H.Alg 2 Notes: Day1: Solving Systems of Equations (Sections ) Activity: Text p. 116

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H.Alg 2 Notes: Day: Solving Systems of Equations (Sections 3.-3.3) Activity: Text p. 6 Systems of Equations: A set of or more equations using the same. The graph of each equation is a line. Solutions of systems of equations is a set of that makes each equation true. Using the Calculator. Solve equations for y. 2. Enter equations into y and y. 2 3. Graph. 4. 2 nd Calculate 5. #5 intersect 6. enter, enter, enter. x + 3y = 2 3x + 3y = -6 2. 4x + y = - -x + 3y = 0 3. 5x + 6y = -24-2x + 3y = 5 ) 3x + y = -9-3x - 2y = 2 2) x + 3y = 2-2x + 4y = 9 3) Jim can buy cd's at a local store for $5.49 each. He can buy the cd's at an online store for $3.99 each plus $6 for shipping. Find the number of cd's Jim can buy for the same amount at the two stores.

) 2x + 4y = -4 3x + 5y = -3 2) 3x - 2y = 5-6x + 4y = 7 3) 2x - 4y = 8 6x - 2y = 24 3) At Renaldi's Pizza, a soda and two slices of the pizza-of-the-day cost $0.25. A soda and four slices of the pizza-of-the-day cost $8.75. Find the cost of each slice. Graphing Systems of Inequalities ) 3x + 2y x + 4y > -2 2) 2 y > x + 2 3 2 y < x - 3 3) A home-based company produces both hand-knitted scarves and sweaters. The scarves take 2 hours of labor to produce, and the sweaters take 4 hours. The labor available is limited to 40 hours per week, and the total production capacity is 5 items per week. Write a system of inequalities representing this situation, where x is the number of scarves and y is the number of sweaters. Use the calculator to find possible answers. 2

Investigation: Using Matrices Lunch Lunch 2 Lunch 3 Cost per lunch $2.50 $.75 $2.00 Number sold 50 00 75. How much money did the cafeteria collect selling lunch? Lunch 2? Lunch 3? 2. How much money did the cafeteria collect selling all three lunches? 3. a. Write a x 3 matrix to represent the cost of the lunches. b. Write a 3 x matrix to represent the number of lunches sold. In order to multiply matrices, STEP : Make sure that the number of columns in the st matrix equals the number of rows in the 2nd matrix. Matrix A is 3 x 2. Matrix B is 2 x 4. The dimensions of AB is 3 x 4. Matrix A is 3 x 4. Matrix B is 2 x 4. Product AB is UNDEFINED. PRACTICE: Determine whether each matrix product is defined. If so, state dimensions. a) A B b) A B 2x5 5x4 x3 4x3 STEP 2: Multiply the elements of each row of the first matrix by the elements of each column in the second matrix. STEP 3: Add the products. Example: a b e f A = B c d = g h ae + bg af + bh AB = ce + dg cf + dh EXAMPLE #: EXAMPLE #2: EXAMPLE #3: 3 0 2 2 4 2 6 0 6 5 2 0 3 9 0 6 0 5 2 3 9 Multiplication of matrices is commutative. 6

EXAMPLE #5: EXAMPLE #6: Example #7: 4 a a 2b 6 b 3 5 6 2 2 3 a 2b 4 a 3 5b 3 5 6 2 Using Matrix Multiplication in Real Life Example #7: Two softball teams submit equipment lists for the season. Women's Team Men's Team 2 bats 5 bats 45 balls 38 balls 5 uniforms 7 uniforms Each bat costs $2, each ball costs $4, and each uniform costs $30. a. Write the equipment lists and the cost per item in matrix form. b. Use matrix multiplication to find the total cost of equipment for each team. Example #8: Two lacrosse teams submit equipment lists to their sponsors. Women's team: 5 sticks, 5 balls, and 6 uniforms. Men's team: 8 sticks, 22 balls, and 7 uniforms. Each stick costs $55, each ball cost $6, and each uniform costs $35. Use matrix multiplication to find the total cost of the equipment for each team. Example #9: Hector owns three fruit farms on which he grows apples, peaches, and apricots. He sells apples for $22 a case, peaches for $25 a case, and apricots for $8 a case. Number of Cases in Stock of Each Type of Fruit Farm Apples Peaches Apricots 290 65 20 2 75 240 90 3 0 75 0 a. Write an inventory matrix for the number of cases for each type of fruit for each farm. b. Write a cost matrix for the price per case for each type of fruit. c. Use matrix multiplication to find the total income for each farm. 7

D ay 7 Section 4.5 & 4.6 Identity and Inverse m atrices w ith D eterm inants Determinants Every square matrix with real number elements has a real number associated with it called its. We use to help us in solving equations. You can only find the determinant of SQUARE MATRICES (i.e. 2X2 and 3X3, etc). The determinant is denoted by straight bars around the matrix instead of brackets. Ex: 2 3 4 5 the straight bars mean find the determinant! You find the determinant differently for different size matrices. For 2X2 matrices it is quicker and easier to find them by hand than on the calculator. Lets see why. Finding the determinant of a 2X2 Matrix Finding the determinant of a 2X2 matrix can be summed up in four words: CROSS MULTIPLY AND SUBTRACT! You will always be subtracting 2 numbers, so when I see a 2X2 determinant I always write: - and then I fill in the blanks. To fill in the blanks we cross multiply. Observe. a b c d = ad bc Find the 2X2 determinants! 2 3 Ex : = 9 8 - = Ex 2: 2 4 5 7 = - = Ex 3: 4 6 = 2 3 - = Ex 4: 3 4 = = Be careful! This has minus a negative! 6 8 8

