 # Systems and Matrices (Chapter5)

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Systems and Matrices (Chapter5)
College Algebra Systems and Matrices (Chapter5) 1

5.1 Systems of Linear Equations
After completing this Section, you should be able to: Know if an ordered pair is a solution to a system of linear equations in two variables or not. Solve a system of linear equations in two variables by graphing. Solve a system of linear equations in two variables by the substitution method. Solve a system of linear equations in two variables by the elimination by addition method. Solve a system of linear equations in three variables by the elimination method.

Linear Systems Any equation of the form
a1x1 + a2x2 +    + anxn = b, for all real numbers a1,a2,…,an (not all of which are 0) and b, is a linear equation or a first-degree equation in n unknowns. A set of equations is called a systems of equations. The solutions of a system of equations must satisfy every equation in the system. If all the equations in a system are linear, the system is a system of linear equations, or a linear system.

Solution of a System There are three possible outcomes that you may encounter when working with these systems: one solution no solution infinite solutions

One Solution Consistent
The graphs of two equations intersect a single point. The coordinates of this point give the only solution of the system. A consistent system is a system that has at least one solution.

No Solution Inconsistent The graphs are distinct parallel lines.
The equations are independent. That is, there is no solution common to both equations. An inconsistent system is a system that has no solution.

Infinite Solutions Dependent The graphs are the same line.
Any solution of one equation is also the solution of the other. Thus there are infinite number of solutions. The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other equation.  In other words, they end up being the same line.

Three Ways to Solve Systems of Linear Equations in Two Variables
There are three ways to solve systems of linear equations in two variables: graphing  substitution method elimination method .

Solve by Graphing Example : Solve the system of equation by graphing.
The solution to this system is (2, 1).

Substitution Method In a system of two equations with two variables, the substitution method involves using one equation to find an expression for one variable in terms of the other, then substituting into the other equation of the system. Example: Solve the system. 4x + 2y = (1) 3x  7y =  (2)

Solution 4x + 2y = 8 (1) 3x  7y = 11 (2)
Begin by solving one of the equations for one of the variables. 4x + 2y = 8 2y = 4x + 8 y = 2x (3) Now replace y with 2x + 4 in the second equation and solve for x. 3x  7(2x + 4) = 11 3x + 14x  28 = 11 17x = 17 x = 1 Replace x with 1 in equation (3) to obtain y = 2(1) + 4 = 2. The solution of the ordered pair is (1,2). Check the solution in both equations (1) and (2). 4x + 2y = 8 4(1) + 2(2) = 8 8 = 8 3x  7y = 11 3(1) 7(2) = 11 11 =  11

Solve by the Elimination by Addition Method
The elimination method uses multiplication and addition to eliminate a variable from one equation. To eliminate a variable, the coefficients of that variable in the two equations must be additive inverses. To achieve this, we use properties of algebra to change the system to an equivalent system, one with the same solution set.

Equivalent Systems Transformations of a Linear System
Interchange any two equations of the system. Multiply or divide any equation of the system by a nonzero real number. Replace any equation of the system by the sum of that equation and a multiple of another equation in the system.

Example Solve the system using the elimination method. 6x + 2y = 4
If we multiply the first equation by 5 and the second equation by 3, we will be able to eliminate the x variable. 30x + 10y = Substituting: x + 2y = 4 30x  21y = x + 2(4) = 4  11y = x  8 = 4 y =  x = 12 The solution is (2,  4) x = 2

Solving an Applied Problem by Writing a System of Equations
Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3 Write a system of equations that relates the unknowns. Step 4 Solve the system of equations. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem.

Solving Systems with Three Unknowns (Variables)
The graph of a linear equation in three unknowns requires a three-dimensional coordinate system. Some of the possible intersections of planes representing three equations are shown below.

Systems of Three Equations with Three Variables
To solve a system with three unknowns, first eliminate a variable from any two of the equations. Then eliminate the same variable from a different pair of equations. Eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Find the values of the remaining variables by substitution. Solutions of the systems are written as ordered triples. Example: Solve the system. 3x + 9y + 6z = 3 (1) 2x + y  z = (2) x + y + z = (3)

Solution 3x + 9y + 6z = 3 (1) 2x + y  z = 2 (2) x + y + z = 2 (3)
Eliminate z by adding equations (2) and (3) to get 3x + 2y = 4 (4) To eliminate z from another pair of equations, multiply both sides of equations (2) by 6 and add the result to equation (1). 3x + 9y + 6z = 3 12x + 6y  6z = 12 (1) 15x + 15y = 15 (5) To eliminate x from the equations (4) and (5), multiply both sides of equation (4) by 5 and add the result to equation (5). Solve the new equation for y. 15x  10y = 20 15x + 15y = 15 5y = 5 y = 1

Solution continued Using y = 1, find x from equation (4) by substitution. 3x + 2(1) = 4 (4) x = 2 Substitute 2 for x and 1 for y in equation (3) to find z. 2 + (1 ) + z = (3) z = 1 Verify that the ordered triple (2, 1, 1) satisfies all three equations in the original system. The solution set is {(2, 1, 1)}.

