Determinants and Their Properties
Determinants and Their Properties
Determinants and Their Properties
Module
Three
DETERMINANTS
37
Introduction
In the last module, multiplicative inverses of 2 by 2 matrices discussed
and showed that a matrix A has a multiplicative inverse A-1 if a certain real
number associated with the matrix is non – zero.
Another very important number associated with a square matrix A is the
determinant of A. however, the use of determinants in the solutions of systems
of linear equations is merely one of the numerous applications of determinants.
a11 a12
|A| = a21 a22 = a11 a22 - a12 a21
Example 1:
Compute the value of determinant.
4 -8
A=
-3 10
Solution:
|A| = (4) (10) – (-3) (- 8)
= 40 – 24
= 16
Example 2:
Compute the value of determinant
B= 7 2
5 0
Solution:
| B| = 7 (0) – (5)(2)
= 0 – 10
= -10
38
Example 3:
A = t–1 2
3 t-2
Solution:
= t2 – 3t + 2 - 6
= t2 – 3t – 4
To remember this algebraic sum, rewrite the first and second columns.
Form the sum of the products of the entries in the lines from left to right and
subtract from this number of the products of entries on the lines from right to
left.
a11 a12 a13 a11 a12
|A| = a21 a22 a23 a21 a22
a31 a32 a33 a31 a32
Example 4:
Compute the value of determinant
1 2 3
A= 2 1 3
3 1 2
Solution:
|A| = 6
39
We could obtain the same result by using the easy method illustrated
above:
1 2 3 1 2
2 1 3 2 1
3 1 2 3 1
|A| = 6
Example 5:
Compute the value of the determinant
4 3 1
A= 15 7 0
-3 -2 0
Solution:
|A| = (4)(7)(0) + (3)(0)(-3) + (1)(15)(-2) – (-3)(7)(1) – (-2)(0)(4) -
(0)(15)(3)
|A| = -9
PROPERTIES OF DETERMINANTS
Verification:
1 2 3
2 1 3
Let A =
3 1 2
Solution:
|A| = (1)(1)(2) + (2)(3)(3) + (3)(2)(1) – (3)(1)(3) – (1)(3)(1) - (2)(2)(2)
|A| = 6
1 2 3
AT = 2 1 1
3 3 2
Verification:
Let A = 2 -1 B= 3 2
3 2 2 -1
Solution:
= 7 = -7
Verification:
Let A = 1 2 3
-1 0 7
1 2 3
Solution:
Verification:
Let A = 1 2 3
4 5 6
0 0 0
Solution:
In particular, if all the entries in one row (column) are zero, then the
determinant is zero.
Verification:
Let B= 2 6
1 12
det (B) = 2 ( 12 ) – 1 ( 6 ) = 18
To simplify the computation of det (A) factor out the common factors of
rows and columns of A.
c det (A) = 6 ( 4 -1 ) = 18
Verification:
1 2 3 5 0 9
We have 2 -1 3 = 2 -1 3
1 0 1 1 0 1
It is obtained by adding twice the second row to the first row. By applying
the definition of determinant to the second determinant, both are seen to have
the value 4.