Ex 5: 2 2 6 4 = = Determinant of a 3 x 3 matrix. To find the determinant of a 3 x 3 matrix, use diagonals.. Begin by writing the first two columns on the right side of the determinant. 2. Draw diagonals from each element of the top row of the determinant downward to the right. Find the product of the elements on each diagonal. Then draw diagonals from the elements in the third row of the determinant upward to the right. Find the product of the elements on each diagonal. 3. To find the value of the determinant, add the products of the first set of diagonals and then subtract the products of the second set of diagonals. The answer is the determinant. Evaluate: 3 3 4 2 0 5 2 2 7 2 4 5 2 0 Solve for x: 2 3 6 9 x 4 = 403 0 2x Finding the area of a triangle using determinants: The area of the triangle in a coordinate plane can be found using the following formula: x y A = ± x2 y2 2 x y 3 3 9

Find the area of a triangle with vertices (3, 6), (-, 2) and (4, 9). Find the area of a triangle with vertices (6, 6), (6, 6) and (4, ). Day 8: Identity Matrices Find each product: 6 8 0 = 4 3 0 What pattern do you see? 0 0 6 0 = 0 9 3 4 6 0 0 3 2 0 0 = 2 29 6 0 0 The square n x n matrix with one s along the main is the identity matrix for multiplication. (Any number times the identity element is itself.) The symbol for the identity matrix is I. I _ = I _ = _ 2x 2 3x 3 Multiplicative Inverse What is a multiplicative inverse? The number we multiply a number by to get (the identity). 5. / 5 = / 5 is the inverse of 5. For matrices, this means the matrix we multiply by another matrix to get the identity matrix! Yet again, only SQUARE MATRICES have inverses since only SQUARE MATRICES have identities. How do we find the inverse for a 2X2 matrix?. Find the determinant of the matrix. 2. Switch the numbers in the matrix accordingly. 3. Multiply the matrix by the scalar /determinant Formula: For Matrix A = a b c d L NM O QP the inverse A- = A L NM d c b a O QP Notice inside the matrix a and d switch places and b and c change signs. 0

Let s do a problem: Ex: Find the inverse of A = L NM 4 2 6 2 O QP = Determinant = -8 - -2 = 4 L NM Switch = 2 2 6 4 O QP Multiply = 2 2 = 4 6 4 Summary:. Suppose A is a square matrix and X is another square matrix such that. 2. X is the of A. 3. We write the inverse of A as, so = and = 4. Some square matrices have no if the determinant is zero. Using inverse matrices to solve matrix equations If A, B, and X are matrices such that AX = B, and matrix A has an inverse, you can multiply each side of the equation by A - to find X. AX = B (A - A)X = A - B x = A - B EXAMPLES: SOLVE EACH EQUATION FOR X.. 2 5 2 x = 3 2 2. 3 4 0 22 4 5 x = 0 28

Day2 ~ Honors Algebra: Word Problems in two variables A. How to solve a standard word problem using 2 variables and 2 equations:. Write a let statement to represent the variables. 2. Write 2 equations by stating the relationships between the variables. 3. Solve the system by: elimination, substitution, or graphing. 4. Check your answer. *IF you have 2 variables, you need 2 equations. B. Distance Problems *Remember: Distance = x Tailwind - wind blowing in the direction as the plane Headwind - wind blowing in the direction as the plane Upstream - the current Downstream - the current. A coal barge on the Ohio River travels 24 miles upstream in 3 hours. The return trip takes the barge only 2 hours. Find the rate of the barge in still water. Let b = rate of the barge in still water Let c = the rate of the current rate time distance rt = d Downstream Upstream 2. With the wind, a plane few 400 km in 4 hours. On the return trip, the pilot was forced to land after.5 hours, having traveled only 450 km. Find the rate of the plane in still air and the rate of the wind. Let p = the rate of the plane in still air Let w = the rate of the wind rate time distance rt = d With the wind Against the wind 3

Honors Algebra 2: Word Problems in two Variables Mixture: (Price x Amount = Total) Central Perk Coffee shop sells a cup of expresso for $2.00 and a cup of cappuccino for $2.50. On Friday, Rachel sold 30 more cups of cappuccino than expresso, and she sold $78.50 worth of espresso and cappuccino. How many cups of each were sold? Type Price Amount Total Chemistry: (Percentage x Amount = Total) Cara is preparing an acid solution. She needs 200 milliliters of 48% concentration solution. Cara has 60% and 40% concentration solutions in her lab. How many milliliters of 40% acid solution should be mixed with 60% acid solution to make the required amount of 48% acid solution? Solution Percentage Amount Total Interest: (Principal x rate = Interest) Last year Zach received $469.75 interest from two investments. The interest rates were 7.5% on one account and 8% on the other. The total amount invested was $6000. How much was invested at each rate? Account Principal Rate Total Chemistry: (Percentage x Amount = Total) Snapp s Lab needs to make 000 gallons of 34% acid solution. The only solutions available are 25% acid and 50% acid. How many gallons of each solution should be mixed to make the 34% acid solution? Solution Percentage Amount Total 4