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Unformatted text preview: LINEAR ALGEBRA
W W L CHEN
c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.
It is available free to all individuals, on the understanding that it is not to be used for ﬁnancial gain,
and may be downloaded and/or photocopied, with or without permission from the author.
However, this document may not be kept on any information storage and retrieval system without permission
from the author, unless such system is not accessible to any individuals other than its owners. Chapter 3
DETERMINANTS 3.1. Introduction
In the last chapter, we have related the question of the invertibility of a square matrix to a question of
solutions of systems of linear equations. In some sense, this is unsatisfactory, since it is not simple to
ﬁnd an answer to either of these questions without a lot of work. In this chapter, we shall relate these
two questions to the question of the determinant of the matrix in question. As we shall see later, the
task is reduced to checking whether this determinant is zero or nonzero. So what is the determinant?
Let us start with 1 × 1 matrices, of the form
A = (a).
Note here that I1 = ( 1 ). If a = 0, then clearly the matrix A is invertible, with inverse matrix
A−1 = ( a−1 ) .
On the other hand, if a = 0, then clearly no matrix B can satisfy AB = BA = I1 , so that the matrix A
is not invertible. We therefore conclude that the value a is a good “determinant” to determine whether
the 1 × 1 matrix A is invertible, since the matrix A is invertible if and only if a = 0.
Let us then agree on the following deﬁnition.
Definition. Suppose that
A = (a)
is a 1 × 1 matrix. We write
det(A) = a,
and call this the determinant of the matrix A.
Chapter 3 : Determinants page 1 of 23 c Linear Algebra W W L Chen, 1982, 2005 Next, let us turn to 2 × 2 matrices, of the form
ab
cd A= . We shall use elementary row operations to ﬁnd out when the matrix A is invertible. So we consider the
array
ab
cd (AI2 ) = (1) 1
0 0
1 , and try to use elementary row operations to reduce the left hand half of the array to I2 . Suppose ﬁrst
of all that a = c = 0. Then the array becomes
0
0 b
d 1
0 0
1 , and so it is impossible to reduce the left hand half of the array by elementary row operations to the
matrix I2 . Consider next the case a = 0. Multiplying row 2 of the array (1) by a, we obtain
a
ac b1
ad 0 0
a . Adding −c times row 1 to row 2, we obtain
a
b
10
0 ad − bc −c a (2) . If D = ad − bc = 0, then this becomes
ab
00 10
−c a , and so it is impossible to reduce the left hand half of the array by elementary row operations to the
matrix I2 . On the other hand, if D = ad − bc = 0, then the array (2) can be reduced by elementary row
operations to
10
01 −b/D
a/D d/D
−c/D , so that
A−1 = d −b
−c a 1
ad − bc . Consider ﬁnally the case c = 0. Interchanging rows 1 and 2 of the array (1), we obtain
cd
ab 0
1 1
0 . Multiplying row 2 of the array by c, we obtain
c
ac d0
bc c 1
0 . Adding −a times row 1 to row 2, we obtain
c
0
Chapter 3 : Determinants d
bc − ad 0
c 1
−a .
page 2 of 23 c Linear Algebra W W L Chen, 1982, 2005 Multiplying row 2 by −1, we obtain
(3) c
0 d
01
ad − bc −c a . Again, if D = ad − bc = 0, then this becomes
c
0 d
0 0
−c 1
a , and so it is impossible to reduce the left hand half of the array by elementary row operations to the
matrix I2 . On the other hand, if D = ad − bc = 0, then the array (3) can be reduced by elementary row
operations to
10
01 −b/D
a/D d/D
−c/D , so that
A−1 = 1
ad − bc d −b
−c a . Finally, note that a = c = 0 is a special case of ad − bc = 0. We therefore conclude that the value ad − bc
is a good “determinant” to determine whether the 2 × 2 matrix A is invertible, since the matrix A is
invertible if and only if ad − bc = 0.
Let us then agree on the following deﬁnition.
Definition. Suppose that
A= ab
cd is a 2 × 2 matrix. We write
det(A) = ad − bc,
and call this the determinant of the matrix A. 3.2. Determinants for Square Matrices of Higher Order
If we attempt to repeat the argument for 2 × 2 matrices to 3 × 3 matrices, then it is very likely that
we shall end up in a mess with possibly no ﬁrm conclusion. Try the argument on 4 × 4 matrices if you
must. Those who have their feet ﬁrmly on the ground will try a diﬀerent approach.
Our approach is inductive in nature. In other words, we shall deﬁne the determinant of 2× 2 matrices in
terms of determinants of 1 × 1 matrices, deﬁne the determinant of 3 × 3 matrices in terms of determinants
of 2 × 2 matrices, deﬁne the determinant of 4 × 4 matrices in terms of determinants of 3 × 3 matrices,
and so on.
Suppose now that we have deﬁned the determinant of (n − 1) × (n − 1) matrices. Let (4) a11
.
.
A=
.
an1 Chapter 3 : Determinants ... a1n
.
.
. . . . ann
page 3 of 23 c Linear Algebra W W L Chen, 1982, 2005 be an n × n matrix. For every i, j = 1, . . . , n, let us delete row i and column j of A to obtain the
(n − 1) × (n − 1) matrix (5) a11
.
.
. a(i−1)1 Aij = • a(i+1)1 .
.
.
an1 ... a1(j −1)
.
.
. . . . a(i−1)(j −1)
...
•
. . . a(i+1)(j −1)
.
.
.
...
an(j −1) •
.
.
.
•
•
•
.
.
.
• a1(j +1)
.
.
. ... a(i−1)(j +1)
•
a(i+1)(j +1)
.
.
.
an(j +1) ...
...
...
... a1n .
. . a(i−1)n • . a(i+1)n .
. .
ann Here • denotes that the entry has been deleted.
