1. L9Matrix and linear equation

2. Matrix and linear equation

3. From Linear Equations to Coefficient Matrices
Example 1: 3x - 2y = 10 -6x + 4y = -7
A =
You Try 1:
-2x - 2y = 10
7x + 5y = -3
B =
7 5
Show where coefficients are
Look at signs!!
Note X and Y position
No constants used!

4. From Linear Equations to Constant Matrices
Example 2: 3x - 2y = 10 -6x + 4y = -7
A = 10 -7
You Try 2:
-2x - 2y = 10
7x + 5y = -3
B = 10 -3
Note only constants here (no variables!)

5. From Linear Equations to Augmented Matrices
Example 3:
3x - 2y = 10
-6x + 4y = -7
A =
B =
You Try 3:
-2x - 2y = 10
7x + 5y = -3
The augmented matrix is a combination of the coefficient and constant matrices
Note the X, Y and constant positions

6. Using Matrices to Solve Systems of Equations
Augmented Matrix – a matrix that is used to solve a system of equations.
𝑥+𝑦+𝑧=0 3𝑥−2𝑦+4𝑧=9 𝑥−𝑦−𝑧=0
2𝑥+𝑦=5 −4𝑥+6𝑦=−2
Augmented matrix
Augmented matrix

7. Using Matrices to Solve Systems of Equations
Given the augmented matrix, write the system of equations.
System of Equations
System of Equations
5𝑥−𝑦=9
3𝑥+6𝑦−2𝑧=−8
2𝑥+8𝑦=7
2𝑥+0𝑦+5𝑧=13
𝑥+3𝑦−7𝑧=12
3𝑥+6𝑦−2𝑧=−8
2𝑥+5𝑧=13
𝑥+3𝑦−7𝑧=12

8. Using Matrices to Solve Systems of Equations
The use of Elementary Row Operations is required when solving a system of equations using matrices.
Elementary Row Operations
I. Interchange two rows.
II. Multiply one row by a nonzero number.
III. Add a multiple of one row to a different row.
𝑅 1 ↔ 𝑅 3
𝑅 3 → 2𝑟 3
𝑅 2 → −2𝑟 1 + 𝑟 2

9. Using Matrices to Solve Systems of Equations
The solution to the system of equations is complete when the augmented matrix is in Row Echelon Form.
Row Echelon Form
A matrix is in row echelon form (ref) when it satisfies the following conditions.
The first non-zero element in each row, called the leading entry, is 1.
Each leading entry is in a column to the right of the leading entry in the previous row.
Rows with all zero elements, if any, are below rows having a non-zero element.

10. Using Matrices to Solve Systems of Equations
Reduced Row Echelon Form
A matrix is in reduced row echelon form (rref) when it satisfies the following conditions.
The matrix is in row echelon form (i.e., it satisfies the three conditions listed for row echelon form.
The leading entry in each row is the only non-zero entry in its column.

11. Use matrices to solve the following systems of equations. (pg. 569 #58)
𝐴𝑢𝑔𝑚𝑒𝑛𝑡𝑒𝑑 𝑚𝑎𝑡𝑟𝑖𝑥
𝑥−𝑦+𝑧=−4
2𝑥−3𝑦+4𝑧=−15
𝑅 2 → −2𝑟 1 + 𝑟 2
5𝑥+𝑦−2𝑧=12
𝑅 3 → −5𝑟 1 + 𝑟 3
𝑅 2 → −𝑟 2
𝑅 3 → −6𝑟 2 + 𝑟 3
𝑅 3 → 𝑟 3
𝑥−𝑦+𝑧=−4
𝑦−2(−2)=7
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑦−2𝑧=7
𝑦=3
(𝑥, 𝑦,𝑧)
𝑧=−2
𝑥−(3)+(−2)=−4
(1, 3,−2)
𝑟𝑜𝑤 𝑒𝑐ℎ𝑒𝑙𝑜𝑛 𝑓𝑜𝑟𝑚
𝑥=1

12. Use matrices to solve the following systems of equations.
𝐴𝑛𝑜𝑡ℎ𝑒𝑟 𝑜𝑝𝑡𝑖𝑜𝑛:
𝑥−𝑦+𝑧=−4
𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑒 𝑡𝑜 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑅𝑜𝑤 𝐸𝑐ℎ𝑒𝑙𝑜𝑛 𝐹𝑜𝑟𝑚
2𝑥−3𝑦+4𝑧=−15
5𝑥+𝑦−2𝑧=12
𝑅 1 → 𝑟 2 + 𝑟 1
𝑅 1 → 𝑟 3 + 𝑟 1
𝑅 2 → 2𝑟 3 + 𝑟 2
𝑥=1
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑦=3
(𝑥, 𝑦,𝑧)
𝑧=−2
(1, 3,−2)
𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑟𝑜𝑤 𝑒𝑐ℎ𝑒𝑙𝑜𝑛 𝑓𝑜𝑟𝑚

