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Slide 1 - The Pedaled Boxcar Challenge

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					EXAMPLE 1     Add and subtract matrices


 Perform the indicated operation, if possible.
      3 0     –1 4     3 + (– 1)   0+4       2      4
 a. – 5 – 1 + 2 0 =      –5+2 –1+0 = –3            –1

     7 4  –2    5       7 – (– 2)          4–5        9 –1
 b. 0 –2 – 3 – 10 =         0–3      – 2 – (– 10) = – 3 8
    –1 6  –3    1      – 1 – (– 3)          6–1       2 5
EXAMPLE 2     Multiply a matrix by a scalar


 Perform the indicated operation, if possible.
        4–1     – 2(4) – 2(– 1) –8      2
 a. – 2 1 0 = – 2(1) – 2(0) = – 2       0
        2 7     – 2(2) – 2(7)   – 4 – 14


      –2 –8   –3
                                           + –3 8
                  8          4(– 2) 4(– 8)
 b. 4       +            =
       5 0     6 –5            4(5)   4(0)    6 –5

                           – 8 – 32   –3       8
                         =          +
                           20     0    6      –5
EXAMPLE 2   Multiply a matrix by a scalar


                           – 8 + (– 3)      – 32 + 8
                       =
                               20 + 6       0 + (– 5)

                           – 11 – 24
                       =
                             26   –5
EXAMPLE 3     Solve a multi-step problem


 Manufacturing
 A company manufactures small and large steel DVD
 racks with wooden bases. Each size of rack is
 available in three types of wood: walnut, pine, and
 cherry. Sales of the racks for last month and this
 month are shown below.
EXAMPLE 3     Solve a multi-step problem


 Organize the data using two matrices, one for last
 month’s sales and one for this month’s sales. Then
 write and interpret a matrix giving the average
 monthly sales for the two month period.
 SOLUTION
 STEP 1 Organize the data using two 3 X 2 matrices, as
        shown.
          Last Month (A)        This Month (B)
         Small      Large        Small     Large
 Walnut 125          100            95      114
 Pine     278        251           316      215
 Cherry 225          270           205      300
EXAMPLE 3     Solve a multi-step problem


 STEP 2 Write a matrix for the average monthly sales
        by first adding A and B to find the total sales
        and then multiplying the result by 1 .
                                             2
                    125 100       95       114
    1 (A + B) = 1   278 251    + 316       215
    2           2   225 270      205       300

                    220 214
             = 1    594 466
               2
                    430 570
EXAMPLE 3     Solve a multi-step problem


                  110 107
                = 297 233
                  215 285


  STEP 3 Interpret the matrix from Step 2. The
         company sold an average of 110 small walnut
         racks, 107 large walnut racks, 297 small pine
         racks, 233 large pine racks, 215 small cherry
         racks, and 285 large cherry racks.
EXAMPLE 4      Solve a matrix equation


 Solve the matrix equation for x and y.
        5x – 2      3   7     – 21    15
    3          +            =
         6 –4      –5 –y          3 – 24


 SOLUTION

 Simplify the left side of the equation.

     5x   –2    3    7       – 21   15     Write original
 3           +           =                 equation.
      6   –4   –5   –y          3 – 24
EXAMPLE 4        Solve a matrix equation


       5x + 3          5         – 21   15 Add matrices inside
   3                         =
             1     –4 –y            3 – 24 parentheses.

       15x + 9          15       – 21   15 Perform scalar
                             =
        3        – 12 – 3y          3 – 24 multiplication.

 Equate corresponding elements and solve the two
 resulting equations.
      15x + 9 = – 21          – 12 – 3y = -24
            x=–2                      y=4


  ANSWER
 The solution is x = – 2 and y = 4.
EXAMPLE 1     Describe matrix products


 State whether the product AB is defined. If so, give
 the dimensions of AB.
  a. A: 4 3, B: 3 2                  b. A: 3 4, B: 3 2


  SOLUTION
  a. Because A is a 4 3 matrix and B is a 3 2 matrix,
  the product AB is defined and is a 4 2 matrix.

 b. Because the number of columns in A (four) does
 not equal the number of rows in B (three), the
 product AB is not defined.
EXAMPLE 2     Find the product of two matrices




                     1 4                5 –7
    Find AB if A =           and B =
                     3 –2               9  6

   SOLUTION


  Because A is a 2 2 matrix and B is a 2 2 matrix, the
  product AB is defined and is a 2 2 matrix.
EXAMPLE 2      Find the product of two matrices



  STEP 1

  Multiply the numbers in the first row of A by the
  numbers in the first column of B, add the products,
  and put the result in the first row, first column of AB.

