Transcript Section 5.3

Section 5.3
Matrices
And
Systems of Equations
Systems of Equations in Two Variables
Matrices
A rectangular array of numbers is called a
matrix (plural, matrices).
Example: 4 2 3


1 5 4


The matrix shown above is an augmented
matrix because it contains not only the
coefficients but also the constant terms.
4 2
The matrix 1 5  is called the coefficient


matrix.
Matrices continued
The rows of a matrix are horizontal.
The columns of a matrix are vertical.
The matrix shown has 2 rows and 3 columns.
1 2 3
4 5 6


A matrix with m rows and n columns is said to
be of order m  n.
When m = n the matrix is said to be square.
Gaussian Elimination with Matrices
Row-Equivalent Operations
1. Interchange any two rows.
2. Multiply each entry in a row by the same
nonzero constant.
3. Add a nonzero multiple of one row to
another row.
Example
Solve the following system:
2x  y  z  8
x  3 y  2 z  1
4x
 5 z  23
.
Example continued
First, we write the augmented matrix,
writing 0 for the missing y-term in the last equation.
2x  y  z  8
x  3 y  2 z  1
4x
 5 z  23
 2 1 1 8 
1 3 2 1


 4 0 5 23
Our goal is to find a row-equivalent matrix of the form
1
0

0
a
1
0
b
d
1
c

e

f 
Example continued
 2 1 1 8 
1 3 2 1


 4 0 5 23
1 3 2 1
 2 1 1 8 


 4 0 5 23
New row 1 = row 2
New row 2 = row 1
Example continued
1 3 2 1
 2 1 1 8 


 4 0 5 23
We multiply the first row
by 2 and add it to the
second row.
We also multiply the first
row by 4 and add it to
the third row.
Row 1 is unchanged
1 3 2 1
0 5 5 10  New row 2=  2(row 1) + row 2


0 12 13 27  New row 3=  4(row 1)+row3
Example continued
1 3 2 1
0 5 5 10 


0 12 13 27 
We multiply the second row
by 1/5 to get a 1 in the
second row, second column.
1 3 2 1
0 1 1 2  New row 2= 1 (row 2)
5


0 12 13 27 
Example continued
1 3 2 1
0 1 1 2 


0 12 13 27 
We multiply the second row
by 12 and add it to the
third row.
1 3 2 1
0 1 1 2 


0 0 1 3  New row 3=  12(row 2) + row 3
Example continued
1 3 2 1
0 1 1 2 


0 0 1 3 
Now, we can write the
system of equations
that corresponds to our
last matrix.
x  3 y  2 z  1
yz2
z 3
Example continued
We back-substitute 3 for z in equation (2)
and solve for y.
yz2
y3 2
y  1
Example continued
Next, we back-substitute 1 for y and 3
for z in equation (1) and solve for x.
x  3 y  2 z  1
x  3( 1)  2(3)  1
x  3  6  1
x  3  1
x2
The triple (2, 1, 3) checks in the original
system of equations, so it is the solution.
Row-Echelon Form
1. If a row does not consist entirely of 0’s, then the
first nonzero element in the row is a 1 (called a
leading 1).
2. For any two successive nonzero rows, the
leading 1 in the lower row is farther to the right
than the leading 1 in the higher row.
3. All the rows consisting entirely of 0’s are at the
bottom of the matrix.
If a fourth property is also satisfied, a matrix is
said to be in reduced row-echelon form:
Example
Which of the following matrices are in row-echelon form?
a)
1 6 7 5 
0 1 3 4 


0 0 1 8 
b)
0 2 4 1 
0 0 0 0 


c)
1 2 7 6 
0 0 0 0 


0 1
4 2 
d)
1 0 0 3.5
0 1 0 0.7 


0 0 1 4.5 
Gauss-Jordan Elimination
We perform row-equivalent operations on a
matrix to obtain a row-equivalent matrix in
row-echelon form. We continue to apply these
operations until we have a matrix in reduced
row-echelon form.
.
Gauss-Jordan Elimination Example
Example: Use
Gauss-Jordan
elimination to solve
the system of
equations from the
previous example.
1 3 2 1
0 1 1 2 


0 0 1 3 
Gauss-Jordan Elimination continued
1 3 2 1
0 1 1 2 


0 0 1 3 
1 3 0 5 
0 1 0 1


0 0 1 3 
We continue to perform
row-equivalent operations
until we have a matrix in
reduced row-echelon form.
New row 1 = 2(row 3) + row 1
New row 2 =  1(row 3) + row 2
Gauss-Jordan Elimination continued
1 3 0 5 
0 1 0 1


0 0 1 3 
1 0 0 2 
0 1 0 1


0 0 1 3 
Next, we multiply the
second row by 3 and
add it to the first row.
New row 1 = 3(row 2) + row 1
Gauss-Jordan Elimination continued
Writing the system of
equations that
corresponds to this
matrix, we have
1 0 0 2 
0 1 0 1


0 0 1 3 
x
y
z
2
 1
3
We can actually read the solution, (2, 1, 3), directly
from the last column of the reduced row-echelon matrix.
Special Systems
When a row consists entirely of 0’s, the
equations are dependent and the system is
equivalent.
1 0 4 6 
0 1 4 8 


0 0 0 0 
Special Systems
When we obtain a row whose only nonzero entry
occurs in the last column, we have an inconsistent
system of equations.
For example, in the matrix
1 0 4 6 
0 1 4 8 


0 0 0 9 
the last row corresponds to the false equation
0 = 9, so we know the original system has no solution.
Another Example
Solve the system of equations using Gaussian elimination.
x + 3y – 6z = 7
2x – y + 2z = 0
x + y + 2z = -1
Yet Another Example
Solve the system of equations using Gaussian elimination.
x + 2y = 1
2x + 4y = 3