Definition. The number Cij = (−1)i+j det(Aij ) is called the cofactor of the entry aij of A. In other
words, the cofactor of the entry aij is obtained from A by ﬁrst deleting the row and the column containing
the entry aij , then calculating the determinant of the resulting (n − 1) × (n − 1) matrix, and ﬁnally
multiplying by a sign (−1)i+j .
Note that the entries of A in row i are given by
( ai1 ... ain ) . Definition. By the cofactor expansion of A by row i, we mean the expression
n (6) aij Cij = ai1 Ci1 + . . . + ain Cin .
j =1 Note that the entries of A in column j are given by a1j
. . .
.
anj Definition. By the cofactor expansion of A by column j , we mean the expression
n (7) aij Cij = a1j C1j + . . . + anj Cnj .
i=1 We shall state without proof the following important result. The interested reader is referred to
Section 3.8 for further discussion.
PROPOSITION 3A. Suppose that A is an n × n matrix given by (4). Then the expressions (6) and
(7) are all equal and independent of the row or column chosen.
Definition. Suppose that A is an n × n matrix given by (4). We call the common value in (6) and (7)
the determinant of the matrix A, denoted by det(A).
Chapter 3 : Determinants page 4 of 23 c Linear Algebra W W L Chen, 1982, 2005 Let us check whether this agrees with our earlier deﬁnition of the determinant of a 2 × 2 matrix.
Writing
a11
a21 A= a12
a22 , we have
C11 = a22 , C12 = −a21 , C21 = −a12 , C22 = a11 . It follows that by row 2 :
by column 1 : a11 C11 + a12 C12 = a11 a22 − a12 a21 ,
a21 C21 + a22 C22 = −a21 a12 + a22 a11 ,
a11 C11 + a21 C21 = a11 a22 − a21 a12 , by column 2 : a12 C12 + a22 C22 = −a12 a21 + a22 a11 . by row 1 : The four values are clearly equal, and of the form ad − bc as before.
Example 3.2.1. Consider the matrix 2
A = 1
2 3
4
1 5
2.
5 Let us use cofactor expansion by row 1. Then
C11 = (−1)1+1 det 4
1 2
5 = (−1)2 (20 − 2) = 18, C12 = (−1)1+2 det 1
2 2
5 = (−1)3 (5 − 4) = −1, C13 = (−1)1+3 det 1
2 4
1 = (−1)4 (1 − 8) = −7, so that
det(A) = a11 C11 + a12 C12 + a13 C13 = 36 − 3 − 35 = −2.
Alternatively, let us use cofactor expansion by column 2. Then
C12 = (−1)1+2 det 1
2 2
5 = (−1)3 (5 − 4) = −1, C22 = (−1)2+2 det 2
2 5
5 = (−1)4 (10 − 10) = 0, C32 = (−1)3+2 det 2
1 5
2 = (−1)5 (4 − 5) = 1, so that
det(A) = a12 C12 + a22 C22 + a32 C32 = −3 + 0 + 1 = −2. When using cofactor expansion, we should choose a row or column with as few nonzero entries as
possible in order to minimize the calculations.
Chapter 3 : Determinants page 5 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.2.2. Consider the matrix 2
1
A=
5
2 3
4
4
1 5
2
.
5
5 0
0
8
0 Here it is convenient to use cofactor expansion by column 3, since then det(A) = a13 C13 + a23 C23 + a33 C33 + a43 C43 = 8C33 2
= 8(−1)3+3 det 1
2 3
4
1 5
2 = −16,
5 in view of Example 3.2.1. 3.3. Some Simple Observations
In this section, we shall describe two simple observations which follow immediately from the deﬁnition
of the determinant by cofactor expansion.
PROPOSITION 3B. Suppose that a square matrix A has a zero row or a zero column. Then det(A) =
0.
Proof. We simply use cofactor expansion by the zero row or zero column.
Definition. Consider an n × n matrix a11
.
.
A=
. ... a1n
.
.
.
. an1 ... ann If aij = 0 whenever i > j , then A is called an upper triangular matrix. If aij = 0 whenever i < j , then
A is called a lower triangular matrix. We also say that A is a triangular matrix if it is upper triangular
or lower triangular.
Example 3.3.1. The matrix 1
0
0 3
5
6 2
4
0 is upper triangular.
Example 3.3.2. A diagonal matrix is both upper triangular and lower triangular.
PROPOSITION 3C. Suppose that the n × n matrix a11
.
A= .
. ... a1n
.
.
. an1 ... ann is triangular. Then det(A) = a11 a22 . . . ann , the product of the diagonal entries.
Chapter 3 : Determinants page 6 of 23 c Linear Algebra W W L Chen, 1982, 2005 Proof. Let us assume that A is upper triangular – for the case when A is lower triangular, change the
term “leftmost column” to the term “top row” in the proof. Using cofactor expansion by the leftmost
column at each step, we see that a22
.
det(A) = a11 det .
.
an2 ... a2n
a33
.
.
.
= a11 a22 det .
.
.
an3 . . . ann ... a3n
.
.
= . . . = a11 a22 . . . ann
. . . . ann as required. 3.4. Elementary Row Operations
We now study the eﬀect of elementary row operations on determinants. Recall that the elementary row
operations that we consider are: (1) interchanging two rows; (2) adding a multiple of one row to another
row; and (3) multiplying one row by a nonzero constant.
PROPOSITION 3D. (ELEMENTARY ROW OPERATIONS) Suppose that A is an n × n matrix.
(a) Suppose that the matrix B is obtained from the matrix A by interchanging two rows of A. Then
det(B ) = − det(A).
(b) Suppose that the matrix B is obtained from the matrix A by adding a multiple of one row of A to
another row. Then det(B ) = det(A).
(c) Suppose that the matrix B is obtained from the matrix A by multiplying one row of A by a nonzero
constant c. Then det(B ) = c det(A).
