Chapter III
Determinants of Square Matrices
Associated with every square matrix A is a scalar called the determinant of the matrix.
Notation: Determinant of A is denoted by either det(A) or A .
1) Definition of the Determinant:
We can define the determinant iteratively as follows:
If A is the 1by1 matrix A  a11 , where a is a scalar, then A  a.
Example 1: Find det(A) if :
(i) A  3
(ii) A   5
(iii) A  0.
Solution:
(i) A  3
(ii) A  5
(iii) A  0.
Note that the determinant of a square matrix can be positive, negative or zero.
a b 
If A is the 2by2 matrix A  
 , then A  ad  bc.
 c d  22
Example 2: Find A if:
1 2
(i) A  

3 4
2 1 
(ii) A  

 3 4
1 2 
(iii) A  
.
 2 4
Solution:
(i) A  1(4)  2(3)  2
(ii) A  2(4)  1(3)  5
(iii) A  1(4)  2(2)  0.
 a11 a12 a13 
If A is the 3by3 matrix A  a21 a22 a23  , then
a31 a32 a33 
A  (a11a22a33  a12a23a31  a13a32a21)  (a11a23a32  a22a13a31  a33  a12  a21).
1 2 3
Example 3: Find A if: A  4 5 6.
7 8 9
Solution: Applying the above short-cut we get A  0.
2) The Determinant of an n by n Matrix
Let
 a11
a 21

 .
A
 ai1
 .

 an1
... a1n 
a 22 ... a 2 j ... a 2n 
.
.
.
.
. 
.
ai 2 ... aij ... ain 
.
.
.
.
. 

an2 ... anj ... ann 
a12 ... a1 j
Minors:
The Minor M ij is the (n-1) by (n-1) sub-matrix obtained by removing the ith row and jth
column of A. Note that there are n 2 minors for an n by n matrix.
Cofactors:
A cofactor of the element aij is denoted by Cij and defined as: Cij  (1) i  j M ij .
Cofactor Expansion:
To find det(A), we find the cofactor expansion by any row or by any column.
If we expand by the ith row, then:
n
A   aijCij  ai1Ci1  ai 2Ci 2  ...  ainCin =
j 1
n
 (1)
i j
j 1
aij M ij .
If we expand by the jth column, then:
n
A   aijCij  a1 j C1 j  a2 j C2 j  ...  anjCnj =
i 1
n
 (1)
i 1
i j
aij M ij .
1 2 3 
Example 4: For A  4 5 6 , find all of its minors and all of its cofactors. Hence, evaluate
7 8 9
det(A) using cofactor expansion of a row or column of your choice.
Solution: We can expand by any row or any column. Suppose we expand by the first column.
Then:
Cofactor of element 1 is: (1)11
5 6
 3.
8 9
Cofactor of element 4 is: (1) 21
2 3
 6.
8 9
Cofactor of element 7 is: (1) 31
2 3
 3.
5 6
Thus: det(A) = 1(cofactor of 1) + 4(cofactor of 4) + 7(cofactor of 7) = 1(-3)+4(6)+7(-3) = 0.
Note: In order to minimize the number of calculations in computing the determinant, we expand
by the row or column with the largest number of zeroes. The value of the determinant is the same
regardless of the row or column by which we expand.
3) Properties of the Determinant:
If a square matrix is large then calculation of its determinant is time consuming. Therefore, we
try to take advantage of the structure of the matrix and properties of the determinant to minimize
the computations.
Group 1:
(1)
A  AT .
Illustration:
a b 
A
 ; A  ad  bc.
c d 
a c 
T
AT  
 ; A  ad  bc.
b
d


1 2 3
Example 5: In Example 4, we had: A  4 5 6 , A  0. With your knowledge that A  0 ,
7 8 9
find determinant of B, without further calculations, if
1 4 7 
B  2 5 8 .
3 6 9
Solution: Since B  AT , B  AT  A  0.
(2)
AB  A B .
Illustration:
a b 
A
 ; A  ad  bc.
c d 
e f 
B
 ; B  eh  fg.
g h 
A B  (ad  bc)(eh  fg)  adeh  bcfg  bceh  adfg
a b   e
AB  

c d   g
f  ae  bg

h  ce  dg
af  bh
.
cf  dh 
AB  (ae  bg )(cf  dh)  (af  bh)(ce  dg )
 aecf  aedh  bgcf  bgdh  afce  afdg  bhce  bhdg  adeh  bcfg  bceh  adfg.
Clearly, AB  A B .
 1 4 7
1 4 7 