13. Consistent or Inconsistent System?
𝑥=1
𝑦=5
𝑂𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:𝐶𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
𝑧=2
𝑥=4
𝑦=7
𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:𝐼𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
0=2
𝑥+5𝑧=2
𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠:𝐶𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
𝑦+3𝑧=2
𝑧=𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
(𝑥=2−5𝑧,𝑦=2−3𝑧,𝑧=𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟) 𝑜𝑟
{(𝑥, 𝑦,𝑧)|𝑥=2−5𝑧,𝑦=2−3𝑧,𝑧=𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟}

14. Use matrices to solve the following systems of equations. (pg. 569 #66)
𝐴𝑢𝑔𝑚𝑒𝑛𝑡𝑒𝑑 𝑚𝑎𝑡𝑟𝑖𝑥
𝑥+2𝑦−𝑧=3
2𝑥−𝑦+2𝑧=6
𝑅 2 → −2𝑟 1 + 𝑟 2
𝑥−3𝑦+3𝑧=4
𝑅 3 → −𝑟 1 + 𝑟 3
𝑅 2 → − 𝑟 2
𝑅 3 → 5𝑟 2 + 𝑟 3
0=1
𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡 𝑠𝑦𝑠𝑡𝑒𝑚

15. Check this answer by multiplying
Check this answer by multiplying. We should get the identity matrix if we’ve found the inverse.

16. We can use A-1 to solve a system of equations
To see how, we can re-write a system of equations as matrices.
coefficient matrix
variable matrix
constant matrix

17. left multiply both sides by the inverse of A
This is just the identity
but the identity times a matrix just gives us back the matrix so we have:
This then gives us a formula for finding the variable matrix: Multiply A inverse by the constants.

18. x y find the inverse -2r1+r2 r1-3r2 -r2
This is the answer to the system y

19. An example:
Use matrix inverses to solve the system below:
1. Determine the matrix of coefficients, A, the matrix X, containing the variables x, y, and z. and the column matrix B, containing the numbers on the right hand side of the equal sign.

20. Continuation:
2. Form the matrix equation AX=B . Multiply the 3 x 3 matrix A by the 3 x 1 matrix X to verify that this multiplication produces the 3 x 3 system on the left:

21. Problem continued:
If the matrix A inverse exists, then the solution is determined by multiplying A inverse by the column matrix B. Since A inverse is 3 x 3 and B is 3 x 1, the resulting product will have dimensions 3 x1 and will store the values of x , y and z.
The inverse matrix A can be determined by the methods of a previous section or by using a computer or calculator. The display is shown below:

22. Solution
When the product of A inverse and matrix B is found the result is as follows:
The solution can be interpreted from the X matrix: x = 0, y = 2 and z = -1/2 . Written as an ordered triple of numbers, the solution is
(0 , 2 , -1/2)

23. Another example: Using matrix techniques to solve a linear system
Solve the system below using the inverse of a matrix
The coefficient matrix A is displayed to the left:: The inverse of A does not exist. We cannot use the technique of multiplying A inverse by matrix B to find the variables x, y and z. Whenever, the inverse of a matrix does not exist, we say that the matrix is singular.
There are two cases were inverse methods will not work:
1. if the coefficient matrix is singular
2. If the number of variables is not the same as the number of equations.

24. Application A $30 $20 B $40 Guitar model Labor cost Material cost
Production scheduling: Labor and material costs for manufacturing two guitar models are given in the table below: Suppose that in a given week $1800 is used for labor and $1200 used for materials. How many of each model should be produced to use exactly each of these allocations?
Guitar model
Labor cost
Material cost A
$30
$20 B
$40

25. Solution
Let A be the number of model A guitars to produce and B represent the number of model B guitars. Then, multiplying the labor costs for each guitar by the number of guitars produced, we have
30x + 40y = 1800
Since the material costs are $20 and $30 for models A and B respectively, we have 20A + 30B = 1200.
This gives us the system of linear equations:
30A+ 40B = 1800
20A+30B=1200
We can write this as a matrix equation:

26. solution Produce 60 model A guitars and no model B guitars.
Using the result
The inverse of matrix A is
Produce 60 model A guitars and no model B guitars.