Systems of Linear Algebraic Equations
Nizar Salim / Lecture 3
1.2.3. Determinants:
The term determinant of square matrix A, denoted det(A) or |A|,
refers to both the collection of the elements of the square matrix, enclosed
in vertical lines, and the scalar value represented by that array. Thus
n nn
n
n
a a
a a a
a a a
A A
... ... ...
... ... ... ... ...
... ... ... ... ...
... ...
... ...
det( )
1
21 22 2
11 12 1
___________(1.11)
Only square matrix have determinant
The scalar value of the determinant of a 2×2 matrix is the product of
the elements on the major diagonal minus the product of the elements on
the minor diagonal. Thus
11 22 21 12
21 22
11 12 det( ) a a a a
a a
a a
A A ____________(1.12)
The scalar value of the determinant of a 3×3 matrix composed of the sum
of six triple products which can be obtained from the augmented
determinant:
31 32
21 22
11 12
31 32 33
21 22 23
11 12 13
a a
a a
a a
a a a
a a a
a a a
_______________(1.13)
The 3×3 is augmented by repeating the first two columns of the
determinant on the right-hand-side of the determinant. Three triple
products are formed, starting with the elements of the first row multiplied
by the two remaining elements on the right-downward-sloping diagonals.
Three more triple products are formed, starting with the element of the
third row multiplied by the row remaining elements on the right-upwardsloping
diagonals. The value of the determinant is the sum of the first
three triple products minus the sum of the last three triple products. Thus
31 22 13 32 23 11 33 21 12
11 22 33 12 23 31 13 21 32 det( )
a a a a a a a a a
A A a a a a a a a a a
______(1.14)
Systems of Linear Algebraic Equations
Nizar Salim / Lecture 3
Example 5:
Evaluate the determinant of the coefficient matrix of equation (1.2) by
the diagonal method.
Solution:
20 20 130
20 40 20
80 20 20
A
The augmented determinant is
20 20
20 40
80 20
20 20 130
20 40 20
80 20 20
Applying Eq. (1.14)
det(A) = |A| =(80)(40)(130)+(-20)(-20)(-20)+(-20)(-20)(-20)
-(-20)(40)(-20)-(-20)(-20)(80)-(130)(-20)(-20)
=416000-8000-8000-16000-32000-52000
= 300000
The diagonal method of evaluating determinants applies only to 2x2 and 3x3
determinants.
Cofactor: the method of cofactors in not recommended for anything
larger than a 4×4 determinant, the minor Mij is the determinant of the
(n-1) × (n-1) sub matrix of the n×n matrix A obtained by deleting the ith
row and the jth column. The cofactor Aij associated with the minor Mij is
Using cofactors, the determinant of matrix A is the sum of the products of
the elements of any row or column, multiplied by their corresponding
cofactors. Thus expanding across any fixed row i yield.
n
j
n
j
ij ij
i j
ij ij A A a A a M
1 1
det( ) ( 1) ___________(1.15)
Example 6:
Evaluate of a 3×3 determinant by the cofactor method.
20 20 130
20 40 20
80 20 20
A
Systems of Linear Algebraic Equations
Nizar Salim / Lecture 3
Sol.
20 20
20 40
( 20)
20 130
20 20
( 20)
20 130
40 20
A (80)
|A| = 80(5200+400)-(-20)(-2600+400)+(-20)(400+800)
=384000-60000-24000 = 300000
1.3 Direct Elimination Methods:
There are a number of methods for the direct solution of systems of
linear algebraic equations. One of the more well-known is
1.3.1 Cramer s Rule:
It is not an elimination method, Cramer rule is a direct method for
solving systems of linear algebraic equations.
Consider the system of linear algebraic equations, Ax=b, which
represents n equations. Cramer rule states that the solution for Xj
(j=1,2, ,n) is given by
( 1,2,....., )
det( )
det( )
j n
A
A
X
j
j ______________(1.16)
Where Aj is the n×n matrix, obtain by replacing column j in matrix A by
the column vector b. for example consider the system of two linear
algebraic equations.
a11x1 + a12x2 = b1 _________________(1.17a)
a21x1 + a22x2 = b2 _________________(1.17b)
Applying Cramer rule yield
21 22
11 12
21 2
11 1
2
21 22
11 12
2 22
1 12
1
a a
a a
a b
a b
and x
a a
a a
b a
b a
x ___________(1.18)
The determinants in equation (1.18) can be evaluated by the diagonal
method.
Systems of Linear Algebraic Equations
Nizar Salim / Lecture 3
Example 7:
Illustrate Cramer Rule by solving equation (1.2)
80x1 20x2 20x3 = 20
- 20x1 + 40x2 20x3 = 20
- 20x1 - 20x2 + 130x3 = 20
Solution:
First, calculate det(A).
300000
20 20 130
20 40 20
80 20 20
det( A)
180000
20 20 130
20 40 20
20 20 20
det( A1 )
300000
20 20 130
20 20 20
80 20 20
det( A2 )
120000
20 20 20
20 40 20
80 20 20
det( A3 )
det( )
det( )
A
A
X
j
j
X1=180,000/300,000=0.6
X2=300,000/300,000=1
X3=120,000/300,000=0.4
Systems of Linear Algebraic Equations
Nizar Salim / Lecture 3
1.3.2 Gaussian Elimination Method.
We can find the unknowns in a system of two or more equations also
by the Gaussian elimination method.
This method is best illustrated with the following example.
Example 8:
Use the Gaussian elimination method to find v1,v2 and v3 of
2v1 - v2 + 3v3 = 5
-4v1- 3v2 - 2v3 = 8 __________(1.19)
3v1 + v2 - v3= 4
Solution:
First step, we add the first equation with the third to eliminate the
unknown v2 and we obtain the following equation.
5v1 + 2v3= 9 ________________(1.20)
Next we multiply the third equation by 3, and we add it with the second
to eliminate v2, then, we obtain the following rquation.
5v1 - 5v3= 20 _______________(1.21)
Subtraction of (1.21) from (1.20) yields
7v3=-11
v3=-11/7 ______________(1.22)
Now we can find the unknown v1 from either (1.20) or (1.21) by
substitution of (1.22) into (1.20) we obtain
5v1 + 2(-11/7) = 9
v1 = 17/7
Finally we can find the last unknown v2 from any of three equations of
(1.19) by substitution in to the first equation we obtain
v2 = 2v1 +3v3 -5 =34/7 33/7 - 35/7 = -34/7
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