Example 6: Let A   9 5 0 , B  2 5 8 . Find AB .
 3 6 9
3 6 9
Solution: AB  A B  A (0)  0. (Note that B  0 from Example 5).
(3)
If A is a diagonal matrix, det(A) is the product of the diagonal elements.
Illustration:
a 0 
A
 ; A  ad  0  ad .
0 d 
 a 0 0
b 0
A  0 b 0; A  a
 0  0  abc.
0 c
0 0 c 
For a general n by n diagonal matrix, expand by the first row to get the desired result.
Example 7: Find det(A) and det(B) if:
10 0 0
  1 0 0
A   0 5 0 , B   0 5 0 .
 0 0 0
 0 0 9
Solution: det(A) = (-1)(5)(9)=-45.
det(B) = 10(5)(0) = 0.
Note that if a diagonal matrix has a zero as a diagonal element, its determinant is zero.
(4)
I  1.
Using Result 3, above, we see that:
1 0
 1(1)  1
0 1
1 0 0
0 1 0  1(1)(1)  1
0 0 1
For an n by n Identity matrix, I n  1.
(5)
If A is an upper triangular matrix (or a lower triangular matrix), det(A) is the product of
the diagonal elements.
Illustration:
a 0
 ac  0b  ac.
b c
a d
 ac  0d  ac.
0 c
For a general n by n upper triangular matrix, we expand by the first column to establish the
result.
For a general n by n lower triangular matrix, we expand by the first row to establish the result.
Example 8: Find det(A) if:
7 100 9 
0 
1) A  0 2
0 0  1
Solution: Since A is upper triangular, det(A) = 7(2)(-1)= -14.
3 0 0 0 0 
0 4 0 0 0 


2) 3 0 6 0 0


3 0 0 2 0 
3 5 8 8 2
Solution: Since A is lower triangular, det(A) = 3(4)(6)(2)(2)= 288.
Group 2:
(1)
(2)
If A has a row consisting entirely of zeroes, then det(A) = 0.
If A has a column consisting entirely of zeroes, then det(A) = 0.
Illustration:
a b
Let A  d e
 0 0
c
f  . Expanding by Row 3, we get: det(A)=0.
0 
a
Let B   c
 e
(3)
(4)
b
d
f
If A has two rows that are the same, then det(A) = 0.
If A has two columns that are the same, then det(A) = 0.
Illustration:
a b
Let A  d e
 a b
a b
Let B   c d
 e f
(5)
(6)
0
0 . Expanding by Column 3, we get: det(B)=0.
0
c
f  . Expanding by Row 2, we get: det(A)=0.
c 
a
c  . Expanding by Column 2, we get: det(B)=0.
e 
If A has two rows that are proportional to each other, then det(A) = 0.
If A has two columns that are proportional to each other, then det(A) = 0.
Illustration:
a b c
e
f  . Expanding by Row 2, we get: det(A)=0.
Let A   d
2a 2b 2c 
 a b 2a 
Let B   c d 2c  . Expanding by Column 2, we get: det(B)=0.
 e f 2e 
(7)
(8)
If a diagonal matrix (or an upper triangular or a lower triangular) has a zero element
along the diagonal, the determinant is zero.
A matrix is SINGULAR iff its determinant is zero (that is, if the determinant of a
matrix is zero then the matrix is singular, and if the matrix is singular then its
determinant is zero).
2
0

Example 9: Decide if the matrix A  0

0
0
3  1 0 4
5 5  1 2
0 0
1 1  is singular or non-singular.