    1    4     5 –7         1(5) + 4(9)
                        =
    3   –2     9 6
EXAMPLE 2     Find the product of two matrices



  STEP 2

  Multiply the numbers in the first row of A by the
  numbers in the first column of B, add the products,
  and put the result in the first row, second column of
  AB.

    1    4   5 –7      1(5) + 4(9)     1( – 7) + 4(6)
                  =
    3   –2   9 6
EXAMPLE 2      Find the product of two matrices



  STEP 3

  Multiply the numbers in the second row of A by the
  numbers in the first column of B, add the products,
  and put the result in the second row, first column of
  AB.

    1    4   5 –7     1(5) + 4(9)       1( – 7) + 4(6)
                  =
    3   –2   9 6      3(5) + (– 2)(9)
EXAMPLE 2       Find the product of two matrices



  STEP 4

  Multiply the numbers in the second row of A by the
  numbers in the second column of B, add the
  products, and put the result in the second row,
  second column of AB.

    1    4   5 –7    1(5) + 4(9)          1( – 7) + 4(6)
                  = 3(5) + (– 2)(9)
    3   –2   9 6                         3( –7) + ( –2)(6)
EXAMPLE 2       Find the product of two matrices


  STEP 5

   1(5) + 4(9)         1(– 7) + 4(6)          41 17
                                         =
   3(5) + (–2)(9)      3(–7) + (–2)(6)        –3 –33
EXAMPLE 3         Use matrix operations


   Using the given matrices, evaluate the expression.

      4        3          –3     0        1     4
  A= –1       –2 ,B=        1   –2 , C = –3    –1
      2        0

  a. A(B + C)                                 b. AB + AC

   SOLUTION

                    4     3
  a. A(B + C) =    –1    –2       –3       0 + 1     4
                    2     0        1      –2   –3   –1
EXAMPLE 3       Use matrix operations



                 4     3        –2    4        –14         7
            =   –1    –2        –2   –3   =     6         2
                 2     0                       –4         8

               4       3   –3    0    4        3       1 4
 b. AB + AC = –1      –2    1   –2 + –1       –2      –3 –1
               2       0              2        0

                –9    –6        –5       13        – 14   7
                 1     4   +     5      –2 =        6     2
                –6     0         2       8         –4     8
EXAMPLE 4      Use matrices to calculate total cost


    Sports Two hockey
 teams submit equipment
 lists for the season as
 shown.

 Each stick costs $60, each
 puck costs $2, and each
 uniform costs $35. Use matrix
 multiplication to find the total
 cost of equipment for each
 team.
EXAMPLE 4     Use matrices to calculate total cost


   SOLUTION

  To begin, write the equipment lists and the costs
  per item in matrix form. In order to use matrix
  multiplication, set up the matrices so that the
  columns of the equipment matrix match the rows
  of the cost matrix.
EXAMPLE 4    Use matrices to calculate total cost


                         Equipment
                    Sticks Pucks Uniforms
  Women’s team       14     30    18
  Men’s team         16     25    20

                  Cost
                 Dollars
  Sticks           60
  Pucks            2
  Uniforms         35
EXAMPLE 4     Use matrices to calculate total cost


  The total cost of equipment for each team can be
  found by multiplying the equipment matrix by the
  cost matrix. The equipment matrix is 2 3 and the
  cost matrix is 3 1. So, their product is a 2 1
  matrix.

        14    30      18           60
        16    25      20           2
                                   35

   = 14(60) + 30(2) + 18(35) = 1530
     16(60) + 25(2) + 20(35)   1710
EXAMPLE 4    Use matrices to calculate total cost


 The labels for the product matrix are shown below.


                        Total Cost
                         Dollars
  Women’s team           1530
  Men’s team             1710



 ANSWER     The total cost of equipment for the women’s
            team is $1530, and the total cost for the
            men’s team is $1710.

				
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posted:3/21/2013
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