# 10.1 Systems of Linear Equations: Substitution and Elimination

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1 10.1 Systems of Linear Equations: Substitution and Elimination What does it mean to be a solution to a system of equations? - It is the set of all ordered pairs (x, y) that satisfy the two equations. You can have one solution, multiple solutions, infinite solutions or no solutions. Three methods of solution Graphing: Graph the two equations and find the intersection. Example Solve the following system of equations by graphing. { x y = 0 (1) 3x y = 0 () Substitution Solve one of the equations for x or y and substitute into the other equation. Example Use substitution to solve the following system of equations. { x y = 0 (1) 3x y = 0 () In this case we solve equation (1) for x and substitute into equation (). 1

2 Elimination Multiply the equations by appropriate constants so that, when you add the two equations, one variable cancels. Example Solve by elimination. { x+7y = 1 3x 5y = 10 Example Solve by any method. { 6x+5y = 3 x 5 6 y = 7

3 Example Solve by any method. { 3x+y = x 3y = 6 Example Solve by any method. { x+3y = 18 5x y = 11 3

4 Example Solve by any method. { 3 x 1 5 y = 8 x+3y = Multivariable Linear Systems We saw in section that we could solve systems of equations using the method of substitution. We can do the same thing for equations with three variables. Example In this first example we will look at a method known as back-substitution. In this method we know the value of one of the variables and we can use that to solve for the other two by substituting into the other equations. 5x 8z = 5y 5z = 10 z = 4 4

5 We saw in section that we could solve systems of equations using the method of elimination. We can do the same thing for equations with three variables. Example x+4y +z = 1 x y 3z = x+ y z = 1 If we want to solve these three equations we need a more complicated set of rules. We want to be able to do things which don t change the value of the system. It turns out that we can do three things that we call row operations. 1. Interchange two equations.. Multiply an equation by a nonzero constant. 3. Add a multiple of one equation to a multiple of any other equation and replace either with the result. We will start by using these row operations to eliminate one of the variables from two of the equations. 5

6 These equations don t always have nice neat answers. Example x 3y +6z = 6 x+y z = 5 5x 8y +13z = 7 6

7 10. Matrices and systems of equations We saw in sections 10.1 that we could solve systems of equations using the method of elimination. We saw that if we wanted to solve systems of three equations we needed a more complicated set of rules that we call row operations. 1. Interchange two equations.. Multiply an equation by a nonzero constant. 3. Add a multiple of one equation to a multiple of any other equation and replace either with the result. We don t want to have to do these calculations with all these variables so we use Matrices. An m n matrix is a rectangular array with m rows and n columns that looks like: a 11 a 1 a 1n A = a 1 a a m1 a m a mn Each entry a ij is a number. Example A = 4 4 A = A = 4 A = x 1 matrix x 1 matrix 1 x matrix x matrix Q. How is this going to help us solve systems of equations? A. We will take a system of equations and write it as an augmented matrix which is easier to solve. Example Write the system of equations as an augmented matrix: x 8y +5z = 8 7x 15z = 38 3x y 8z = There are two parts to this matrix: The coefficient matrix and the constant matrix. 7

8 Since this matrix simply represents a set of equations we can perform the same operations that we used to solve the systems in section We will reduce the matrix with elementary row operations. 1. Interchange two rows.. Multiply a row by a nonzero constant. 3. Add a multiple of one row to a multiple of any other row and replace either with the result. When we are done with this process we MUST have the matrix in reduced row echelon form a b c We want it in this form because now we can covert back into a system of equations and we get x = a the answers immediately: y = b z = c We ll begin with a small example: Example { 3x y = x+y = 10 8

9 If we have more rows and columns the procedure is the same but we have to work longer. Example Solve using Gauss Jordan Elimination: x y +z = 3 3x+y z = 7 x 3y +z =

10 Example Solve using Gauss Jordan Elimination: x+3z = 3 4x 3y +7z = 5 8x 9y +15z = Example Solve using Gauss Jordan Elimination: { x 3y = 4x+6y = 9 10

11 10.4 Matrix Algebra Addition and Subtraction Multiplication by a number Multiplication by another matrix Addition and Subtraction Q. What does it mean for two matrices to be equal? A. It means they are the same size and have the EXACT same entries. We can only add and subtract matrices that are the same size. Q. How do we add matrices? A. We add corresponding entries. a b c d e f + g h = (a+e) (b+f) (c+g) (d+h) Example = What about standard addition properties? Matrix addition is: 1. Commutative: A+B = B +A.. Associative: (A+B)+C = A+(B +C) Definition The Zero Matrix is a matrix with all entries zero. We often use 0 to represent it. Example = =