0 0  7 6
0 0
0 7 
Solution: Since A is upper triangular, its determinant is the product of diagonal elements.
Since there is a zero element along the diagonal, det(A)=0.
Since det(A) = 0, then A is singular.
1 0 0 
Example 10: Find det(A) if A  0 0  1 .
1  1 1 
Solution: Expand by Row1 to get: det(A) = -1.
Note that although there is a zero element along the diagonal of A, its determinant is not zero. In
other words, property 7, above, does not apply in this case since the matrix is neither diagonal
nor upper/lower triangular.
Group 3:
(1)
If two rows in a matrix are interchanged, the determinant changes sign.
Illustration:
a b 
Let A  
 . Then A  ad  bc.
c d 
Now, let B be the matrix obtained by interchanging Rows 1 and 2 of matrix A. Then,
c d 
B
 and B  bc  ad  (ad  bc)   A .
a b 
b
If matrix C is obtained by interchanging Columns 1 and 2 or matrix A, then C  
d
C  bc  ad  (ad  bc)   A .
a
and
c 
(2)
If a row (or a column) in a matrix is multiplied by a constant, say α, then
the determinant is multiplied by α.
Illustration:
a b 
Let A  
 . Then A  ad  bc.
c d 
Now, let B be the matrix obtained by multiplying Row 1 of matrix A by 7. Then,
7a 7b
B
 and B  7ad  7bc  7(ad  bc)  7 A .
c d
If C is the matrix obtained by multiplying Column 2 of matrix A by -3. Then,
a  3b 
C
 and C  3ad  3bc  3(ad  bc)  3 A .
 c  3d 
(3)
A   n A if A is an n by n matrix.
Note that multiplying a matrix by a constant translates into multiplying each row (or each
column) by the same constant. If a matrix has n rows (or n columns) and the constant
multiplies each row, then by property 2 the determinant is multiplied by the same constant
“n” times.
Illustration:
a b 
Let A  
 . Then A  ad  bc.
c d 
5a 5b 
2
Now, let B = 5A. Then, B  
 , and det(B) = 25ad-25bc = 25det(A) = 5 A .
5
c
5
d


(4)
If a constant multiple of one row in a matrix is added to another row, the determinant
remains the same.
Illustration:
a b 
Let A  
 . Then A  ad  bc.
c d 
Let matrix B be obtained by adding 3 times row 1 to row 2, then:
b 
 a
B
 . Then B  a(d  3b)  b(c  3a)  ad  bc  A .
c  3a d  3b
1 0 0 
Example11: In Example 10, we had A  0 0  1 and det(A) = -1.
1  1 1 
8 0
(a) With your knowledge of det(A), what is det(B) if B  0 0
7  7
(b) With your knowledge of det(A), what is det(B) if B  5 A ?
1 0
(c) With your knowledge of det(A), what is det(B) if B  3 0
1  1
0
 4 ?
7 
0
 1 ?
1 
0 0  1
(d) With your knowledge of det(A), what is det(B) if B  1  1 1  ?
1 0 0 
Solution:
(a) Matrix B is obtained from A by multiplying Row 1 or A by 8, multiplying Row 2 of A by
4, and multiplying Row 3 of A by 7. Thus, B  8(4)(7) A  224 A  224(1)  224.
(b) B  (5) 3 A  125.
(c) Matrix B is obtained by adding 3 times Row 1 of A to Row 2. Thus, the determinant does
not change and B  A  1.
(d) Matrix B is obtained by interchanging Rows 1 and 2 of A then interchanging Rows 2 and
3. Each interchange changes the sign of the determinant of A (that is, multiplies it by -1).
Thus, B  (1)(1) A  A  1.
Activity 1
One way of computing the determinant of a matrix is to use row operations to reduce the matrix
to an upper triangular matrix whose determinant is the product of the diagonal elements. Now, if
we keep track of the row operations, and utilize the above properties of the determinant, we can
easily find determinant of the original matrix.
Use this process to reduce matrix A to upper triangular then calculate its determinant, where
8  1 7 
A  4 3  4 .
7  7 2 
Assignment #3
Q1. Compute the determinant of the given matrix:
5  3 2
B  1 0 2
2  1 3
12 3
A

 1 5
2  3
0 3

D  0 0

0 0
0 0
1
2
1
0
0
0 4
5 7 
6 0

5 2
0  1
a b
Q2: Assuming that d e
g h
 2 3 0
5
4 2
C
 1 1 0

 2 1 0
cos  sin  tan  
E   0
cos   sin  
 0
sin  cos  
c
f  4 , find:
i
1
0
3

0
0 a 0 
F  b c d 
0 e 0 
2a 2b 2c
i)
d
g
e
h
a
iv) 2d  3g
g
3a
b
a g bh ci
2c
e 2f
3g  h 2i
ii) 3d
f
i
b
c
2e  3h 2 f  3i
h
i
iii)
d
g
e
h
d e
v) a b
f
c
2c
vi) 2 f
g h
i
2i
f
i
b a
e d.
h g
Q3: Let A and B be n by n matrices with A  3, B  2. Find:
AB ; A 3 ; 2 A ; 3B T ; AAT .
Q4: The equation of a straight line through the points ( x1 , y1 ) & ( x2 , y2 ) is given by:
x
x1
x2
y 1
y1 1 . Find the equation of the straight line through (3,2) and (5,-7).
y2 1
Q5: What is the determinant of an orthogonal matrix of order n by n?