chapter 4.4 solving systems of equations by matrices
TRANSCRIPT
Objectives
Use matrices to solve a system of two equations.
Use matrices to solve a system of three equations.
What is a matrix?
A matrix (plural: matrices) is a rectangular array of numbers. The following are examples of matrices:
11 12
21 22
2 11 0
0 1 40 1
3 1
3
6
a a
a a
Matrices
The numbers aligned horizontally in a matrix are in the same row. The numbers aligned vertically are in the same column.
2 1 0
1 6 2
row 1
row 2
column 2
column1
column 3
This matrix has 2 rows and 3 columns.It is called a 2 X 3(read “two by three”) matrix.
How do equations and matrices relate?
System of Equations Corresponding
(in standard form) Matrix
2 3 6
0
x y
x y
2 3 6
1 1 0
The rows of the matrix correspond to the equations in the system. The Coefficients of each variable are placed to the left of the vertical dashed line.Matrices are a shorthand notation for representing systems of equations.
Elementary Row Operations Any two rows in a matrix may be interchanged. The elements of any row may be multiplied (or
divided) by the same nonzero number. The elements of any row may be multiplied (or
divided) by a nonzero number and added to their corresponding elements in any other row. Notice that these row operations are the same
operations that we can perform on equations in a system.
Solving a system of 2 equations by matrices
Matrix System of Equations
1 2 3
0 1 5
1 2 3 2 3
0 1 5 5
x y x yor
x y y
In the second equation, we have y = 5. Substituting this in the first equation, we have x + 2(5) = -3 or x = - 13. The solution of the system in ordered pair is ( -13, 5)
Example 1:
Use matrices to solve the system:
Step 1: Set up the corresponding matrix
3 5
2 4
x y
x y
1 3 5
2 1 4
Example 1: Continued
Step 2: Use elementary row operations to write an equivalent matrix that looks like:
In the given matrix, the element in the first row is already 1, as desired. Next we write an equivalent matrix with a 0 below the 1. To do this, we multiply row 1 by -2 and add to row 2. We will change only row 2.
1 3 5
2 1 4
1
0 1
a b
c
1 3 5
2(1) 2 2(3) 1 2(5) 4
Row 1element
Row 2element
Row 1element
Row 1element
Row 2element
Row 2element
Simplifies to: 1 3 5
0 7 14
Example 1: Continued
Step 3: Now we change -7 to a 1 by use of an elementary row operation. We dived row 2 by -7
1 3 5
0 7 14
7 7 7
1 3 51 3 5
0 7 140 1 2
simplifies to
Example 1: Continued
The last matrix corresponds to the system:
To find x, we let y = 2 in the first equation.x + 3(2) = 5
x = -1
The ordered pair solution is (-1, 2). Check to see that this ordered pair satisfies the equations.
3 5
2
x y
y
1 3 5
0 1 2
Let’s try another one!
Use matrices to solve:
Step 1: Set up matrix
Step 2: Write an equivalent matrix that looks like:
To do this, multiply the first row by -2 and add to row 2. Leave row 1 the same!
2 4
2 3 13
x y
x y
1 2 4
2 3 13
Simply take the coefficientfrom each variable…
1
0 1
a b
c
2 2 2
1 2 4
(1) 2 (2) 3 ( 4) 13
1 2 4
0 7 21simplifies to
Let’s try another one!
Step 3: Now change the -7 to 1 by using elementary row operation. We divide row 2 by -7
Step 4: Write the system that corresponds to the matrix:
Step 5: Substitute to find the unknown variable.
x + 2y = -4
x + 2 (-3) = -4
x – 6 = -4
x = 2
1 2 41 2 4
0 7 21
70
71
73
simplifies to
2 4
3
x y
y
The ordered pair solutionis (2, -3)
Example 2: Use matrices to solve the system:
Step 1: Set up a corresponding matrix
Step 2: To get 1 in the row 1, column 1 position, we divide the elements of row 1 by 2.
2 3
4 2 5
x y
x y
2 1 3 1 31
2 24 2 54 2 5
2 2 2 simplifies to
2 1 3
4 2 5
Example 2 Continued Step 3: To get 0 under 1, we multiply the elements of
row 1 by -4 and add the new elements to the elements of row 2.
1 31
2 21 3
(1) 4 ( ) 2 (4 4 52
4 )2
1 31
2 20 0 1
simplifies to
Step 4: Write a correspondingsystem:
1 3
2 20 1
x y
The equation 0 = -1 is false for all y or x values; hence the system has no solution!
Concept Check:
Consider the system
What is wrong with the corresponding matrix shown below?
2 3 8
5 3
x y
x y
2 3 8
0 5 3
Answer bottom left on page 251
Solving a system of three equation in three variables using matrices The matrix must be written in the following
form: 1
0 1
0 0 1
a b d
c e
f
Example 3 Use matrices to solve the system:
Step 1: Set up a corresponding matrix
2 2
2 2 5
3 2 8
x y z
x y z
x y z
1
1
2
2
1 3
2 1 2
2 5
8
Our goal is to write an equivalent matrix with 1’s along the diagonal…seethe numbers in red and 0’s below the 1’s. The element in row 1, column 1 is already 1. Next we get 0’s for each element in the rest of column 1. The numbers in blue need to get changed to zeros.
Example 3: Continued Step 2: Multiply the elements in row 1 by 2 and add the
new elements to row 2. Multiply the elements of row 1 by -1 and add the new elements to the elements of row 3. We do not change row 1!
2 1 2 5
1
2 2 2 2
1 1
1 2 1 2
(1) 2 (1) (2)
(1) (2) 1 13 21 (2) 8( )
Green = Row 1Pink = # multiplied by
1 2 1 2
0 3 4 9
0 1 3 10
simplifies to
Example 3 Continued
Step 4: We continue down the diagonal and use elementary row operations to get 1 where the element 3 is now. To do this, we interchange rows 2 and 3.
1 2 1 2
0 3 4 9
0 1 3 10
1 2 1 2
0 1 3 10
0 3 4 9
is equivalent to
Example 3: Continued
Step 5: Next we want the new row 3, column 2 element to be 0. We multiply the elements of row 2 by -3 and add the result to the elements of row 3.
0 (1) ( 3)
1 2 1 2
0 1 3 10
0 3 4 9( 103 3 3 3 )
Green = Row 2Pink = # multiplied by
1 2 1 2
0 1 3 10
0 0 13 39
simplifies to
Example 3: Continued
Step 6: Finally, we divide the elements of row 3 by 13 so that the final diagonal element is 1.
13 13 13
1 2 1 2
0 1 3 10
0 9
1
0 13 3
3
1 2 1 2
0 1 3 10
0 0 1 3
simplifies to
Step 7: Write the system that correspondsto the matrix.
2 2
3 10
3
x y z
y z
z
Step 8: Substitute z = 3 into the 2nd equationy – 3(3) = -10
y – 9 = -10 y = -1
Step 9: Substitute z =3 and y = -1 into the 1st equationx + 2(-1) + 3 = 2
x – 2 + 3 = 2x + 1 = 2
x = 1
Ordered triplesolution:(1, -1, 3)