Mat1511 Quiz 1

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Precalculus B
1
QUIZ

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Precalculus B
Due date N/A
Time limit 1 hours 30 minutes
Grade out of 50.00


QUESTION 1
Use Descartes Rule of signs to find the possible number of negative
real zeros of
𝑷(𝒙) = 𝟑𝒙𝟔 + 𝟓𝒙𝟓 + 𝟒𝒙𝟒 + 𝟕𝒙𝟑 + 𝟔𝒙𝟐 − 𝟐𝒙 + 𝟏 ?
Choose the correct answer:
o a. 6 𝑜𝑟 4 𝑜𝑟 2 𝑜𝑟 0
o b. 5 𝑜𝑟 3 𝑜𝑟 1
o c. 2 𝑜𝑟 0
o d. 4 𝑜𝑟 2 𝑜𝑟 0
o e. None of the options


𝑃(−𝑥) = 3(−𝑥)6 + 5(−𝑥)5 + 4(−𝑥)4 + 7(−𝑥)3 + 6(−𝑥)2 − 2(−𝑥) + 1
= 3(𝑥 6 ) + 5(−𝑥 5 ) + 4(𝑥 4 ) + 7(−𝑥 3 ) + 6(𝑥 2 ) + 2𝑥 + 1
= +3𝑥 6 − 5𝑥 5 + 4𝑥 4 − 7𝑥 3 + 6𝑥 2 + 2𝑥 + 1


Number of sign changes: 4
∴ The number of negative real zeros is: 4 or 2 or 0

QUESTION 2

Find 𝑨𝑩 ?
1 2 0 −1
−2 −1 3
𝐴=[ ] 𝐵 = [5 −1 2 0]
4 0 1
2 3 4 −3
Choose the correct answer:
−1 14 −7
o a. [ 6 11 3 ]
0 5 −2

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−1 6 10 −7
o b. [ ]
6 11 4 −7
−1 14 −7 0
o c. [ ]
6 11 −2 −3
o d. 𝐴𝐵 is impossible to compute

o e. None of the options


1 2 0 −1
−2 −1 3
=[ ] ∙ [5 −1 2 0 ]
4 0 1
2 3 4 −3
−2(1) − 1(5) + 3(2) −2(2) − 1(−1) + 3(3) −2(0) − 1(2) + 3(4) −2(−1) − 1(0) + 3(−3)
=[ ]
4(1) + 0(5) + 1(2) 4(2) + 0(−1) + 1(3) 4(0) + 0(2) + 1(4) 4(−1) + 0(0) + 1(−3)

−1 6 10 −7
=[ ]
6 11 4 −7

QUESTION 3

4 2
If [ ] 𝑋 = 𝐼 , find the value of the bottom-right element of 𝑋.
1 3
Choose the correct answer:
2
o a.
5
3
o b. −
10
o c. 4
1
o d. − 10
o e. None of the options


1 0
𝐼=[ ]
0 1
4 2
• Let 𝐴 = [ ] , 𝑋 = 𝐴−1
1 3

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4 2 1 0
[ ⋮ ]
1 3 0 1
1 3 0 1
𝑅1 ↔ 𝑅2 [ ⋮ ]
4 2 1 0
1 3 0 1
𝑅2 −4𝑅1 →𝑅2 [ ⋮ ]
→ 0 −10 1 −4
0 1
1 3
1
− 𝑅2 [ ⋮ 1 2]
10 0 1 −
→ 10 5
3 1
𝑅1 −3𝑅2 →𝑅1 1 0 −
→ [ ⋮ 10 5]
0 1 1 2
−
10 5

3 1
−
𝐴−1 = [ 10 5]
1 2
−
10 5
2
∴ The value of the bottom-right element of 𝑋 is .
5

QUESTION 4

Solve for 𝒙 in the given matrix equation
1 2 1
𝑑𝑒𝑡 [3 𝑥 0] = −3
2 1 4
Choose the correct answer:
o a. 𝑥 = −2
o b. 𝑥 =4
o c. 𝑥 = −6
o d. 𝑥 =9
o e. None of the options

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𝑎21 𝐴21 + 𝑎22 𝐴22 + 𝑎23 𝐴23 = −3
3[(−1)2+1 ∙ 𝑀21 ] + 𝑥[(−1)2+2 ∙ 𝑀22 ] + (0)𝐴23 = −3
1 3 1 1
−3 ∙ | |+𝑥∙| | = −3
−2 4 2 4
−3[2(4) − (1)(1)] + 𝑥[1(4) − (1)(2)] = −3
−3(8 − 1) + 𝑥(4 − 2) = −3
−3(7) + 𝑥(2) = −3
−21 + 2𝑥 = −3
2𝑥 = −3 + 21
2𝑥 18
=
2 2
𝑥=9

