algebra 2 unit 3 – chapter 4 section 4.1 – matrix operations day 1
TRANSCRIPT
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
#11 – 35 ODD
#37 – 41 ALL