MAT1503 Assignment 3 Semester 1 With Complete Solution 2023.

MAT1503 Assignment 3 Semester 1 With Complete Solution 2023.




QUESTION 1


1 −1 2 1
[ 3 −1 5 | −2 ]
−4 2 𝑥2 − 8 𝑥 + 2

↓ 3𝑅1 − 𝑅 2 → 𝑅2

1 −1 2 1
[ 0 −2 1 | 5 ]
−4 2 𝑥2 − 8 𝑥 + 2

↓ 4𝑅1 + 𝑅 3 → 𝑅3

1 −1 2 1
[0 −2 1 | 5 ]
0 −2 𝑥2 𝑥 + 6

↓ 𝑅2 + 𝑅3 → 𝑅3

1 −1 2 1
[0 −2 1 | 5 ] 𝑟𝑜𝑤 𝑒𝑛𝑐ℎ𝑙𝑜𝑛 𝑓𝑜𝑟𝑚
0 0 𝑥2 + 1 𝑥 + 11



i).

𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥2 + 1 = 0 𝑎𝑛𝑑 𝑥 + 11 ≠ 0

𝑥2 + 1 ≥ 1, 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑥2 + 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.

, ii).

𝐸𝑥𝑎𝑐𝑡𝑙𝑦 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥2 + 1 ≠ 0 𝑎𝑛𝑑 𝑥 + 11 ∈ ℝ

𝑥2 + 1 ≠ 0 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥, 𝑠𝑜 𝑥 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:

𝑥∈ℝ

𝑥 ∈ (−∞, ∞)



iii).

𝐼𝑛𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑚𝑎𝑛𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛:

𝑥2 + 1 = 0 𝑎𝑛𝑑 𝑥 + 11 = 0

𝑥2 + 1 = 0 𝑎𝑛𝑑 𝑥 = −11

𝑥2 + 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

𝑎 𝑎12 𝑎13 … … 𝑎𝑛1
𝖥𝑎1121 𝑎22 𝑎23 … … 𝑎𝑛2 1
I ⋮ ⋮ I
⋮ ⋮ ⋮ ⋮
𝑇=I 0 0 0 ⋮ ⋮ 0 II
I
I ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ I
[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) (0 | ⋮
𝑖+𝑗 ⋱ ⋮ |)
𝑎𝑛1 ⋯ 𝑎𝑛𝑛 𝑎𝑛1 ⋯ 𝑎𝑛𝑛