MAT1511 Assignment 7 (Solutions) 2023

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MAT1511 QUIZ
2023 Assignment 7




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, MAT1511 Assignment 7
Due date 30 September 2023
Time limit 2 hours
Grade 100.00 out of 100.00


QUESTION 1
Use Descartes Rule of signs to find the possible number of positive
real zeros of
𝑷(𝒙) = 𝟐𝒙𝟔 − 𝟑𝒙𝟓 − 𝟏𝟑𝒙𝟒 + 𝟐𝟗𝒙𝟑 − 𝟐𝟕𝒙𝟐 + 𝟑𝟐𝒙 − 𝟔 ?
Choose the correct answer:
o a. 7 𝑜𝑟 1
o b. 6 𝑜𝑟 3 𝑜𝑟 2 𝑜𝑟 1
o c. 5 𝑜𝑟 3 𝑜𝑟 1
o d. 4 𝑜𝑟 2
o e. None of the options

P(x) = 2x 6 − 3x 5 − 13x 4 + 29x 3 − 27x 2 + 32x − 6
TERM SIGN
6
2𝑥 +
5
−3𝑥 −
−13𝑥 4 −
3
29𝑥 +
2
−27𝑥 −
32𝑥 +
−6 −

Number of sign changes: 5
∴ The number of positive real zeros is: 5 or 3 or 1

QUESTION 2

Which of the following is equal to 𝑨𝑩 , for matrices 𝑨 and 𝑩 below?
1 −2
−2 1 5
𝐴=[ ] 𝐵 = [3 4]
3 1 1
0 −1
Choose the correct answer:


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, −5 1
o a. [ ]
12 −4
−5 12
o b. [ ]
3 −18
−5 3
o c. [ ]
15 −4
o d. 𝐴𝐵 is impossible to compute

o e. None of the options

1 −2
−2 1 5
=[ ] ∙ [3 4 ]
3 1 1
0 −1
−2(1) + 1(3) + 5(0) −2(−2) + 1(4) + 5(−1)
=[ ]
3(1) + 1(3) + 1(0) 3(−2) + 1(4) + 1(−1)
1 3
=[ ]
6 −3
∴ 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑜𝑛𝑠

QUESTION 3

−2 4 𝑤 𝑥 1 0
If [ ][ ] = [ ] , then 𝑧 = ⋯
6 −8 𝑦 𝑧 0 1
Choose the correct answer:
o a. 𝑧 = −2
o b. 𝑧=0
1
o c. 𝑧=4
o d. 𝑧=2
o e. None of the options

−2(𝑤) + 4(𝑦) −2(𝑥) + 4(𝑧) 1 0
[ ]=[ ]
6(𝑤) − 8(𝑦) 6(𝑥) − 8(𝑧) 0 1


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, −2𝑥 + 4𝑧 = 0 6𝑥 − 8𝑧 = 1
4𝑧 2𝑥 6(2𝑧) − 8𝑧 = 1
=
2 2 12𝑧 − 8𝑧 = 1
𝑥 = 2𝑧 4𝑧 = 1
1
∴𝑧=
4
QUESTION 4

Solve for 𝒙 in the given matrix equation
1 −2 3
𝑑𝑒𝑡 [−1 𝑥 1] = −26
−2 3 4
Choose the correct answer:
o a. 𝑥 = −2
23
o b. 𝑥=− 5
o c. 𝑥 = −1
13
o d. 𝑥= 5
o e. None of the options


−2 3 1 3
−1 {(−1)2+1 ∙ | |} + 𝑥 {(−1)2+2 ∙ | |}
3 4 −2 4
1 −2
+ 1 {(−1)2+3 ∙ | |} = −26
−2 3

−2 3 1 3 1 −2
−1 (−1 | |) + 𝑥 (1 | |) + 1 (−1 | |) = −26
3 4 −2 4 −2 3

1{−2(4) − (3)3} + 𝑥{1(4) − (3)(−2)} − 1{1(3) − (−2)(−2)} = −26

−17 + 10𝑥 + 1 = −26
10𝑥 = −26 + 16
10𝑥 −10
=
10 10
𝑥 = −1


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,QUESTION 5

Use the Cramer’s rule to solve the unknowns of the system
𝑥 + 2𝑦 = 2 − 𝑧
3𝑥 − 6𝑦 = 2 − 2𝑧
2𝑥 = 8 + 𝑧

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

𝑥 + 2𝑦 + 𝑧 = 2
3𝑥 − 6𝑦 + 2𝑧 = 2
2𝑥 −𝑧 =8
1 2 1
|𝐷| = |3 −6 2 |
2 0 −1
= 𝑎31 𝐴31 + 𝑎32 𝐴32 + 𝑎33 𝐴33
2 1 1 2
= 2 {(−1)3+1 ∙ | |} + (0)𝐴32 − 1 {(−1)3+3 ∙ | |}
−6 2 3 −6
= 2{2(2) − (1)(−6)} − 1{1(6) − (2)(3)}
= 2(10) − 1(−12) = 32

2 2 1
|𝐷𝑥 | = |2 −6 2 |
8 0 −1
= 𝑎31 𝐴31 + 𝑎32 𝐴32 + 𝑎33 𝐴33
2 1 2 2
= 8 {(−1)3+1 ∙ | |} + (0)𝐴32 − 1 {(−1)3+3 ∙ | |}
−6 2 2 −6
= 8{2(2) − (1)(−6)} − 1{2(−6) − (2)(2)}
= 8(10) − 1(−16) = 96




Page 4 of 22

, 1 2 1
|𝐷𝑦 | = |3 2 2 |
2 8 −1
= 𝑎11 𝐴11 + 𝑎12 𝐴12 + 𝑎13 𝐴13
2 2 3 2
= 1 {(−1)1+1 ∙ | |} + 2 {(−1)1+2 ∙ | |}
8 −1 2 −1
3 2
+ 1 {(−1)1+3 ∙ | |}
2 8
= 1{2(−1) − (2)(8)} − 2{3(−1) − (2)(2)} + 1{3(8) − (2)(2)}
= −18 − 2(−7) + 20
= 16
1 2 2
|𝐷𝑧 | = |3 −6 2|
2 0 8
= 𝑎31 𝐴31 + 𝑎32 𝐴32 + 𝑎33 𝐴33
2 2 1 2
= 2 {(−1)3+1 ∙ | |} + (0)𝐴32 + 8 {(−1)3+3 ∙ | |}
−6 2 3 −6
= 2{2(2) − (2)(−6)} + 8{1(−6) − (2)(3)}
= 2(16) + 8(−12)
= 32 − 96 = −64
|𝐷𝑥 | |𝐷𝑦 | |𝐷𝑧 |
∴𝑥= ∴ 𝑦 = ∴𝑧=
|𝐷| |𝐷| |𝐷|
96 16 −64
= = =
32 32 32
=3 1 = −2
=
2
QUESTION 6
Which of the following is the second row of the inverse of matrix 𝑨
shown below?
1 0 2
𝐴 = [0 2 −1]
2 5 2
Choose the correct answer:
o a. (−2 ; −2 ; 1)
o b. (−4 ; −5 ; 2)
o c. (0 ; 0 ; 1)
o d. (2 ; 5 ; 2)
o e. None of the options

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