Sketch of Proof. (a) The proof is by induction on n. It is easily checked that the result holds when
n = 2. When n > 2, we use cofactor expansion by a third row, say row i. Then
n aij (−1)i+j det(Bij ). det(B ) =
j =1 Note that the (n − 1) × (n − 1) matrices Bij are obtained from the matrices Aij by interchanging two
rows of Aij , so that det(Bij ) = − det(Aij ). It follows that
n det(B ) = − aij (−1)i+j det(Aij ) = − det(A)
j =1 as required.
(b) Again, the proof is by induction on n. It is easily checked that the result holds when n = 2. When
n > 2, we use cofactor expansion by a third row, say row i. Then
n aij (−1)i+j det(Bij ). det(B ) =
j =1 Note that the (n − 1) × (n − 1) matrices Bij are obtained from the matrices Aij by adding a multiple of
one row of Aij to another row, so that det(Bij ) = det(Aij ). It follows that
n aij (−1)i+j det(Aij ) = det(A) det(B ) =
j =1 as required.
Chapter 3 : Determinants page 7 of 23 c Linear Algebra W W L Chen, 1982, 2005 (c) This is simpler. Suppose that the matrix B is obtained from the matrix A by multiplying row i of
A by a nonzero constant c. Then
n caij (−1)i+j det(Bij ). det(B ) =
j =1 Note now that Bij = Aij , since row i has been removed respectively from B and A. It follows that
n caij (−1)i+j det(Aij ) = c det(A) det(B ) =
j =1 as required.
In fact, the above operations can also be carried out on the columns of A. More precisely, we have
the following result.
PROPOSITION 3E. (ELEMENTARY COLUMN OPERATIONS) Suppose that A is an n × n matrix.
(a) Suppose that the matrix B is obtained from the matrix A by interchanging two columns of A. Then
det(B ) = − det(A).
(b) Suppose that the matrix B is obtained from the matrix A by adding a multiple of one column of A
to another column. Then det(B ) = det(A).
(c) Suppose that the matrix B is obtained from the matrix A by multiplying one column of A by a
nonzero constant c. Then det(B ) = c det(A).
Elementary row and column operations can be combined with cofactor expansion to calculate the
determinant of a given matrix. We shall illustrate this point by the following examples.
Example 3.4.1. Consider the matrix 2
1
A=
5
2 3
4
4
2 2
1
4
0 5
2
.
5
4 Adding −1 times column 3 to column 1, we have 3
4
4
2 2
1
4
0 5
2
.
5
4 3
4
3
2 0
0
det(A) = det 1
2 2
1
4
0 5
2
.
3
4 Adding −1/2 times row 4 to row 3, we have 0
0
det(A) = det 0
2
Using cofactor expansion by column 1, we have 3
det(A) = 2(−1)4+1 det 4
3
Adding −1 times row 1 to row 3, we have 2
1
4 5
3
2 = −2 det 4
3
3 3
det(A) = −2 det 4
0
Chapter 3 : Determinants 2
1
2 2
1
4 5
2.
3 5
2 .
−2
page 8 of 23 c Linear Algebra W W L Chen, 1982, 2005 Adding 1 times column 2 to column 3, we have 3
det(A) = −2 det 4
0 7
3.
0 2
1
2 Using cofactor expansion by row 3, we have
3
4 det(A) = −2 · 2(−1)3+2 det 7
3 = 4 det 3
4 7
3 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = 4(9 − 28) = −76.
Let us start again and try a diﬀerent way. Dividing row 4 by 2, we have 3
4
4
1 2
1
4
0 5
2
.
5
2 3
3
4
1 2
1
det(A) = 2 det 5
1 2
1
4
0 5
0
.
5
2 Adding −1 times row 4 to row 2, we have 2
0
det(A) = 2 det 5
1
Adding −3 times column 3 to column 2, we have 2
0
det(A) = 2 det 5
1 −3
0
−8
1 2
1
4
0 5
0
.
5
2 Using cofactor expansion by row 2, we have 2
det(A) = 2 · 1(−1)2+3 det 5
1 −3
−8
1 5
2
5 = −2 det 5
2
1 −3
−8
1 5
5.
2 Adding −2 times row 3 to row 1, we have −5
−8
1 0
det(A) = −2 det 5
1 1
5.
2 Adding −5 times row 3 to row 2, we have 0
det(A) = −2 det 0
1 −5
−13
1 1
−5 .
2 Using cofactor expansion by column 1, we have
det(A) = −2 · 1(−1)3+1 det −5
−13 1
−5 = −2 det −5
−13 1
−5 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = −2(25 + 13) = −76.
Chapter 3 : Determinants page 9 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.4.2. Consider the matrix 2
2 A = 4 1
2 1
3
7
0
1 0
1
2
1
0 3
5 7. 3
0 1
2
3
1
2 Here we have the least number of nonzero entries in column 3, so let us work to get more zeros into this
column. Adding −1 times row 4 to row 2, we have 1
3
7
0
1 0
0
2
1
0 1
1
3
1
2 3
2 7. 3
0 1
3
7
0
1 2
1 det(A) = det 4 1
2 0
0
0
1
0 1
1
1
1
2 3
2 1. 3
0 Adding −2 times row 4 to row 3, we have 2
1 det(A) = det 2 1
2
Using cofactor expansion by column 3, we have 2
1
4+3
det(A) = 1(−1)
det 2
2 1
3
7
1 3
2
2
1 = − det 1
2
0
2 1
1
1
2 1
3
7
1 1
1
1
2 3
2
.
1
0 Adding −1 times column 3 to column 1, we have 1
0
det(A) = − det 1
0 1
3
7
1 1
1
1
2 3
2
.
1
0 Adding −1 times row 1 to row 3, we have 1
0
det(A) = − det 0
0 1
3
6
1 1
1
0
2 3
2
.
