Chapter 8 Systems of Equations and Inequalities

⎡ 7 1 0 50 ⎤
4
→ ⎢ 1 ⎥ ( R1 = r1−2r2) ⎢⎣ 0 1 − 75 8 ⎥⎦
The matrix in the last step represents the system
⎧ 7
⎪x+
z = 50 4
⎨ 1
⎪⎩
y− z = 75 8
Therefore the solution is
7
x = 50 − z ,
4
1
y = 75 + z , z is any real number.
8
Possible combinations:
Supplement 1 Supplement 2 Supplement 3
50mg 75mg 0mg
36mg 76mg 8mg
22mg 77mg 16mg
8mg 78mg 24mg
88. Let x = the amount of powder 1.
Let y = the amount of powder 2.
Let z = the amount of powder 3.
⎧0.20x+
0.40y+ 0.30z = 12 Vitamin B12
⎨
⎩0.30x+
0.20y+ 0.40z = 12 Vitamin E
Multiplying each equation by 10 yields
⎧2x+
4y+ 3z = 120
⎨
⎩3x+
2y+ 4z = 120
Set up a matrix and solve:
⎡2 4 3120⎤
⎢ ⎥
⎣3 2 4120⎦
⎡243120 ⎤
3
→ ⎢ ⎥ ( R2 = r2 − r 2 1)
⎣0 −4 −0.5 −60⎦
⎡2 0 2.5 60 ⎤
→ ⎢ ⎥ ( R1 = r1+ r2)
⎣0 −4 −0.5 −60⎦
The matrix in the last step represents the system
⎧ 2x+ 2.5z = 60
⎨
⎩−4y−
0.5z =−60
Thus, the solution is x = 30 − 1.25z
,
y = 15 − 0.125z
, z is any real number.
Possible combinations:
Section 8.3: Systems of Linear Equations: Determinants
815
Powder 1 Powder 2 Powder 3
30 units 15 units 0 units
20 units 14 units 8 units
10 units 13 units 16 units
0 units 12 units 24 units
89 – 91. Answers will vary.
Section 8.3
1. determinants
2. ad − bc
3. False
4. False
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5.
6.
7.
8.
9.
10.
11.
3 1
= 3(2) − 4(1) = 6 − 4 = 2
4 2
6 1
= 6(2) − 5(1) = 12 − 5 = 7
5 2
6 4
= 6(3) −( − 1)(4) = 18 + 4 = 22
−1
3
8 − 3
= 8(2) −4( − 3) = 16 + 12 = 28
4 2
−3 − 1
= −3(2) −4( − 1) =− 6 + 4 =− 2
4 2
− 4 2
= −4(3) −( − 5)(2) =− 12 + 10 =−2
−5
3
3 4 2
1 5 1 5
1 − 1 5 = 3
−
− 4 + 2
1 −1
2
2 2 1 2
1 2 −
− − 1 2
( ) [ ]
+ 2[ 1(2) −1( −1)
]
= 3⎡⎣ −1 ( −2) −2(5) ⎤⎦−41(
−2) −1(5)
= 3( −8) −4( − 7) + 2(3)
=− 24 + 28 + 6
=
10