Add and subtract matrices EXAMPLE 1 Add and subtract matrices Perform the indicated operation, if possible. 3 0 – 5 – 1 a. – 1 4 2 0 + 3 + (– 1) 0 + 4 – 5 + 2 – 1 + 0 = = 2 4 – 3 – 1 – 2 5 3 – 10 –3 1 7 4 0 –2 –1 6 b. – 7 – (– 2) 4 – 5 0 – 3 – 2 – (– 10) – 1 – (– 3) 6 – 1 = 9 –1 – 3 8 2 5 =

EXAMPLE 2 Multiply a matrix by a scalar Perform the indicated operation, if possible. a. 4 – 1 1 0 2 7 – 2 – 2(4) – 2(– 1) – 2(1) – 2(0) – 2(2) – 2(7) = – 8 2 – 2 0 – 4 – 14 = b. 4 – 2 – 8 5 0 – 3 8 6 – 5 + 4(– 2) 4(– 8) 4(5) 4(0) – 3 8 6 –5 = + – 8 – 32 20 0 – 3 8 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. Walnut Pine Cherry 125 100 278 251 225 270 95 114 316 215 205 300 Last Month (A) This Month (B) Small Large Small Large

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 (A + B) = 1 2 125 100 278 251 225 270 95 114 316 215 205 300 + 220 214 594 466 430 570 1 2 =

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.

Solve a matrix equation EXAMPLE 4 Solve a matrix equation Solve the matrix equation for x and y. 5x – 2 6 – 4 3 7 – 5 – y – 21 15 3 – 24 = + 3 SOLUTION Simplify the left side of the equation. 5x – 2 6 – 4 3 7 – 5 – y – 21 15 3 – 24 = 3 + Write original equation.

Solve a matrix equation EXAMPLE 4 Solve a matrix equation 5x + 3 1 5 – 4 – y – 21 15 3 – 24 = 3 Add matrices inside parentheses. 15x + 9 15 3 – 12 – 3y – 21 15 3 – 24 = Perform scalar multiplication. Equate corresponding elements and solve the two resulting equations. – 12 – 3y = -24 y = 4 15x + 9 = – 21 x = – 2 The solution is x = – 2 and y = 4. ANSWER

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

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 3 – 2 5 – 7 9 6 1(5) + 4(9) = 1( – 7) + 4(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 3 – 2 5 – 7 9 6 1(5) + 4(9) 1( – 7) + 4(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 3 – 2 5 – 7 9 6 = 1(5) + 4(9) 1( – 7) + 4(6) 3(5) + (– 2)(9) 3( –7) + ( –2)(6)

EXAMPLE 2 Find the product of two matrices STEP 5 1(5) + 4(9) 1(– 7) + 4(6) 3(5) + (–2)(9) 3(–7) + (–2)(6) 41 17 –3 –33 =

Using the given matrices, evaluate the expression. EXAMPLE 3 Use matrix operations Using the given matrices, evaluate the expression. A = 4 3 – 1 – 2 2 0 , B = – 3 0 1 –2 , C = 1 4 –3 –1 a. A(B + C) b. AB + AC SOLUTION a. A(B + C) = 4 3 – 1 – 2 2 0 –3 0 1 –2 + 1 4 –3 –1

EXAMPLE 3 Use matrix operations 4 3 – 1 – 2 2 0 = – 2 4 – 2 –3 –14 7 6 2 – 4 8 = b. AB + AC = 4 3 –1 –2 2 0 4 3 –1 –2 2 0 – 3 0 1 –2 + 1 4 –3 –1 – 9 – 6 1 4 – 6 0 – 5 13 5 –2 2 8 + – 14 7 6 2 – 4 8 =

EXAMPLE 4 Use matrices to calculate total cost Two hockey teams submit equipment lists for the season as shown. Sports 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 Men’s team 14 30 18 16 25 20 Cost Dollars Sticks Pucks Uniforms 60 2 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 16 25 20 60 2 35 = 14(60) + 30(2) + 18(35) 16(60) + 25(2) + 20(35) 1530 1710

EXAMPLE 4 Use matrices to calculate total cost The labels for the product matrix are shown below. Total Cost Dollars Women’s team Men’s team 1530 1710 The total cost of equipment for the women’s team is $1530, and the total cost for the men’s team is $1710. ANSWER