−2
0 Using cofactor expansion by column 1, we have 3
det(A) = −1(−1)1+1 det 6
1 1
0
2 2
3
−2 = − det 6
0
1 1
0
2 2
−2 .
0 Adding 1 times row 1 to row 2, we have 3
det(A) = − det 9
1
Chapter 3 : Determinants 1
1
2 2
0.
0
page 10 of 23 c Linear Algebra W W L Chen, 1982, 2005 Using cofactor expansion by column 3, we have
9
1 det(A) = −2(−1)1+3 det 1
2 = −2 det 9
1 1
2 . Using the formula for the determinant of 2 × 2 matrices, we conclude that det(A) = −2(18 − 1) = −34.
Example 3.4.3. Consider the matrix 1
2 4
A=
3 2
1 0
4
6
5
4
0 2
5
1
0
5
2 4
7
9
1
3
5 1
6
2
2
6
1 Here note that rows 1 and 6 are almost identical. Adding 102
2 4 5 4 6 1
det(A) = det 3 5 0 245
000
Adding −1 times row 5 to row 2, we have 1
0 4
det(A) = det 3 2
0
Adding −4 times row 6 to row 2, we have 1
0 4
det(A) = det 3 2
0 0
2 1
.
5 2
0 −1 times row 1 to row 6, we have 410
7 6 2 9 2 1
.
1 2 5 362
100 0
0
6
5
4
0 2
0
1
0
5
0 4
4
9
1
3
1 1
0
2
2
6
0 0
0 1
.
5 2
0 0
0
6
5
4
0 2
0
1
0
5
0 4
0
9
1
3
1 1
0
2
2
6
0 0
0 1
.
5 2
0 It follows from Proposition 3B that det(A) = 0. 3.5. Further Properties of Determinants
Definition. Consider the n × n matrix a11
.
.
A=
. ... a1n
.
.
.
. an1 ... ann By the transpose At of A, we mean the matrix a11
t
.
.
A=
.
a1n ... an1
.
.
. ... ann obtained from A by transposing rows and columns.
Chapter 3 : Determinants page 11 of 23 c Linear Algebra Example 3.5.1. Consider the matrix 1
A = 4
7 2
5
8 3
6.
9 4
5
6 W W L Chen, 1982, 2005 7
8.
9 Then 1
At = 2
3 Recall that determinants of 2 × 2 matrices depend on determinants of 1 × 1 matrices; in turn, determinants of 3 × 3 matrices depend on determinants of 2 × 2 matrices, and so on. It follows that determinants
of n × n matrices ultimately depend on determinants of 1 × 1 matrices. Note now that transposing a
1 × 1 matrix does not aﬀect its determinant (why?). The result below follows in view of Proposition 3A.
PROPOSITION 3F. For every n × n matrix A, we have det(At ) = det(A).
Example 3.5.2. We have 2
1 det 0 1
3 2
3
1
2
5 4
7
2
3
7 1
0
1
1
3 2
2
1
2 0 = det 4 2
1
0
2 1
3
7
0
1 0
1
2
1
0 1
2
3
1
2 3
5 7 = −34. 3
0 Next, we shall study the determinant of a product. In Section 3.8, we shall sketch a proof of the
following important result.
PROPOSITION 3G. For every n × n matrices A and B , we have det(AB ) = det(A) det(B ).
PROPOSITION 3H. Suppose that the n × n matrix A is invertible. Then
det(A−1 ) = 1
.
det(A) Proof. In view of Propositions 3G and 3C, we have det(A) det(A−1 ) = det(In ) = 1. The result follows
immediately.
Finally, the main reason for studying determinants, as outlined in the introduction, is summarized by
the following result.
PROPOSITION 3J. Suppose that A is an n × n matrix. Then A is invertible if and only if det(A) = 0.
Proof. Suppose that A is invertible. Then det(A) = 0 follows immediately from Proposition 3H.
Suppose now that det(A) = 0. Let us now reduce A by elementary row operations to reduced row
echelon form B . Then there exist a ﬁnite sequence E1 , . . . , Ek of elementary n × n matrices such that
B = Ek . . . E1 A.
It follows from Proposition 3G that
det(B ) = det(Ek ) . . . det(E1 ) det(A).
Chapter 3 : Determinants page 12 of 23 c Linear Algebra W W L Chen, 1982, 2005 Recall that all elementary matrices are invertible and so have nonzero determinants. It follows that
det(B ) = 0, so that B has no zero rows by Proposition 3B. Since B is an n × n matrix in reduced row
echelon form, it must be In . We therefore conclude that A is row equivalent to In . It now follows from
Proposition 2N(c) that A is invertible.
Combining Propositions 2Q and 3J, we have the following result.
PROPOSITION 3K. In the notation of Proposition 2N, the following statements are equivalent:
(a) The matrix A is invertible.
(b) The system Ax = 0 of linear equations has only the trivial solution.
(c) The matrices A and In are row equivalent.
(d) The system Ax = b of linear equations is soluble for every n × 1 matrix b.
(e) The determinant det(A) = 0. 3.6. Application to Curves and Surfaces
A special case of Proposition 3K states that a homogeneous system of n linear equations in n variables
has a nontrivial solution if and only if the determinant if the coeﬃcient matrix is equal to zero. In this
section, we shall use this to solve some problems in geometry. We illustrate our ideas by a few simple
examples.
Example 3.6.1. Suppose that we wish to determine the equation of the unique line on the xy plane that
passes through two distinct given points (x1 , y1 ) and (x2 , y2 ). The equation of a line on the xy plane is
of the form ax + by + c = 0. Since the two points lie on the line, we must have ax1 + by1 + c = 0 and
ax2 + by2 + c = 0. Hence
xa + yb + c = 0,
x1 a + y1 b + c = 0,
x2 a + y2 b + c = 0.
Written in matrix notation, we have x x1
x2 y
y1
y2 1
a
0
1b = 0.
