MAC1105-College Algebra. Chapter 5-Systems of Equations & Matrices

Transcription

1 MAC05-College Algebra Chapter 5-Systems of Equations & Matrices 5. Systems of Equations in Two Variables Solving Systems of Two Linear Equations/ Two-Variable Linear Equations A system of equations is composed of two or more equations considered simultaneously. The solution set of the system consists of all ordered pairs that make both equations true. I. Solving Systems of Equations Graphically A. The system has one solution if exactly one point of intersection exists in a system of two linear equations. B. There is no solution between two parallel lines (no common points). C. There are infinitely many solutions if two lines described are identical... same line (infinitely many common points). D. If a system of two linear equations in two variables has one solution, it is consistent. E. If a system has no solution, it is inconsistent. F. If a system of two linear equations in two variables has infinite number of solutions, it is dependent. Otherwise, it is independent. Example: Consider the system: x y, x y 7

2 Consider the system: x y, y x 8 II. Solving Systems of Equations Using the Substitution Example: Solve the system x y, 3x y 5 First, consider the x y equation and solve for y. y x Then, consider the 3x y 5equation and replace y by the x. So, 3x y 5 becomes 3x ( x) 5. So, 3x ( x) 5becomes 3x x 5. So, 3x x 5becomes 4x 5 or 4x 6.. So, x 4. Now, substitute 4 for x in either x y or 3x y 5, the result will be y 7.

3 Example: Solve the system x 5y 4, y 7 x. III. Solving Systems of Equations Using Elimination Example: Solve the System x y, x y 7 Try to eliminate one of the two variables by :. adding two equations if one of the two variables have same coefficients and opposite signs in two equations. OR. subtracting two equations if one of the two variables have same coefficients and same signs in two equations. So, x y + ( x y 7) 3x 0 9, so x 3 Then, use back-substitution to solve for y, into x y 7, so 3 y 7, so y 4. 3

4 Example: Solve the system x 3y, x 6y 5 Solve the system: 5 x 7 y 3 3x y 5 x y Solve the system: 4 5 x y 4 3 4

5 Solving A System of Linear Equations Using Determinants I. The solution of a x b y c a x b y c follows: may be written in determinant form as x c c a a b b b b y a a a a c c b b *If the determinant of the numerator is not zero and determinant of the denominator is zero, the system is inconsistent. If the determinant of numerator and denominator are both zero, the system is dependent. If the determinant of the denominator is not zero, there is a unique solution and the system is independent and consistent. II. Cramer's Rule: To solve a system of two linear equations in two variables. * To find the determinant of the numerator: a. For x, take the determinant of the denominator and replace the coefficients of x, a' s, by the corresponding constants, c' s. b. For y, take the determinant of the denominator and replace the coefficients of y, b' s, by the corresponding constants, c' s. III. Examples: Try solving this system of linear equations, using determinants: A. x 3 y 5x 4y 4 B. 3 x 4 y 0 6x 8y 5 5

6 IV. To solve a system of three linear equations with three variables using determinants: A. The general system of three equations in three variables is given as a x b y c z d a x b y c z d a x b y c z d The general solution may be written in terms of determinants as follows: x d b c d b c d b c a b c a b c a b c y a d c a d c a d c a b c a b c a b c z a b d a b d a b d a b c a b c a b c B. Examples: Try to solve:. 3x y 4z 5 x 5y 9 x 6y z 4 6

7 V. Solve a system of equations with a unique solution, using inverse of matrix: AX If A is an invertible matrix, the system of linear equations represented by B has a unique solution given by X A B Example: Use an inverse matrix to solve this system of equations: x 3y z 3x 3y z x 4y z 3 So, A 3 3, B 4 A solution X 0 A B the solution is: x, y, z 3x y z 0 Example: Find the solution of: x y z 5 4x 4y 3z 7

8 Matrices For linear systems involving three or more variables, the technique of using a rectangular array of real numbers is called a matrix (the plural form of matrix is matrices). I. Definitions The system x 3y 7, x 4y can be written as This rectangular array of numbers is called a matrix, more specifically an augmented matrix. 3 is the coefficient matrix of the system. 4 rows: (horizontal) column: (vertical) order: size; m x n square matrix: m = n;number of rows = number of columns main diagonal elements/entries: elements/entries at the location where position of row = position of column. Augmented Matrix:: For a system of equations, it is a table of numbers enclosed in brackets, where the rows represent the equations, all columns, but the last column hold the coefficients of the variables in the equations, and the last column is the right side of the equations. 8

