100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Exam (elaborations) mat1503_assignment_3_ MAT1503 ASSIGNMENT 3 SEMESTER 1 2021 R139,08   Add to cart

Exam (elaborations)

Exam (elaborations) mat1503_assignment_3_ MAT1503 ASSIGNMENT 3 SEMESTER 1 2021

 5 views  0 purchase
  • Course
  • Institution

Exam of 16 pages for the course mat1503_assignment_3_ at Chamberlain College Of Nursng (i).)

Preview 3 out of 16  pages

  • August 5, 2021
  • 16
  • 2021/2022
  • Exam (elaborations)
  • Questions & answers
avatar-seller
MAT1503 ASSIGNMENT 3 2021

written by

jctutor0814378595




www.stuvia.com




Downloaded by: mkhaliphit | mkhaliphit@gmail.com
Distribution of this document is illegal

, Stuvia.com - The Marketplace to Buy and Sell your Study Material




MAT1503


ASSIGNMENT 3 SEMESTER 1 2021




QUESTION 1


1 −1 2 1
[ 3 −1 5 | −2 ]
−4 2 𝑥 2 − 8 𝑥 + 2

↓ 3𝑅1 − 𝑅́ 2 → 𝑅2

1 −1 2 1
[ 0 −2 1 | 5 ]
−4 2 𝑥 2 − 8 𝑥 + 2

↓ 4𝑅1 + 𝑅́ 3 → 𝑅3

1 −1 2 1
[0 −2 1 | 5 ]
0 −2 𝑥 2 𝑥 + 6

↓ 𝑅2 + 𝑅́3 → 𝑅3

1 −1 2 1
[0 −2 1 | 5 ] 𝑟𝑜𝑤 𝑒𝑛𝑐ℎ𝑙𝑜𝑛 𝑓𝑜𝑟𝑚
0 0 𝑥 2 + 1 𝑥 + 11



i).

𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥 2 + 1 = 0 𝑎𝑛𝑑 𝑥 + 11 ≠ 0

𝑥 2 + 1 ≥ 1, 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑥 2 + 1 𝑖𝑠 𝑛𝑒𝑣𝑒𝑟 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑧𝑒𝑟𝑜 𝑡ℎ𝑢𝑠 𝑡ℎ𝑒𝑟𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑛𝑜 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥 𝑤ℎ𝑒𝑟𝑒

𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 ℎ𝑎𝑠 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.



𝑻𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒙 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕.




Downloaded by: mkhaliphit | mkhaliphit@gmail.com
Distribution of this document is illegal

, Stuvia.com - The Marketplace to Buy and Sell your Study Material




ii).

𝐸𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥 2 + 1 ≠ 0 𝑎𝑛𝑑 𝑥 + 11 ∈ ℝ

𝑥 2 + 1 ≠ 0 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥, 𝑠𝑜 𝑥 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:

𝑥∈ℝ

𝑥 ∈ (−∞, ∞)



iii).

𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥 2 + 1 = 0 𝑎𝑛𝑑 𝑥 + 11 = 0

𝑥 2 + 1 = 0 𝑎𝑛𝑑 𝑥 = −11

𝑥 2 + 1 𝑤𝑖𝑙𝑙 𝑎𝑙𝑤𝑎𝑦𝑠 𝑏𝑒 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑥 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑜𝑟 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.



𝑻𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒙 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕.




QUESTION 2

𝑎11 𝑎12 𝑎13 … … 𝑎𝑛1
𝑎21 𝑎22 𝑎23 … … 𝑎𝑛2
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
𝑇= 0 0 0 ⋮ ⋮ 0
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
[𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 … … 𝑎𝑛𝑛 ]

𝐿𝑒𝑡𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑜𝑓 𝑇 𝑏𝑦 𝑒𝑥𝑝𝑎𝑛𝑑𝑖𝑛𝑔 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠.
𝑎12 ⋯ 𝑎𝑛1 𝑎11 ⋯ 𝑎(𝑛−1)1
𝑑𝑒𝑡(𝑇) = (−1)𝑖+𝑗 (0 | ⋮ ⋱ ⋮ |) + ⋯ +(−1)𝑖+𝑗 (0 | ⋮ ⋱ ⋮ |)
𝑎𝑛1 ⋯ 𝑎𝑛𝑛 𝑎𝑛1 ⋯ 𝑎𝑛𝑛




Downloaded by: mkhaliphit | mkhaliphit@gmail.com
Distribution of this document is illegal

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller GRADUATEnurse. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for R139,08. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

75323 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 14 years now

Start selling
R139,08
  • (0)
  Buy now