1
c
0 Clearly there is a nontrivial solution (a, b, c) to this system of linear equations, and so we must have xy1
det x1 y1 1 = 0,
x2 y2 1
the equation of the line required.
Example 3.6.2. Suppose that we wish to determine the equation of the unique circle on the xy plane
that passes through three distinct given points (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ), not all lying on a straight
line. The equation of a circle on the xy plane is of the form a(x2 + y 2 ) + bx + cy + d = 0. Since the three
2
2
points lie on the circle, we must have a(x2 + y1 ) + bx1 + cy1 + d = 0, a(x2 + y2 ) + bx2 + cy2 + d = 0, and
1
2
2
2
a(x3 + y3 ) + bx3 + cy3 + d = 0. Hence
(x2 + y 2 )a + xb + yc + d = 0,
2
(x2 + y1 )a + x1 b + y1 c + d = 0,
1
2
(x2 + y2 )a + x2 b + y2 c + d = 0,
2
2
(x2 + y3 )a + x3 b + y3 c + d = 0.
3
Chapter 3 : Determinants page 13 of 23 c Linear Algebra W W L Chen, 1982, 2005 Written in matrix notation, we have x2 + y 2
2 x2 + y1
1
2
2
x2 + y2
2
2
x3 + y3 x
x1
x2
x3 y
y1
y2
y3 1
a
0
1b 0 = .
c
0
1
1
d
0 Clearly there is a nontrivial solution (a, b, c, d) to this system of linear equations, and so we must have x2 + y 2
2 x2 + y1
det 1
2
2
x2 + y2
2
2
x3 + y3 x
x1
x2
x3 y
y1
y2
y3 1
1 = 0,
1
1 the equation of the circle required.
Example 3.6.3. Suppose that we wish to determine the equation of the unique plane in 3space that
passes through three distinct given points (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ), not all lying on a
straight line. The equation of a plane in 3space is of the form ax + by + cz + d = 0. Since the
three points lie on the plane, we must have ax1 + by1 + cz1 + d = 0, ax2 + by2 + cz2 + d = 0, and
ax3 + by3 + cz3 + d = 0. Hence
xa + yb + zc + d = 0,
x1 a + y1 b + z1 c + d = 0,
x2 a + y2 b + z2 c + d = 0,
x3 a + y3 b + z3 c + d = 0.
Written in matrix notation, we have x x1 x2
x3 y
y1
y2
y3 z
z1
z2
z3 1
a
0
1b 0 = .
1
c
0
1
d
0 Clearly there is a nontrivial solution (a, b, c, d) to this system of linear equations, and so we must have x x1
det x2
x3 y
y1
y2
y3 z
z1
z2
z3 1
1 = 0,
1
1 the equation of the plane required.
Example 3.6.4. Suppose that we wish to determine the equation of the unique sphere in 3space that
passes through four distinct given points (x1 , y1 , z1 ), (x2 , y2 , z2 ), (x3 , y3 , z3 ) and (x4 , y4 , z4 ), not all lying
on a plane. The equation of a sphere in 3space is of the form a(x2 + y 2 + z 2 ) + bx + cy + dz + e = 0.
Since the four points lie on the sphere, we must have
2
2
a(x2 + y1 + z1 ) + bx1 + cy1 + dz1 + e = 0,
1
2
2
a(x2 + y2 + z2 ) + bx2 + cy2 + dz2 + e = 0,
2
2
2
a(x2 + y3 + z3 ) + bx3 + cy3 + dz3 + e = 0,
3 and
2
2
a(x2 + y4 + z4 ) + bx4 + cy4 + dz4 + e = 0.
4
Chapter 3 : Determinants page 14 of 23 c Linear Algebra W W L Chen, 1982, 2005 Hence
(x2 + y 2 + z 2 )a + xb + yc + zd + e = 0,
2
2
(x2 + y1 + z1 )a + x1 b + y1 c + z1 d + e = 0,
1
2
2
(x2 + y2 + z2 )a + x2 b + y2 c + z2 d + e = 0,
2
2
2
(x2 + y3 + z3 )a + x3 b + y3 c + z3 d + e = 0,
3
2
2
(x2 + y4 + z4 )a + x4 b + y4 c + z4 d + e = 0.
4 Written in matrix notation, we have x2 + y 2 + z 2
2
2 x2 + y1 + z1
1
2
2
2 x2 + y2 + z2
2
2
2
x3 + y3 + z3
2
2
2
x4 + y4 + z4 x
x1
x2
x3
x4 y
y1
y2
y3
y4 z
z1
z2
z3
z4 1
a
0
1b 0 1c = 0. d
0
1
1
e
0 Clearly there is a nontrivial solution (a, b, c, d, e) to this system of linear equations, and so we must have x2 + y 2 + z 2
2
2 x2 + y1 + z1
1
2
2
2
det x2 + y2 + z2
2
2
2
x3 + y3 + z3
2
2
2
x4 + y4 + z4 x
x1
x2
x3
x4 y
y1
y2
y3
y4 z
z1
z2
z3
z4 1
1 1 = 0, 1
1 the equation of the sphere required. 3.7. Some Useful Formulas
In this section, we shall discuss two very useful formulas which involve determinants only. The ﬁrst one
enables us to ﬁnd the inverse of a matrix, while the second one enables us to solve a system of linear
equations. The interested reader is referred to Section 3.8 for proofs.
Recall ﬁrst of all that for any n × n matrix a11
.
.
A=
. ... a1n
.
.
,
. an1 ... ann the number Cij = (−1)i+j det(Aij ) is called the cofactor of the entry aij , and the (n − 1) × (n − 1) matrix ...
a1(j −1)
•
a1(j +1)
...
a1n
a11 .
.
.
.
.
.
.
.
. .
.
.
.
. . a(i−1)1 . . . a(i−1)(j −1) • a(i−1)(j +1) . . . a(i−1)n Aij = •
...
•
•
•
...