9 Ex. x y z 3 x y z x 3y z *The main diagonal of an augmented matrix is the set of all augmented matrix entries that lie to the left of the vertical line and have identical row and column positions. *An augmented matrix is upper triangular if all entries below the main diagonal are zero. A system of equations is ready for solution by back-substitution when its augmented matrix is upper triangular and when all nonzero elements on the main diagonal are ones. Ex Main diagonal elements are:, 4, 6 Ex Main diagonal elements are:, 0 Ex This augmented matrix is Upper Triangular 9

10 S. Nunamker II. Elementary Row Operations: Operations on equations that change the look of the equations without changing the solution of the system. A. Interchange the positions of any two equations. B. Multiply an equation by a nonzero number. C. Add to one equation a nonzero constant times another equation. III. Gaussian Elimination The process of using elementary row operations to place zeros below the main diagonal of an augmented matrix. A. Pivot is a nonzero element in an augmented matrix that is used to transform elements below it to zero. Pivot must be nonzero. If an element on the main diagonal is zero when we need it as a pivot, we use the first elementary row operation to interchange the row containing the zero diagonal entry with any row below it that has a nonzero entry in the same column as the zero diagonal entry. Ex by interchanging the first and second row by multiplying the second row by / / R R 0

11 IV. Gauss-Jordan Reduction: Step : Use Gaussian elimination to transform an augmented matrix into the augmented matrix that is upper triangular matrix with all nonzero elements on the main diagonal set to one. Step : Beginning with the last column of the upper triangular matrix having a one on its main diagonal and progressing backward sequentially to the second column, use only the third elementary row operation to transform all elements in the upper triangular matrix above the main diagonal to zero. Complete all work on column before moving to another column, and apply all operations to the entire augmented matrix. Ex. x + y - z = -3 x y z -x - y + z = x + 3y - z = R R R R ( ) R R

12 R R ( ) R 3 R 3 This is in the upper triangular matrix form: This completes Step of Gauss-Jordan reduction. We now begin with column 3 and use elementary row operations to place zeros above the one on main diagonal. Once this is done, we move to column and place a zero above the one on the main diagonal in that column. We have: R ( ) R R Which eliminates z from the second equation, and R R R Which eliminates z from the first equation. R ( ) R R

13 The last step eliminates y from the first equation. The system of equations corresponding to this augmented matrix is x =, y =, z = 3, which also gives the solution to the original system without requiring any back-substitution. *If one of the equations associated with the augmented matrix in upper triangular form is false, then a system of linear equations has no solution. Example: Solve the system: x y 4z 3 x y 0z 6 3x 4z 7 The augmented matrix for this system: The goal is to find a row-equivalent matrix of the form: a b c 0 d e 0 0 f 3

14 R R ----> R R R --> R R 3 R 3 --> R R ---> *Work on elements in each column at a time, starting from the element/entry in the main diagonal, go R R 3 R 3 --->

15 R R ---> At this point, the system of equations that corresponds to the last matrix above is: x y 0z 6 () y 8z 3 () z (3) Now, we may back-substitute for z in equation () and solve for y: y 8( ) 3 y 4 3 y 7 Now, we back-substitute 7 for y and for z in equation ( ) and solve for x: x 7 0( ) 6 x x 9 6 x 3 5

16 *The last matrix is in row-echelon form. V. Row-Echelon Form To be in this form, a matrix must have the following properties:. If a row does not consist entirely of 0's, then the first nonzero element in the row is a (called a leading ).. For any two successive nonzero rows, the leading in the lower row is farther to the right than the leading in the higher row. 3. All the rows consisting entirely of 0's are at the bottom of the matrix. If a fourth property is also satisfied, a matrix is said to be in reduced row-echelon form: 4. Each column that contains a leading has 0's everywhere else (if every column that has a leading has zeros in every position above and below its leading. VI. Gauss-Jordan Elimination This method is named for Karl Friedrich Gauss and Wilhelm Jordan (84-899). This is the continuation steps following the Gaussian Elimination (as a an alternative to backsubstitution ) to solve the system of equations. Using previous example, at the end of Gaussian Elimination, 6

17 we have: We need to continue performing row-equivalent operations until we have a matrix in reduced row-echelon form. We need to work from the third column first, and then the second column R 3 R R ---> R3 R R --> R R R ---> At this point, we have x 3, y 7, z read directly from the reduced row-echelon matrix. 7

18 example; x 0y z 0x 3y z 6 x y 0z 9 Set up augmented matrix and solve for x, y, z 8