1. 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)
– – 1 + 0 = =
– 3 – 1 –
3 – 10 –
0 –2 – b. –
7 – (– 2) – 5
0 – – 2 – (– 10)
– 1 – (– 3) – 1 =
9 –1 –
2 5 =

2. 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) = – –
– 4 – 14 =
b. 4
– 2 – 8 –
6 – 5 +
4(– 2) 4(– 8)
4(5) (0) –
6 –5 = +
– 8 – 32 –
6 – 5 = +

3. EXAMPLE 2
Multiply a matrix by a scalar
– 8 + (– 3) –
(– 5) =
– 11 – 24
– 5 =

4. 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.

5. 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
Last Month (A) This Month (B)
Small Large Small Large

6. 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 + 1 2 =

7. EXAMPLE 3
Solve a multi-step problem =
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.

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

9. Solve a matrix equation
EXAMPLE 4
Solve a matrix equation
5x + 3 1 5
– 4 – y –
3 – 24 = 3
Add matrices inside
parentheses.
15x
– 12 – 3y –
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

10. EXAMPLE 1
Describe matrix products
State whether the product AB is defined. If so, give the dimensions of AB.
a. A: 4 3, B: b. A: 3 4, B:
SOLUTION
a. Because A is a matrix and B is a matrix, the product AB is defined and is a 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.

11. EXAMPLE 2
Find the product of two matrices
Find AB if A =
3 – 2
and B =
5 – 7
SOLUTION
Because A is a matrix and B is a matrix, the product AB is defined and is a matrix.

12. 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.
3 – 2
5 – 7
1(5) + 4(9) =

13. 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.
3 – 2
5 – 7
1(5) + 4(9) =
1( – 7) + 4(6)

14. 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.
3 – 2
5 – 7
1(5) + 4(9)
1( – 7) + 4(6)
3(5) + (– 2)(9) =

15. 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.
3 – 2
5 – 7 =
1(5) + 4(9)
1( – 7) + 4(6)
3(5) + (– 2)(9)
3( –7) + ( –2)(6)

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

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

18. EXAMPLE 3
Use matrix operations
– – 2 = –
– –3 – – =
b. AB + AC =
– –2
– –2 –
1 –2 +
–3 –1
– – 6 – – –2 + – – =

19. 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.

20. 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.

21. EXAMPLE 4
Use matrices to calculate total cost
Equipment
Sticks Pucks Uniforms
Women’s team
Men’s team
Cost
Dollars
Sticks
Pucks
Uniforms 60 2 35

22. 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 and the cost matrix is So, their product is a matrix. 60 2 35 =
14(60) + 30(2) + 18(35)
16(60) + 25(2) + 20(35)
1530
1710

23. 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