• a(i+1)1 . . . a(i+1)(j −1) • a(i+1)(j +1) . . . a(i+1)n .
.
.
.
.
.
.
.
. .
.
.
.
.
.
an1 ... an(j −1) • an(j +1) ... ann is obtained from A by deleting row i and column j ; here • denotes that the entry has been deleted.
Chapter 3 : Determinants page 15 of 23 c Linear Algebra Definition. The n × n matrix Cn1
.
.
. ... C11
.
adj(A) = .
.
C1n W W L Chen, 1982, 2005 . . . Cnn is called the adjoint of the matrix A.
Remark. Note that adj(A) is obtained from the matrix A ﬁrst by replacing each entry of A by its
cofactor and then by transposing the resulting matrix.
PROPOSITION 3L. Suppose that the n × n matrix A is invertible. Then
1
adj(A).
det(A) A−1 = Example 3.7.1. Consider the matrix 1
A = 0
2 0
2.
3 −1
1
0 Then 1 det 0 0
adj(A) = − det 2 0
det
2 2
3
2
3
1
0 On the other hand, adding 1 times
we have 1
det(A) = det 0
2 −1 0
03
10
det
23
1 −1
− det
20 −1 0
12
10
− det
02
1 −1
det
01 − det det 3 = 4 −2 3
3
−2 −2
−2 .
1 column 1 to column 2 and then using cofactor expansion on row 1, −1 0
1
1 2 = det 0
03
2 0
1
2 0
2 = det
3 1
2 2
3 = −1. It follows that A−1 −3
= −4
2 −3
−3
2 2
2 .
−1 Next, we turn our attention to systems of n linear equations in n unknowns, of the form
a11 x1 + . . . + a1n xn = b1 ,
.
.
.
an1 x1 + . . . + ann xn = bn ,
represented in matrix notation in the form
Ax = b,
Chapter 3 : Determinants page 16 of 23 c Linear Algebra W W L Chen, 1982, 2005 where a11
.
.
A=
. (8) an1 a1n
.
.
. ... b1
.
b= . . and bn . . . ann represent the coeﬃcients and x1
.
x= .
. (9) xn
represents the variables.
For every j = 1, . . . , k , write a11
.
Aj (b) = .
.
an1 (10) ... a1(j −1)
.
.
.
an(j −1) ... b1
.
.
.
bn a1(j +1)
.
.
.
an(j +1) ...
... a1n
. ;
.
.
ann in other words, we replace column j of the matrix A by the column b.
PROPOSITION 3M. (CRAMER’S RULE) Suppose that the matrix A is invertible. Then the unique
solution of the system Ax = b, where A, x and b are given by (8) and (9), is given by
x1 = det(A1 (b))
,
det(A) xn = ..., det(An (b))
,
det(A) where the matrices A1 (b), . . . , A1 (b) are deﬁned by (10).
Example 3.7.2. Consider the system Ax = b, where 1
A = 0
2 0
2
3 −1
1
0 1
b = 2.
3 and Recall that det(A) = −1. By Cramer’s rule, we have 1 −1 0
det 2 1 2 303
x1 =
= −3,
det(A) 110
det 0 2 2 233
x2 =
= −4,
det(A) 1 −1 1
det 0 1 2 203
x3 =
= 3.
det(A) Let us check our calculations. Recall from Example 3.7.1 that A−1 −3
= −4
2 −3
−3
2 2
2 .
−1 We therefore have x1
−3 x2 = −4
2
x3
Chapter 3 : Determinants −3
−3
2 2
1
−3
2 2 = −4 .
−1
3
3
page 17 of 23 c Linear Algebra W W L Chen, 1982, 2005 3.8. Further Discussion
In this section, we shall ﬁrst discuss a deﬁnition of the determinant in terms of permutations. In order
to do so, we need to make a digression and discuss ﬁrst the rudiments of permutations on nonempty
ﬁnite sets.
Definition. Let X be a nonempty ﬁnite set. A permutation φ on X is a function φ : X → X which is
onetoone and onto. If x ∈ X , we denote by xφ the image of x under the permutation φ.
It is not diﬃcult to see that if φ : X → X and ψ : X → X are both permutations on X , then
φψ : X → X , deﬁned by xφψ = (xφ)ψ for every x ∈ X so that φ is followed by ψ , is also a permutation
on X .
Remark. Note that we use the notation xφ instead of our usual notation φ(x) to denote the image
of x under φ. Note also that we write φψ to denote the composition ψ ◦ φ. We shall do this only for
permutations. The reasons will become a little clearer later in the discussion.
Since the set X is nonempty and ﬁnite, we may assume, without loss of generality, that it is
{1, 2, . . . , n}, where n ∈ N. We now let Sn denote the set of all permutations on the set {1, 2, . . . , n}. In
other words, Sn denotes the collection of all functions from {1, 2, . . . , n} to {1, 2, . . . , n} that are both
onetoone and onto.
PROPOSITION 3N. For every n ∈ N, the set Sn has n! elements.
Proof. There are n choices for 1φ. For each such choice, there are (n − 1) choices left for 2φ. And so
on.
To represent particular elements of Sn , there are various notations. For example, we can use the
notation
1
2 ... n
1φ 2φ . . . nφ
to denote the permutation φ.
Example 3.8.1. In S4 ,
1
2 2
4 3
1 4
3 denotes the permutation φ, where 1φ = 2, 2φ = 4, 3φ = 1 and 4φ = 3. On the other hand, the reader
can easily check that
1
2 2
4 3
1 4
3 1
3 2
2 3
4 4
1 1
2 = 2
1 3
3 4
4 . A more convenient way is to use the cycle notation. The permutations
1
2 2
4 3
1 4
3 1
3 and 2
2 3
4 4
1 can be represented respectively by the cycles (1 2 4 3) and (1 3 4). Here the cycle (1 2 4 3) gives the
information 1φ = 2, 2φ = 4, 4φ = 3 and 3φ = 1. Note also that in the latter case, since the image of 2
is 2, it is not necessary to include this in the cycle. Furthermore, the information
1
2 2
4 3
1 4
3 1
3 2
2 3
4 4
1 = 1
2 2
1 3
3 4
4 can be represented in cycle notation by (1 2 4 3)(1 3 4) = (1 2). We also say that the cycles (1 2 4 3),
(1 3 4) and (1 2) have lengths 4, 3 and 2 respectively.
