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Solving System of Equations Using Matrices
In this lesson we will solve systems of equations using matrices.
This method is similar to the elimination method, but using matrices may be quicker because you keep track of equations using a shorter notation.
5 3 1 5 3 1
2 6 50 2 6 50
x y
x y
Numerals in the first equation match the numerals in the first row of the matrix.
Numerals in the second equation match the numerals in the second row of the matrix.
The solution to this system of equations is x = 4 and y = -7.
When we write the solution matrix we want it to represent the equations, therefore, x = 4 and y = -7 would look like this:
0 4 1 0 4
0 7 0 1 7
x y
x y
Numerals in the first equation match the numerals in the first row of the matrix.
Numerals in the second equation match the numerals in the second row of the matrix.
In the elimination method you combined equations and multiplied them by numbers.
In much the same way we can modify the rows of the matrix by performing row operations on each number in those rows.◦ Multiply (or divide) all numbers in a row by a non-
zero number◦ Add all numbers in a row to corresponding
numbers in another row◦ Add a multiple of the numbers in one row to the
corresponding numbers in another row◦ Exchange two rows.
5 3 1 1 0 4
2 6 50 0 1 7
Starting matrixEnding matrix
Solution can be read: x = 4 and y = -7
Solve the system of equations using matrices
2 3
3 23
x y
x y
Copy the numerals from each equation to form the matrix
1 2 3
3 1 23
Add -3 times row 1 to row 2 3 6 9
3 1 23
0 7 141 2 3
0 7 14
Divide row 2 by 7 0 1 2
1 2 3
0 7 14
1 2 3
0 1 2
Add 2 times row 2 to row 1 0 2 4
1 2 3
1 0 7
1 0 7
0 1 2
Therefore x = 7 and y = 2.
In this investigation you will see how to combine row operations in your solution process.
Consider the system of equations:
Write the matrix for this system. Describe how to use row operations to get 0
as the first entry in the second row.
2 11
6 5 9
x y
x y
2 1 11
0 8 24
2 1 11
0 1 3
Use row operations on the matrix from the last step to get 1 as the second number in row 2
Use row operations on the matrix from the last step to get 0 as the second number in row 1
2 0 8
0 1 3
Use row operations on the matrix from the last step to get 1 as the first number in row 1
What does the matrix mean? Give the solution to the system of equations.
2 0 8
0 1 3
1 0 4
0 1 3
x=4 and y = 3.
How do the first three rules for ROW OPERATIONS IN A MATRIX correspond to steps we used in the elimination process?
1. Multiply (or divide) all numbers in a row by a non-zero number
2. Add all numbers in a row to corresponding numbers in another row
3. Add a multiple of the numbers in one row to the corresponding numbers in another row
4. Exchange two rows.
On Friday, 3247 people attended the county fair. The entrance fee for an adult was $5 and for a child 12 and under the fee was $3. The fair collected $14,273. How many of the total attendees were adults and how many were children.
A= number of adult attending C= number of children attending Using total attendance: A+C=3247 Using money collected: 5A+3C=14273 Set up a matrix 1 1 3247
5 3 14,373
Use row operations to simplify the matrix to the read the solution. 1 1 3247
5 3 14,373
Add -5 times row 1 to row 2 to get a new row 2 (-5R1+R2)
1 1 3247
0 2 1962
Divide row 2 by -1 (R2/-2) 1 1 3247
0 1 981
Add -1 times row 2 to row 1 to get a new row 1 (-1R1+R1)
1 0 2266
0 1 981
There were 2266 adults and 981 children attending the fair on Friday.
A+C=3247
5A+3C=14273
Check that 2266 adults and 981 is a solution to both equations:
2266 +981=3247
3247 = 3247
5(2266)+3(981)=14273
11330+2943=14273
Press 2nd Matrix and Edit Matrix A. First set the size and the entries.
Enter the matrix
Return to the home screen and press 2nd Matrix and choose Math B: rref(.
Complete the line to say rref([A]). Press Enter and you will see the solution matrix for this system.
8 7 1
3 1 4
You leaned to represent a linear system with a matrix.
You learned to use Row Operations to solve a system of linear equations.
You learned to solve a matrix on the graphing calculator.