Systems of Linear Equations Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

Slides:



Advertisements
Similar presentations
Gauss Elimination.
Advertisements

Differential Equations Verification Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Differential Equations Separable Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Math 3C Practice Final Problems Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Math 3C Practice Midterm #1 Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Directions: Solve the linear systems of equations by graphing. Use the graph paper from the table. Tell whether you think the problems have one solution,
Differential Equations Solving First-Order Linear DEs Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
3-2: Solving Linear Systems
Solving a System of Equations by ELIMINATION. Elimination Solving systems by Elimination: 1.Line up like terms in standard form x + y = # (you may have.
Math 3C Practice Midterm Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Row Reduction Method Lesson 6.4.
Splash Screen Lesson 3 Contents Example 1Elimination Using Addition Example 2Write and Solve a System of Equations Example 3Elimination Using Subtraction.
Systems of Linear Equations Gaussian Elimination Types of Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
4.4 & 4.5 Notes Remember: Identity Matrices: If the product of two matrices equal the identity matrix then they are inverses.
Math Pacing Elimination Using Addition and Subtraction 1.The sum of two numbers is 31. The greater number is 5 more than the lesser number. What are the.
13.6 MATRIX SOLUTION OF A LINEAR SYSTEM.  Examine the matrix equation below.  How would you solve for X?  In order to solve this type of equation,
3-2 Solving Equations by Using Addition and Subtraction Objective: Students will be able to solve equations by using addition and subtraction.
Differential Equations Separable Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
The Inverse of a Matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Solving Systems of Equations by Elimination (Addition) Section 3.2, Part II.
Elimination Method: Solve the linear system. -8x + 3y=12 8x - 9y=12.
Differential Equations Graphing Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Solving Linear Equations Substitution. Find the common solution for the system y = 3x + 1 y = x + 5 There are 4 steps to this process Step 1:Substitute.
Systems of Linear Equations in Vector Form Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Warm- Up Solve the following systems using elimination or substitution : 1. x + y = 6 -3x + y = x + 4y = 7 x + 2y = 7.
Solve Linear Systems by Substitution January 28, 2014 Pages
GUIDED PRACTICE for Example – – 2 12 – 4 – 6 A = Use a graphing calculator to find the inverse of the matrix A. Check the result by showing.
Multiply one equation, then add
3.4 See if you can figure this out: Can you replace the question marks with math symbols to make the following equation correct: (2 ? 3) ? (6 ? 2) ? (3.
Solve Linear Systems by Elimination February 3, 2014 Pages
Coordinate Systems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Differential Equations Second-Order Linear DEs Variation of Parameters Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
3-2: Solving Linear Systems. Solving Linear Systems There are two methods of solving a system of equations algebraically: Elimination Substitution.
Differential Equations Solving First-Order Linear DEs Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Solving a System of Equations by ELIMINATION. Elimination Solving systems by Elimination: 1.Line up like terms in standard form x + y = # (you may have.
Use the elimination method
Substitution Method: Solve the linear system. Y = 3x + 2 Equation 1 x + 2y=11 Equation 2.
Solve Linear Systems By Multiplying First
Differential Equations
Systems of Linear Equations
Systems of Linear Equations
10.3 Solving Linear Systems
Differential Equations
Differential Equations
Systems of Linear Equations
Systems of Linear Equations
Coordinate Systems Prepared by Vince Zaccone
Least Squares Approximations
10 Real Numbers, Equations, and Inequalities.
3-2: Solving Linear Systems
Linear Independence Prepared by Vince Zaccone
Systems of Differential Equations Nonhomogeneous Systems
Differential Equations
Systems of Linear Equations
Before: December 4, 2017 Solve each system by substitution. Steps:
Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.
3-2: Solving Linear Systems
The Inverse of a Matrix Prepared by Vince Zaccone
Differential Equations
Systems of Linear Equations
Systems of Linear Equations
Coordinate Systems Prepared by Vince Zaccone
Solve the linear system.
3-2: Solving Linear Systems
Example 2B: Solving Linear Systems by Elimination
The student will be able to:
Matrices are identified by their size.
3-2: Solving Linear Systems
Differential Equations
Presentation transcript:

Systems of Linear Equations Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Next Step: Replace R2 with 2R1-3R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Next Step: Divide R2 by -7 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Next Step: Replace R1 with R1-R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Replace R1 with R1-R2 Next Step: Divide R1 by 3 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Replace R1 with R1-R2 Divide R1 by 3 The solution is right there in the matrix: x=-3, y=4 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where This is the same answer we got before: X=-3, y=4 *Note: You might be thinking that the inverse matrix way was a whole lot of work to get an answer that we just got faster using the other method. And you would be correct. However, it is a much more general method that will apply to more complicated problems in the future (that would not be so easily solved by the other methods). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: The solution is x= -2, y= 3 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: The solution is x= -2, y= 3 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The solution is x=1, y=2

Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: The solution is x=-4, y=4 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: The solution is x=-4, y=4 Solve for x and y: The solution is x=1, y=-10 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x and y: The solution is x=-4, y=4 Solve for x and y: The solution is x=1, y=-10 Solve for x and y: The solution is x=6, y=5 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve for x, y and z: The solution is x=0, y=-1, z=1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. We can write out the solution by choosing z as our parameter, call it t: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. We can write out the solution by choosing z as our parameter, call it t: If we separate into homogeneous and particular: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB