Archive for June, 2014

Write the augmented matrix for a system

Write the augmented matrix for a system

 

 To begin, we write an augmented matrix for the system. An augmented matrix has a vertical bar that separates the columns of the matrix into two groups.

 

For example, to solve the system

 

x  –  3y =1

2x +y = –5,

 

start with the augmented matrix

 

\begin{bmatrix}  1 &  -3& \left.\begin{matrix}  &  & \\  &  &  \end{matrix}\right|+1\\  2& 1 & \left.\begin{matrix}  &  & \\  &  &  \end{matrix}\right|-5  \end{bmatrix}. Augmented matrix

Define a matrix

Define a matrix

 

An ordered array of numbers such as

\begin{bmatrix}  2 & 3 &4\\  3& 4 & 5  \end{bmatrix}

 

Is called a matrix. The numbers are called elements of the matrix. Matrices  (the plural of matrix) are named according to the number of rows and columns they contain. The rows are read horizontally, and the columns are read vertically.

For example, the first row in the preceding matrix is 2 3 4 and the first column is  \begin{bmatrix}  2& \\  3&  \end{bmatrix}. This matrix is a 2 3 3 (read “two by three”) matrix because it has 2 rows and 3 columns. The number of rows is given first, and then the number of columns.

 

Solving a Problem Involving Prices

Solving a Problem Involving Prices

 

At Panera Bread, a loaf of honey wheat bread costs $2.59, a loaf of sunflower bread costs $2.99, and a loaf of French bread costs $3.29. On a recent day, three times as many loaves of honey wheat were sold as sunflower. The number of loaves of French bread sold was 5 less than the number of loaves of honey wheat sold. Total receipts for these breads were $66.07. How many loaves of each type of bread were sold? (Source: Panera Bread menu.)

 

Step 1:-  Read the problem again. There are three unknowns in this problem.

 

Step 2:-  Assign variables to represent the three unknowns.

Let x = the number of loaves of honey wheat,

y = the number of loaves of sunflower,

and z = the number of loaves of French bread.

 

Step 3:-  Write a system of three equations. Since three times as many

loaves of honey wheat were sold as sunflower,

 

x = 3y , or x – 3y = 0. (1)

Also,

Number of loaves of French bread = z

equals =

5 less than the number of loaves of honey wheat.

So

x – z = 5. (2)

multiplying the cost of a loaf of a each kind of bread by  the number of loaves of that kind sold adding gives the total receipt.

2.59x + 2.99y + 3.29z = 66.07

 

Multiply each side of this equation by 100 to clear it of decimals.

259x + 299y + 329z = 6607 (3)

 

Step 4:-  Solve the system of three equations .

 

Step 5:-  State the answer. The solution set is {(12, 4, 7)}, meaning that 12 loaves of honey wheat, 4 loaves of sunflower, and 7 loaves of French bread were sold.

 

Step 6:-  Check. Since 12 = 3 .4, the number of loaves of honey wheat is three times the number of loaves of sunflower. Also, 12 – 7 = 5, so the number of loaves of French bread is 5 less than the number of loaves of honey wheat. Multiply the appropriate cost per loaf by the

number of loaves sold and add the results to check that total receipts were $66.07.

 

Solving an Applied Problem by Writing a System of Equations

 

Solving an Applied Problem by Writing a System of Equations

 

Step 1:-  Read the problem, several times if necessary, 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 an Inconsistent System with Three Variables

Solve special systems

 

 Linear systems with three variables may be inconsistent or may include dependent equations. The next examples illustrate these cases.

 

Solving an Inconsistent System with Three Variables

 

Solve the system.

2x – 4y + 6z = 5  (1)

-x + 3y – 2z = -1  (2)

x  – 2y + 3z = 1  (3)

 

eliminate x by adding equations (2) and (3) to get the equation

 

y + z = 0.

 

Now , eliminate x again, using equation (1) and  (3).

 

-2x + 4y – 6z = -2 Multiply each side of (3) by -2.

2x – 4y + 6z = 5 (1)

0  =  3   False

The resulting false statement indicates that equations (1) and (3) have no common solution. Thus, the system is inconsistent and the solution set is ø. The graph of this system would show these two planes parallel to one another.

 

Solve linear systems where some of the equations have missing terms

Solve linear systems where some of the equations have missing terms

 

If a linear system has an equation missing a term or terms, one elimination step can be omitted.

Example

Solve the system

6x – 12y = -5 (1)

8y + z = 0 (2)

9x – z = 12 (3)

Since equation (3) is missing the variable y, a good way to begin the solution is to eliminate y again using equation (1) and (2).

12x – 24y = – 10 Multiply each side of (1) by 2.

24y + 3z = 0 Multiply each side of (2) by 3.

12x + 3z = -10 Add. (4)

 

Use this result, together with equation (3), 9x – z = 12,  to eliminate z. Multiply equation (3) by 3.

This gives

27x – 3z = 36 Multiply each side of (3) by 3.

12x + 3z = – 10 (4)

39x = 26          Add.

 

x =  \fra{26}{39}c = \frac{2}{3}.

 

Substituting into equation (3) gives

9x – z = 12 (3)

9(\frac{2}{3}) – z = 12 Let x = \frac{2}{3}.

Substituting  -6 for z in equation (2) gives

8y + z = 0 (2)

8y – 6 = 0 Let z = – 6.

8y = 6

y = \frac{3}{4}.

 

Thus, x = \frac{2}{3}, y = \frac{3}{4}, and z  = -6. Check these values in each of the original equations of the system to verify that the solution set of the system is {(\frac{2}{3},\frac{3}{4}, -6 )}

 

Solving a Linear System in Three Variables

 

Solving a Linear System in Three Variables

 

Step 1:- Eliminate a variable. Use the elimination method to eliminate any variable from any two of the original equations. The result is an equation in two variables.

 

Step 2:- Eliminate the same variable again. Eliminate the same variable from any other two equations. The result is an equation in the same two variables as in Step 1.

 

Step 3:-  Eliminate a different variable and solve. Use the elimination method to eliminate a second variable from the two equations in two variables that result from Steps 1 and 2. The result is an equation in one variable that gives the value of that variable.

 

Step 4:-  Find a second value. Substitute the value of the variable found in Step 3 into either of the equations in two variables to find the value of the second variable.

 

Step 5 :- Find a third value. Use the values of the two variables from Steps 3 and 4 to find the value of the third variable by substituting into any of the original equations.

 

Step 6:-  Check the solution in all of the original equations. Then write the solution set.

 

 

Graphs of Linear Systems in Three Variables

Graphs of Linear Systems in Three Variables

 

a. The three planes may meet at a single, common point that is the solution of the system. See Figure  (a).

 

b. The three planes may have the points of a line in common so that the infinite set of points that satisfy the equation of the line is the solution of the system. See Figure  (b).

 

 

c. The three planes may coincide so that the solution of the system is the set of all points on a plane. See Figure  (c).

 

d. The planes may have no points common to all three so that there is no solution of the system. See Figures  (d), (e), and (f ).

 

graphs

 

 

Special Cases of Linear Systems

Special Cases of Linear Systems

 

If both variables are eliminated when a system of linear equations is solved, then

 

a. There is no solution if the resulting statement is false;

 

b. There are infinitely many solutions if the resulting statement is true.

 

Solving a Linear System by Elimination

Solving a Linear System by Elimination

 

Step 1:-  Write both equations in standard form Ax + By = C.

 

Step 2:-  Make the coefficients of one pair of variable terms opposites.

Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x– or y-terms is 0.

 

Step 3:-  Add the new equations to eliminate a variable. The sum should be an equation with just one variable.

 

Step 4:-  Solve the equation from Step 3 for the remaining variable.

 

Step 5:-  Find the other value. Substitute the result of Step 4 into either of the original equations and solve for the other variable.

 

Step 6:-  Check the solution in both of the original equations. Then write the solution set.