Chapter 3 : Determinants page 18 of 23 c Linear Algebra W W L Chen, 1982, 2005 Example 3.8.2. In S6 , the permutation
1
2 2
4 3
1 4
3 5
6 6
5 can be represented in cycle notation as (1 2 4 3)(5 6).
Example 3.8.3. In S4 or S6 , we have (1 2 4 3) = (1 2)(1 4)(1 3).
The last example motivates the following important idea.
Definition. Suppose that n ∈ N. A permutation in Sn that interchanges two numbers among the
elements of {1, 2, . . . , n} and leaves all the others unchanged is called a transposition.
Remark. It is obvious that a transposition can be represented by a 2cycle, and is its own inverse.
Definition. Two cycles (x1 x2 . . . xk ) and (y1 y2 . . . yl ) in Sn are said to be disjoint if the elements
x1 , . . . , xk , y1 , . . . , yl are all diﬀerent.
The interested reader may try to prove the following result.
PROPOSITION 3P. Suppose that n ∈ N.
(a) Every permutation in Sn can be written as a product of disjoint cycles.
(b) For every subset {x1 , x2 , . . . , xk } of the set {1, 2, . . . , n}, where the elements x1 , x2 , . . . , xk are distinct, the cycle (x1 x2 . . . xk ) satisﬁes
(x1 x2 . . . xk ) = (x1 x2 )(x1 x3 ) . . . (x1 xk );
in other words, every cycle can be written as a product of transpositions.
(c) Consequently, every permutation in Sn can be written as a product of transpositions.
Example 3.8.4. In S9 , the permutation
1
3 2
2 3
5 4
1 5
7 6
8 7
4 8
9 9
6 can be written in cycle notation as (1 3 5 7 4)(6 8 9). By Theorem 3P(b), we have
(1 3 5 7 4) = (1 3)(1 5)(1 7)(1 4) and (6 8 9) = (6 8)(6 9). Hence the permutation can be represented by (1 3)(1 5)(1 7)(1 4)(6 8)(6 9).
Definition. Suppose that n ∈ N. Then a permutation in Sn is said to be even if it is representable as
the product of an even number of transpositions and odd if it is representable as the product of an odd
number of transpositions. Furthermore, we write
(φ) = +1 if φ is even,
−1 if φ is odd. Remark. It can be shown that no permutation can be simultaneously odd and even.
Chapter 3 : Determinants page 19 of 23 c Linear Algebra W W L Chen, 1982, 2005 We are now in a position to deﬁne the determinant of a matrix. Suppose that a11 . . . a1n
.
.
.
(11)
A= .
.
.
an1 . . . ann
is an n × n matrix.
Definition. By an elementary product from the matrix A, we mean the product of n entries of A, no
two of which are from the same row or same column.
It follows that any such elementary product must be of the form
a1(1φ) a2(2φ) . . . an(nφ) ,
where φ is a permutation in Sn .
Definition. By the determinant of an n × n matrix A of the form (11), we mean the sum
(12) det(A) = (φ)a1(1φ) a2(2φ) . . . an(nφ) ,
φ∈Sn where the summation is over all the n! permutations φ in Sn .
It is be shown that the determinant deﬁned in this way is the same as that deﬁned earlier by row or
column expansions. Indeed, one can use (12) to establish Proposition 3A. The very interested reader
may wish to make an attempt. Here we conﬁne our study to the special cases when n = 2 and n = 3.
In the two examples below, we use e to denote the identity permutation.
Example 3.8.5. Suppose that n = 2. We have the following:
elementary product permutation sign contribution a11 a22
a12 a21 e
(1 2) +1
−1 +a11 a22
−a12 a21 Hence det(A) = a11 a22 − a12 a21 as shown before.
Example 3.8.6. Suppose that n = 3. We have the following:
elementary product permutation sign contribution a11 a22 a33
a12 a23 a31
a13 a21 a32
a13 a22 a31
a11 a23 a32
a12 a21 a33 e
(1 2 3)
(1 3 2)
(1 3)
(2 3)
(1 2) +1
+1
+1
−1
−1
−1 +a11 a22 a33
+a12 a23 a31
+a13 a21 a32
−a13 a22 a31
−a11 a23 a32
−a12 a21 a33 Hence det(A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a11 a23 a32 − a12 a21 a33 . We have the
picture below:
+D Chapter 3 : Determinants D +E
D D
a11D E +E
E E
a12
E E E Ey y
a13
y EE y y −
y y y − ay
az
y 11
z 12
Ey
y
z
D
E
E
D yy E yy E zz
a21
a22
a23
aE
a22
yD D
yE E
z 21 E
y
y
z
E
y
yD D
zE E
E
y
z
y
a31
a32
a33
a31
a32
D z z z − page 20 of 23 c Linear Algebra W W L Chen, 1982, 2005 Next, we discuss brieﬂy how one may prove Proposition 3G concerning the determinant of the product
of two matrices. The idea is to use elementary matrices. Corresponding to Proposition 3D, we can easily
establish the following result.
PROPOSITION 3Q. Suppose that E is an elementary matrix.
(a) If E arises from interchanging two rows of In , then det(E ) = −1.
(b) If E arises from adding one row of In to another row, then det(E ) = 1.
(c) If E arises from multiplying one row of In by a nonzero constant c, then det(E ) = c.