QUESTION 5

Use the Cramer’s rule to solve the unknowns of the system
𝑥 + 𝑦 + 𝑧 = −5
6𝑥 + 𝑦 − 3𝑧 = 1
−𝑥 + 2𝑦 + 2𝑧 = −4

Choose the correct answer:
o a. 𝑥 = 2, 𝑦 = −3 , 𝑧 = −4
o b. 𝑥 = −2, 𝑦 = 1 , 𝑧 = −4
o c. 𝑥 = −3, 𝑦 = 7 , 𝑧 = −4
o d. 𝑥 = −2, 𝑦 = −1 , 𝑧 = 4
o e. None of the options


1 1 1
• |𝐷| = | 6 1 −3|
−1 2 2
= 𝑎11 𝐴11 + 𝑎12 𝐴12 + 𝑎13 𝐴13
1 −3 6 −3 6 1
=1∙| |−1∙| |+1∙| |
2 2 −1 2 −1 2
= [1(2) − (−3)(2)] − 1[6(2) − (−3)(−1)] + [6(2) − (1)(−1)]
= 8 − 1(9) + 13 = 𝟏𝟐

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−5 1 1
• |𝐷𝑥 | = | 1 1 −3|
−4 2 2
= 𝑎11 𝐴11 + 𝑎12 𝐴12 + 𝑎13 𝐴13
1 −3 1 −3 1 1
= −5 ∙ | |−1∙| |+1∙| |
2 2 −4 2 −4 2
= −5[1(2) − (−3)(2)] − 1[1(2) − (−3)(−4)] + [1(2) − (1)(−4)]
= −5(8) − 1(−10) + 6 = −𝟐𝟒

1 −5 1
• |𝐷𝑦 | = | 6 1 −3|
−1 −4 2
= 𝑎11 𝐴11 + 𝑎12 𝐴12 + 𝑎13 𝐴13
1 −3 6 −3 6 1
=1∙| |+5∙| |+1∙| |
−4 2 −1 2 −1 −4
= [1(2) − (−3)(−4)] + 5[6(2) − (−3)(−1)] + [6(−4) − (1)(−1)]
= −10 + 5(9) − 23 = 𝟏𝟐

1 1 −5
• |𝐷𝑧 | = | 6 1 1 |
−1 2 −4
= 𝑎11 𝐴11 + 𝑎12 𝐴12 + 𝑎13 𝐴13
1 1 6 1 6 1
=1∙| |−1∙| |−5∙| |
2 −4 −1 −4 −1 2
= [1(−4) − (1)(2)] − 1[6(−4) − (1)(−1)] − 5[6(2) − (1)(−1)]
= −6 − 1(23) − 5(13) = −𝟒𝟖

|𝐷𝑥 | |𝐷𝑦 | |𝐷𝑧 |
∴𝑥= ∴𝑦= ∴𝑧=
|𝐷| |𝐷| |𝐷|
−24 12 −48
= = =
12 12 12
= −2 =1 = −4

QUESTION 6
Which of the following is the first row of the inverse of matrix 𝑨 shown
below?




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1 1 2
𝐴 = [1 1 1]
2 3 4
Choose the correct answer:
o a. (−2 ; 0 ; 1)
o b. (1 ; −2 ; −1)
o c. (2 ; 1 ; 1)
o d. (1 ; 2 ; −1)
o e. None of the options


1 1 2 1 0 0
[1 1 1 ⋮ 0 1 0]
2 3 4 0 0 1
𝑅2 −𝑅1 →𝑅2
1 1 2 1 0 0
→ [0 0 −1 ⋮ −1 1 0]
𝑅3 −2𝑅1 →𝑅3 0 1 0 −2 0 1
→

1 1 2 1
−1×𝑅2 0 0
→ [0 0 1 ⋮ 1 −1 0]
0 1 0 −2 0 1
1 1 2 1 0 0
[0 1 0 ⋮ −2 0 1]
𝑅2 ↔ 𝑅3 0 0 1 1 −1 0


𝑅1 −2𝑅3 →𝑅1 1 1 0 −1 2 0
→ [0 1 0 ⋮ −2 0 1]
0 0 1 1 −1 0

𝑅1 −𝑅2 →𝑅1 1 0 0 1 2 −1
→ [0 1 0 ⋮ −2 0 1]
0 0 1 1 −1 0



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