Solving A System of Linear Equations: y X y X
Solving A System of Linear Equations: y X y X
Solving A System of Linear Equations: y X y X
Matrices:
Depends on the coefficients of x and y and the constants on the right-hand side of
the equation. The matrix of coefficients for this system is the 2 2 matrix
1 2
3 1
If we insert the constants from the right-hand side of the system into the matrix of
coefficients, we get the 2 3 matrix
1 2 5
3 1 6
We use a vertical line between the coefficients and the constants to represent the
equal signs. This matrix is the augmented matrix of the system also it can be written
as:
1 2 x 5
3 1 y 6
Note:
Two systems of linear equations are equivalent if they have the same solution set.
Two augmented matrices are equivalent if the systems they represent are equivalent.
1) Cramer's rule
The solution to the system
a1 x b1 y c1
a2 x b2 y c2
Dx Dy
Is given by x and y where
D D
a1 b1 c1 b1 a1 c1
D , Dx and Dy
a2 b2 c2 b2 a2 c2
Provided that D 0
1
Mathematics: Lecture 5 مدرس مساعد ازهار مالك
Matrices:
Notes:
1. Cramer's rule works on systems that have exactly one solution.
2. Cramer's rule gives us a precise formula for finding the solution to an
independent system.
3. Note that D is the determinant made up of the original coefficients of x and y .
D is used in the denominator for both x and y . Dx is obtained by replacing the
first (or x ) column of D by the constants c1 and c2 . Dy is found by replacing
the second (or y ) column of D by the constants c1 and c2 .
Sol.:
First find the determinants D, Dx , and Dy :
3 2
D 3 (4) 7
2 1
4 2 3 4
Dx 4 - 6 -2 , Dy -9 - 8 -17
3 1 2 3
2
Mathematics: Lecture 5 مدرس مساعد ازهار مالك
Matrices:
Ex.1:
Use Gaussian elimination method to solve the system (two equations in two
variables):
x 3 y 11
2x y 1
Sol.:
Start with the augmented matrix:
1 3 11
2 1 1
1 3 11
R2 -2R1 R 2
0 7 21
1 3 11 1
R2 R 2
0 1 3 7
1 0 2
R1 3R 2 R
0 1 3
1
3
Mathematics: Lecture 5 مدرس مساعد ازهار مالك
Matrices:
Matrix Inverse
The matrix A has an inverse denoted by A-1 if |A|0 where A.A-1=I. We'll take two
methods to find A-1 where A is an n×n matrix.
2 1
Ex1: Use Row operation to find A-1 if A=
1 4
2 1 1 0 1 1
1 1 0
R R 2 2
1 4 0 1
1 1
1 4 0 1 2
1 1
1 2 2 0
R 2 R 2 R 1
7 1
0 1
2 2
1
2 1 1 2 0
R 2 R 2 2 1 2
7 0 1
7 4
4 1
1 1 0 7
R1 R1 R 2 7
1 2
2 0 1
7 7
4
Mathematics: Lecture 5 مدرس مساعد ازهار مالك
Matrices:
4 1
7
A-1 = 7
1 2
7 7
2 1
Ex.: Use determinant to find A-1 where A =
1 4
1
A-1 = adj ( A)
A
2 1
A 8 1 7
1 4
4 1
Cof(A) =
1 2
C11 = (-1)1+1 |4| = 4
C12 = (-1)1+2 |1| = -1
C21 = (-1)2+1 |1| = -1
C22 = (-1)2+2 |2| = 2
4 1 1
T
4
Adj(A) =
1 2 1 2
4 1
1 4 1 7
1
A 7
7 1 2 1 2
7 7
Problems:
1) write the augment matrix to the following systems then find the solution:
x y z 1
a) 2x 2 y 2z 2
3x 3 y 3z 3
5
Mathematics: Lecture 5 مدرس مساعد ازهار مالك
Matrices:
x yz 2
b) 2x y z 1
3x 3 y 3z 8
2x 2 1
x 2 2x 3 2
c)
x1 2x 4 10
x1 x 4 5
2) Find the inverse of each following matrix
2 1 1
a) 3 5 2
5 2 4
1 1 3
b) 2 1 1
1 2 1
References:
1- Calculus & Analytic Geometry (Thomas).
2- Calculus (Haward Anton).
3- Advanced Mathematics for Engineering Studies ) رياض احمد عزت.(أ