Combining Propositions 3D and 3Q, we can establish the following intermediate result.
PROPOSITION 3R. Suppose that E is an n × n elementary matrix. Then for any n × n matrix B ,
we have det(EB ) = det(E ) det(B ).
Proof of Proposition 3G. Let us reduce A by elementary row operations to reduced row echelon form
A . Then there exist a ﬁnite sequence G1 , . . . , Gk of elementary matrices such that A = Gk . . . G1 A.
Since elementary matrices are invertible with elementary inverse matrices,it follows that there exist a
ﬁnite sequence E1 , . . . , Ek of elementary matrices such that
(13) A = E1 . . . Ek A . Suppose ﬁrst of all that det(A) = 0. Then it follows from (13) that the matrix A must have a zero row.
Hence A B must have a zero row, and so det(A B ) = 0. But AB = E1 . . . Ek (A B ), so it follows from
Proposition 3R that det(AB ) = 0. Suppose next that det(A) = 0. Then A = In , and so it follows from
(13) that AB = E1 . . . Ek B . The result now follows on applying Proposition 3R.
We complete this chapter by establishing the two formulas discussed in Section 3.7.
Proof of Proposition 3L. It suﬃces to show that
(14) A adj(A) = det(A)In , as this clearly implies
A
giving the result. To show (14), note that a11
.
.
(15)
A adj(A) =
.
an1 1
adj(A)
det(A) = In , ... C11
a1n
. .
.
.
.
. ... ann Suppose that the right hand side of (15) is equal to b11 . . .
.
.
B=
.
bn 1 . . . C1n ... Cn1
.
.
.
. . . . Cnn b1n
.
.
.
.
bnn Then for every i, j = 1, . . . , n, we have
(16) bij = ai1 Cj 1 + . . . + ain Cjn . It follows that when i = j , we have
bii = ai1 Ci1 + . . . + ain Cin = det(A).
On the other hand, if i = j , then (16) is equal to the determinant of the matrix obtained from A by
replacing row j by row i. This matrix has therefore two identical rows, and so the determinant is 0
(why?). The identity (14) follows immediately.
Chapter 3 : Determinants page 21 of 23 c Linear Algebra W W L Chen, 1982, 2005 Proof of Proposition 3M. Since A is invertible, it follows from Proposition 3L that
A−1 = 1
adj(A).
det(A) By Proposition 2P, the unique solution of the system Ax = b is given by
x = A−1 b =
Written in full, this becomes x1
C11
1 . . =
.
.
.
.
det(A)
xn
C1n ... 1
adj(A)b.
det(A) b1
Cn1
. . .
.
=
.
.
bn . . . Cnn b1 C11 + . . . + bn Cn1
.
. .
. 1
det(A) b1 C1n + . . . + bn Cnn Hence, for every j = 1, . . . , n, we have
xj = b1 C1j + . . . + bn Cnj
.
det(A) To complete the proof, it remains to show that
b1 C1j + . . . + bn Cnj = det(Aj (b)).
Note, on using cofactor expansion by column j , a11
.
.
. a(i−1)1
n i+j
det(Aj (b)) =
bi (−1) det • a(i+1)1
i=1 .
.
. that an1
n a1(j −1)
.
.
. . . . a(i−1)(j −1)
...
•
. . . a(i+1)(j −1)
.
.
.
...
an(j −1) •
.
.
.
•
•
•
.
.
.
• a1(j +1)
.
.
. ... a(i−1)(j +1)
•
a(i+1)(j +1)
.
.
.
an(j +1) ...
...
...
... a1n .
. . a(i−1)n • a(i+1)n .
. .
ann n bi (−1)i+j det(Aij ) = = ... i=1 bi Cij
i=1 as required. Problems for Chapter 3
1. Compute the determinant of each of the matrices in Problem 2.6.
2. Find the determinant 1
P = 8
2 of each of the following matrices: 32
1
1 −1
4 0,
Q = 1 −1 1 ,
12
−1 1
1 a
R =b
c a2
b2
c2 a3
b3 .
c3 3. Find the determinant of the matrix 3
1 2
7
Chapter 3 : Determinants 4
0
3
2 5
1
6
9 2
0
.
3
4
page 22 of 23 c Linear Algebra W W L Chen, 1982, 2005 4. By using suitable elementary row and column operations as well as row and column expansions,
show that 2
2 det 2 4
5 3
3
3
6
8 7
7
6
2
7 1
1
1
3
4 3
5 9 = 2. 4
5 [Remark: Note that rows 1 and 2 of the matrix are almost identical.]
5. By using suitable elementary row and column operations as well as row and column expansions,
show that 2
2 det 4 4
2 1
1
3
3
1 5
5
2
2
6 1
1
1
0
π 3
2 1 = 2. 1
7 [Remark: The entry π is not a misprint!]
6. If A and B are square matrices of the same size and det A = 2 and det B = 3, ﬁnd det(A2 B −1 ).
7. a) Compute the Vandermonde determinants 1 a a2
det 1 b b2
and
1 c c2 1a
1 b
det 1c
1d a2
b2
c2
d2 a3
b3 .
c3
d3 b) Establish a formula for the Vandermonde determinant
1 a
a2 . . . an−1 1
1
1
n−1 1 a2 a2 . . . a2 2
det .
.
.
. .
.
.
.
.
.
.
.
.
1 an a2
n ... an−1
n 8. Compute the determinant a
b
c
det a + x b + x c + x .
a+y b+y c+y Harder Problems for Chapter 3
9. For each of the matrices below, compute its adjoint and use Proposition 3L to calculate its inverse: 113
354
a) 2 −2 1 b) 2 1 1 010
101
10. Use Cramer’s rule to solve the system of linear equations
2x1 + x2 + x3 = 4,
−x1
+ 2x3 = 2,
3x1 + x2 + 3x3 = −2.
Chapter 3 : Determinants page 23 of 23 ...
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