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Published byCandace Price Modified over 6 years ago

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Warm-Up

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Solving Systems of Equations

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Learning Targets l Refresher on solving systems of equations l Matrices –Operations –Uses –Reduced Row Echelon Form

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Solving Systems of Equations l There are multiple ways to solve systems of equations: –Graphing –Substitution (Equal Values Method) –Elimination

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Solve the System by Graphing

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Solve the System using Algebra

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Algebra Method cont.

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Matrix Equations l We have solved systems using graphing, but now we learn how to do it using matrices. This will be particularly useful when we have equations with three variables.

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Matrix Equation Before you start, make sure: 1. That all of your equations are in standard form. 2. The variables are in the same order (alphabetical usually is best). 3. If a variable is missing use zero for its coefficient.

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Setting up the Matrix Equation l Given a system of equations -2x - 6y = 0 3x + 11y = 4 l Since there are 2 equations, there will be 2 rows. l Since there are 2 variables, there will be 2 columns.

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l There are 3 parts to a matrix equation 1)The coefficient matrix, 2)the variable matrix, and 3)the constant matrix. Setting up the Matrix Equation

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-2x - 6y = 0 3x + 11y = 4 l The coefficients are placed into the coefficient matrix.

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-2x - 6y = 0 3x + 11y = 4 l Your variable matrix will consist of a column.

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-2x - 6y = 0 3x + 11y = 4 l The matrices are multiplied and represent the left side of our matrix equation.

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-2x - 6y = 0 3x + 11y = 4 l The right side consists of our constants. Two equations = two rows.

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-2x - 6y = 0 3x + 11y = 4 l Now put them together. We’ll solve it later!

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Create a matrix equation l 3x - 2y = 7 y + 4x = 8 l Put them in Standard Form. l Write your equation.

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3a - 5b + 2c = 9 4a + 7b + c = 3 2a - c = 12 Create a matrix equation

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l To solve matrix equations, get the variable matrix alone on one side. l Get rid of the coefficient matrix by multiplying by its inverse Solving a matrix equation

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l When solving matrix equations we will always multiply by the inverse matrix on the left of the coefficient and constant matrix. (remember commutative property does not hold!!)

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l The left side of the equation simplifies to the identity times the variable matrix. Giving us just the variable matrix.

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l Using the calculator we can simplify the left side. The coefficient matrix will be A and the constant matrix will be B. We then find A -1 B.

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l The right side simplifies to give us our answer. l x = -6 l y = 2 l You can check the systems by graphing, substitution or elimination.

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Advantages l Basically, all you have to do is put in the coefficient matrix as A and the constant matrix as B. Then find A -1 B. This will always work!!!

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Solve: l Plug in the coeff. matrix as A l Put in the const. matrix as B l Calculate A -1 B.

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Solve: l r - s + 3t = -8 l 2s - t = 15 l 3r + 2t = -7

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Working with Matrices on TI-83, TI-84 Source: Mathbits

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Explore: How many matrices does your calculator have? Use the right arrow key to move to MATH. Scroll down and find rref. We will use this key later. Use the right arrow key once more to highlight EDIT.

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Step 1: Go to Matrix (above the x -1 key)

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Step 2: Arrow to the right to EDIT to allow for entering the matrix. Press ENTER

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Step 3: Type in the dimensions (size) of your matrix and enter the elements (press ENTER).

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Step 4: Repeat this process for a different matrix..

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Step 5: Arrow to the right to EDIT and choose a new name.

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Step 6: Type in the dimensions (size) of your matrix and enter the elements (press ENTER).

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: Using Matrices to Solve Systems of Equations: 1. (using the inverse coefficient matrix) Write this system as a matrix equation and solve: 3x + 5y = 7 and 6x - y = -8 Step 1: Line up the x, y and constant values. 3x + 5y = 7 6x - y = -8

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Step 2: Write as equivalent matrices. Step 3: Rewrite to separate out the variables.

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Step 4: Enter the two numerical matrices in the calculator. Step 5: The solution is obtained by multiplying both sides of the equation by the inverse of the matrix which is multiplied times the variables.

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Step 6: Go to the home screen and enter the right side of the previous equation. The answer to the system, as seen on the calculator screen, is x = -1 and y = 2.

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Method 2 2. (using Gauss-Jordan elimination method with reduced row echelon form ) Solve this system of equations: 2x - 3y + z = -5 4x - y - 2z = -7 -x + 2z = -1

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Step 1: Line up the variables and constants 2x - 3y + z = -5 4x - y - 2z = -7 -x +0y + 2z = -1 Step 2: Write as an augmented matrix and enter into calculator.

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Step 3: From the home screen, choose the rref function. [Go to Matrix (above the x -1 key), move right→ MATH, choose B: rref]

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Step 4: Choose name of matrix and hit ENTER Step 5: The answer to the system, will be the last column on the calculator screen: x = -3 y = -1 z = -2.

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Method 2: Case 1: Unique solution Enter as a 3X4 matrix Diagonal is all ones so there is a solution:

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Method 2: Case 2: No solution Enter as a 3X4 matrix Last row: 0 0 0 1 No solution.

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Method 2: Case 3: Infinitely Many Solutions Enter as a 3X4 matrix Last row: 0 0 Infinitely Many Solutions

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Website to Visualize the Solutions http://www.cpm.org/flash/technology/3dsystems.s wf http://www.cpm.org/flash/technology/3dsystems.s wf

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For Tonight Intro to Matrices Worksheet

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