12 The negative of a matrix M is M. It is the matrix whose entries are the negative of the entries in M. So now we can subtract: Example Example Multiplication by a number Multiply every entry in the matrix by the number. Example = 5( 7) 5(3) 5(0) 5(9) 5(4) 5( 5) 5(6) 5() Example A = Find A B and 4B A B =

13 Multiplication of two matrices An n 1 matrix multiplied by a 1 n matrix is the 1 1 matrix given by: a1 a a n b 1 b. = a 1 b 1 +a b + +a n b n Example b n = ( 1)()+(0)(3)+(3)(4)+()( 1) Larger Matrices If A is an n m matrix and B is a p m matrix then the matrix product of A and B, AB, is an m n matrix whose i th row and j th column entry is the real number obtained from multiplying the i th of A by the j th column of B. THE MUMBER OF COLUMNS OF A MUST BE THE SAME AS THE NUMBER OF ROWS OF B. 1 1 Example =

14 3 0 Example A = Find BA and AB B = Writing a system of equations in matrix form. Example Write the system of equations as the matrix equation AX = B where A is the x 1 coefficient matrix, B is the constants matrix and X = x x 3 x 1 +3x 3 = 3 4x 1 3x +7x 3 = 5 8x 1 9x +15x 3 = 9 14

15 Example Write the system of equations as the matrix equation AX = B where A is the x 1 coefficient matrix, B is the constants matrix and X = x x 3 x 1 x +x 3 = 0 x +4x 3 = 13 3x 1 +4x 5x 3 = 14 15

16 Finding the Inverse of a Square Matrix Definition 10.. The Identity Matrix is a square matrix with 1 s along the main diagonal and zeros everywhere else. Example I = , I = We call it the identity matrix because it behaves like 1 in multiplication. Example Example a b c d e f a b c d a b = c d = a b c d a b c d e f = a b c d Definition If M is a square matrix and if there exists M 1 such that MM 1 = I and M 1 M = I then M 1 is the Multiplicative Inverse of M. We often simply call it The Inverse of M. Example A = other = and A 1 = 3 1. Show that these are inverses of each NOT ALL SQUARE MATRICES HAVE INVERSES. For example 1 4 Q. How do we know if an inverse exists for A and how do we find one if it does? A. We perform Gauss Jordan Elimination on the augmented matrix A I until it looks like I A 1 16

17 1 0 Example A = 3 1 Row reduce this matrix: Find A 1 Example A = Find A 1 17

18 3 9 Example A = 6 Row reduce this matrix: Find A 1 Example M = Find M 1 =

19 Solving a matrix equation Suppose we have a system of equations a 1 x 1 +a x +a 3 x 3 = a 4 b 1 x 1 +b x +b 3 x 3 = b 4 c 1 x 1 +c x +c 3 x 3 = c 4 where a i, b i, and c i are real numbers and x 1,x,x 3 are variables. Then we can write the coefficient matrix a 1 a a 3 A = b 1 b b 3 c 1 c c 3 and (IF it exists) we can find the inverse matrix A 1. The original system can be written in matrix form: a 1 a a 3 b 1 b b 3 c 1 c c 3 } {{ } A x 1 x x 3 }{{} X = a 4 b 4 c 4 }{{} b and we end up with an equation of the form AX = b. If this were an algebraic equation where A and b were numbers we could easily solve this by dividing on both sides by A. WE CAN T divide matrices. What we can do with matrices is to multiply by the inverse of A. Then we get something that looks like this AX = b A 1 AX = A 1 b IX = A 1 b X = A 1 b The nice thing about solving an equation this way is that now we can easily solve many problems that have the same A but different b with one simple matrix multiplication. 19

20 Example Solve 3x 1 x +x 3 = 1 x 1 +x = 1 x 1 +x 3 = 1 by writing the equation in matrix form as AX = b and multiplying by A 1. Example Solve 3x 1 x +x 3 = 1 x 1 +x = x 1 +x 3 = 3 0

21 Example Solve x 1 x = x 1 +5x = 5 and x 1 x = 4 x 1 +5x = 1 by writing the equations in matrix form as AX = b and multiplying by A 1. 1

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