Using Curve Fitting to Find an Equation Through Three Points
Example: Find the equation of the parabola y = ax2 + bx + c that passes through (2,4), (1, 1), and (2,5). Solution: Since the three points lie on the graph of the equation y = ax2 + bx + c, they must satisfy the equation. Substituting each ordered pair into the equation gives three equations with three variables. 4 = a(2)2 + b (2) + c or 4 = 4a + 2b + c (1) 1 = a(1)2 + b(1) + c or 1 = a  b + c (2) 5 = a(2)2 + b(2) + c or 5 = 4a  2b + c (3)

Using Curve Fitting to Find an Equation Through Three Points continued
This system can be solved by the elimination method. First eliminate c using equations (1) and (2). 4 = 4a + 2b + c (1) 1 = a + b  c 4 = 3a + 3b (4) Now, use equations (2) and (3) to eliminate the same variable (c). 1 = a  b + c (2) 5 = 4a + 2b  c 4 = 3a + b (5) Solving systems of equations (4) and (5) in two variables by eliminating a. 3 = 3a + 3b (4) 4 = 3a + b (5) 1 = b Find a by substituting for b in equation (4), which is equivalent to 1 = a + b. 1 = a + b 1 = a

Using Curve Fitting to Find an Equation Through Three Points continued
Finally, find c by substituting a = and b = in equation (2). An equation of the parabola is

5.2 Matrix Solution of Linear Systems

Definitions A matrix is a rectangular array of numbers enclosed in brackets. Each number is called an element of the matrix. The size of a matrix is determined by the number of row (horizontal) and columns (vertical).

Definitions continued
Linear System Augmented matrix columns rows

Matrix Row Transformations
For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system. 1. Interchange any two rows. 2. Multiply or divide the elements of any row by a nonzero real number. 3. Replace any row of the matrix by the sum of the elements of that row and a multiple of the elements of another row.

Using the Gauss-Jordan Method to Put a Matrix into Diagonal Form
Step 1 Obtain 1 as the first element of the first column. Step 2 Use the first row to transform the remaining entries in the first column to 0. Step 3 Obtain 1 as the second entry in the second column. Step 4 Use the second row to transform the remaining entries in the second column to 0. Step 5 Continue in this manner as far as possible.

Example Solve the system 5x  3y = 14 4x + y = 18
Write the augmented matrix. Work with the columns and rows so that it is transformed into the form

Example continued Multiply row 1 by 1/5 to get a 1 in the first position. Introduce 0 in the second row by multiplying each element of R1 by 4 and adding to R2. Obtain 1 in the second row, second column by multiplying each element of the second row by 5/17.

Example continued Finally, to get 0 in the first row, second column multiply each element of the second row by 3/5 and add to the first row. This last matrix corresponds to the system x = 4 y = 2 that has the solution set (4, 2). Check the solution in both equations of the original system.

Example Solve the system.
1 in the first row, first column. Introduce 0 in the second row of the first column by multiplying each element of row 1 by 2 and adding to row 2.

Example continued To change the third element in the first column to 0, multiply the first row by 1 and add. Use the same procedure to transform the second and third columns. For both of these columns, perform the additional step of getting 1 in the appropriate position of each column. Do this by multiplying the elements of the row by the reciprocal of the number in that position.

Example continued The solution set is (1, 3, 2).

Special Systems When a row consists entirely of 0’s, the equations are dependent and the system is equivalent.

Special Systems continued
When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, the last row corresponds to the false equation 0 = 9, so we know the original system has no solution.

Determinant Solution of Linear Systems
5.3 Determinant Solution of Linear Systems Objectives Evaluate 2 by 2 and 3 by 3Determinants Use Cramer’s Rule to Solve a System of Two or Three Equations With Two or Three Variables Know Properties of Determinants

Determinants Every n  n matrix A is associated with a real number called the determinant, of A, written |A|. The determinant of a 2  2 matrix is defined as follows. Note: Matrices are enclosed with square brackets, while determinants are denoted with vertical bars.

Example Let Find |A|.

Determinant of a 3  3 Matrix

A 3 by 3 determinant is symbolized by

When evaluating a determinant, you can expand across any row or down any column you choose.

Cofactor Let Mij be the minor for the element aij in an n  n matrix. The cofactor if aij written Aij, is Find the cofactor of the element (2). The cofactor is

Finding the Determinant of a Matrix
Multiply each element in any row or column of the matrix by its cofactor. The sum of these products give the value of the determinant. Example Evaluate expanding by the third row.

Example continued Now find the cofactor of each element of these minors. Find the determinant by multiplying each cofactor by its corresponding element in the matrix and finding the sum of these products.

End of the Lecture Let Learning Continue