distinct real numbers.
Theorem For any b1 , b2, . . . , bn ∈ R there exists a
unique polynomial p (x ) = a0 +a1 x + · · · +an−1 x n−1
of degree less than n such that p (xi ) = bi ,
1 ≤ i ≤ n. n −1
2 a0 + a1 x1 + a2 x1 + · · · + an−1x1 = b1 n
2
a0 + a1 x2 + a2 x2 + · · · + an−1x2 −1 = b2
············ 2
n
a0 + a1 xn + a2 xn + · · · + an−1xn −1 = bn a0 , a1, . . . , an−1 are unknowns. The coeﬃcient
matrix is the Vandermonde matrix. A general system of n linear equations in n variables: a11x1 + a12x2 + · · · + a1n xn = b1 a21x1 + a22x2 + · · · + a2n xn = b2
⇐⇒ Ax = b,
········· an1 x1 + an2 x2 + · · · + ann xn = bn where a11 a12
a a 21 22
A = .
.
.
.
.
.
an1 an2 . . . a1n
x1
b1
b x . . . a2n 2 2
. . . . , x = . , b = . .
.
.
.
.
.
bn
xn
. . . ann Cramer’s rule a11x1 + a12x2 + · · · + a1n xn = b1 a21x1 + a22x2 + · · · + a2n xn = b2
⇐⇒ Ax = b
········· an1 x1 + an2 x2 + · · · + ann xn = bn
Theorem Assume that the matrix A is invertible.
Then the only solution of the system is given by
det Ai
, i = 1, 2, . . . , n ,
xi =
det A
where the matrix Ai is obtained by substituting the
vector b for the i th column of A. Example. x + 2y + 3z = 0
4x + 5y + 6z = 0 7x + 8y + 13z = 1 Augmented matrix of the system: 12 3 0
(A  b) = 4 5 6 0
7 8 13 1
As obtained earlier in this lecture, det A = −12.
Since det A = 0, there exists a unique solution of
the system. Example. x + 2y + 3z = 0
4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0
4 5 6 0
7 8 13 1 By Cramer’s rule, x= 02 3
05 6
1 8 13
12 3
45 6
7 8 13 23
56
−3
1
=
=
=
−12
−12 4 Example. x + 2y + 3z = 0
4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0
4 5 6 0
7 8 13 1 By Cramer’s rule, y= 10 3
40 6
7 1 13
12 3
45 6
7 8 13 −
= 13
46
6
1
=
=−
−12
−12
2 Example. x + 2y + 3z = 0
4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0
4 5 6 0
7 8 13 1 By Cramer’s rule, z= 120
450
781
12 3
45 6
7 8 13 12
45
−3
1
=
=
=
−12
−12 4 System of linear equations: x + 2y + 3z = 0
4x + 5y + 6z = 0 7x + 8y + 13z = 1
Solution: (x , y , z ) = 11
1
,− ,
.
4
24 Determinants and the inverse matrix
Given an n×n matrix A = (aij ), let Mij denote the
(n − 1)×(n − 1) submatrix obtained by deleting the i th row
and the j th column of A.
The cofactor matrix of A is an n×n matrix A = (αij )
deﬁned by αij = (−1)i +j det Mij .
Theorem AT A = AAT = (det A)I .
Corollary If det A = 0 then the matrix A is invertible and
A−1 = (det A)−1 AT .
AAT = (det A)I means that
n
k +j
akj
j =1 (−1)
n
k +j
amj
j =1 (−1) det Mkj = det A
det Mkj = 0 for all k , for m = k . ...
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This note was uploaded on 11/08/2011 for the course MATH 304 taught by Professor Hobbs during the Spring '08 term at Texas A&M.
 Spring '08
 HOBBS
 Linear Algebra, Algebra

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