Algebra 2 Unit 3 – Chapter 4 Section 4.1 – Matrix Operations Day 1.

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Transcript of Algebra 2 Unit 3 – Chapter 4 Section 4.1 – Matrix Operations Day 1.

Algebra 2Unit 3 – Chapter 4

Section 4.1 – Matrix OperationsDay 1

CONCEPT REVIEW

Solve the system. 4x + 2y + 3z = 1 Equation 1

2x – 3y + 5z = –14 Equation 2

6x – y + 4z = –1 Equation 3

SOLUTION

STEP 1 Rewrite the system as a linear system in two variables.

4x + 2y + 3z = 1

12x – 2y + 8z = –2

Add 2 times Equation 3

to Equation 1.

16x + 11z = –1 New Equation 1

2x – 3y + 5z = –14

–18x + 3y –12z = 3

Add – 3 times Equation 3to Equation 2.

–16x – 7z = –11 New Equation 2

STEP 2 Solve the new linear system for both of its variables.

16x + 11z = –1 Add new Equation 1

and new Equation 2. –16x – 7z = –11

4z = –12

z = –3 Solve for z.

x = 2 Substitute into new Equation 1 or 2 to find x.

CONCEPT REVIEW

6x – y + 4z = –1

STEP 3 Substitute x = 2 and z = – 3 into an original equation and solve for y.

Write original Equation 3.

6(2) – y + 4(–3) = –1 Substitute 2 for x and –3 for z.

y = 1 Solve for y.

CONCEPT REVIEW

WHAT IF THERE WAS A DIFFERENT WAY TO WRITE

AND SOLVE THREE VARIABLE THREE

EQUATION SYSTEMS?ANSWER: MATRICES

(plural of a matrix)

QUESTION O

F THE DAY…

What is a matrix?

SECTION 4.1 – MATRICES Matrix – a rectangular arrangement of

numbers into rows and columns.

Dimensions – tell the number of rows and columns of a matrix, and it is how we define the size of a matrix.

SECTION 4.1 – MATRICES Elements/Entries – the numbers that are

located in a matrix.

Equal Matrices – when two matrices have identical dimensions and identical corresponding elements/entries.

Matrix Dimension & Size

6 2 1

2 0 5A

NAME ROWS

COLUMNS2 x 3 Matrix

Parts of a Matrix

Labeling Elements

6 2 1

2 0 5A

11 12 13

21 22 23

a a aA

a a a

-2 5

3 -10

-3 1

7 4

0 -2

-1 6

b. –

EXAMPLE 1: Add and subtract matrices

Perform the indicated operation, if possible.

3 0 –5 –1a.

–1 4 2 0+

3 + (–1) 0 + 4 –5 + 2 –1 + 0= =

2 4 –3 –1

9 –1

–3 8

2 5

= 7 – (–2) 4 – 5

0 – 3 –2 – (–10)

–1 – (–3) 6 – 1

=

-2 5

3 -10

7 4

0 -2

-1 6

c. –

EXAMPLE 1: Add and subtract matrices (cont.)

Perform the indicated operation, if possible.

NOT POSSIBLE; To add or subtract matrices the dimensions of the matrices must be equivalent. Here we have a 2 x 3 and a 2 x 2. Therefore

EXAMPLE 2: Multiply a matrix by a scalar

Perform the indicated operation, if possible.

4(–2) 4(–8) 4(5) 4(0)

–3 8 6 –5

= +

a.4 –11 02 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

+

–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

=

GUIDED PRACTICE

Perform the indicated operation, if possible.

+ –2 5 11 4 –6 8

1. –3 1 –5 –2 –8 4

–5 6 6 2 –14 12

ANSWER

GUIDED PRACTICE

–4 0 7 –2 –3 1

2 2 –3 0 5 –14

2.–

–6 –2 10 –2 –8 15

ANSWER

3. 2 –1 –3 –7 6 1 –2 0 –5

– 4 –8 4 12 28 –24 –4 8 0 20

ANSWER

GUIDED PRACTICE

4 –1–3 –5

–2 –2 0 6

4.3 +

3 –3 –2 1

ANSWER

EXAMPLE 3 Solve a multi-step problem

ManufacturingA company manufactures small and large steel DVDracks 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

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.

Solve a multi-step problem

WalnutPineCherry

125 100278 251225 270

95 114316 215205 300

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

EXAMPLE 3 Solve a multi-step problem

220 214594 466430 570

12

=

95 114316 215205 300

(A + B) =12

12

125 100278 251225 270

+

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 .

12

EXAMPLE 3 Solve a multi-step problem

110 107297 233215 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

SOLUTION

Simplify the left side of the equation.

Write original equation.

Solve the matrix equation for x and y.5x –2 6 –4

3 7 –5 –y

–21 15 3 –24=+3

5x –2 6 –4

3 7 –5 –y

–21 15 3 –24=3 +

EXAMPLE 4 Solve a matrix equation

Add matrices insideparentheses.

Perform scalar multiplication.

Equate corresponding elements and solve the two resulting equations.

The solution is x = –2 and y = 4.

ANSWER

5x + 3 1

5–4 – y

–21 15 3 –24=3

15x + 9 15 3 –12 – 3y

–21 15 3 –24=

–12 – 3y = 224y = 4

15x + 9 = –21 x = –2

GUIDED PRACTICE

5. In Example 3, find B – A and explain what information this matrix gives.

–30 14 38 –36 –20 30

The difference in the number of DVD racks sold this month compare last month.

ANSWER

GUIDED PRACTICE

6. Solve –2–3x –1 4 y

9 –4 –5 3

12 10 2 –18=+

for x and y.

x = 5 and y = 6

ANSWER

HOMEWORKPage 203-204

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