MAT1503 ASSIGNMENT 4 SEM1 with Complete Solutions 2023

MAT1503 ASSIGNMENT 4 SEM1 with Complete Solutions 2023



QUESTION 1




𝑥 𝑦 1
|𝑎1 𝑏1 1| = 0
𝑎 2 𝑏2 1
𝑏 1 𝑎 1 𝑎 𝑏1
𝑥| 1 |− 𝑦| 1 | + 1| 1 |=0
𝑏2 1 𝑎2 1 𝑎2 𝑏2

𝑥(𝑏1 − 𝑏2) − 𝑦(𝑎1 − 𝑎2) + 1(𝑎1𝑏2 − 𝑎2𝑏1) = 0

𝑦(𝑎1 − 𝑎2) = 𝑥(𝑏1 − 𝑏2) + (𝑎1𝑏2 − 𝑎2𝑏1)

(𝑏1 − 𝑏2) (𝑎1𝑏2 − 𝑎2𝑏1) 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒.
𝑦= 𝑥+
(𝑎1 − 𝑎2) (𝑎1 − 𝑎2)




QUESTION 2


2.1).
−4 2
𝐴=[ ]
3 −3
−4 2
−𝐴 = −1 [ ]
3 −3


4 −2
−𝐴 = [ ]
−3 3
𝑑𝑒𝑡(−𝐴) = (4)(3) − (−2)(−3)

𝑑𝑒𝑡(−𝐴) = 12 − 6

𝑑𝑒𝑡(−𝐴) = 6

, −4 3
𝐴𝑇 = [ ]
2 −3

𝑑𝑒𝑡(𝐴𝑇) = (−4)(−3) − (2)(3)

𝑑𝑒𝑡(𝐴𝑇) = 12 − 6

𝑑𝑒𝑡(𝐴𝑇) = 6



𝑊𝑒 𝑐𝑎𝑛 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡: 𝑑𝑒𝑡(−𝐴) = 𝑑𝑒𝑡(𝐴𝑇)



2.2).


3 1 −2
𝐴 = [−5 3 −6]
−1 0 −4
3 1 −2
−𝐴 = −1 [−5 3 −6]
−1 0 −4
−3 −1 2
−𝐴 = [ 5 −3 6]
1 0 4
−3 6 5 6 5 −3
𝑑𝑒𝑡(−𝐴) = −3 | | − (−1) | |+ 2| |
0 4 1 4 1 0

= −3(−12 − 0) + 1(20 − 6) + 2(0 − (−3))

= 56



3 −5 −1
𝐴𝑇 = [ 1 3 0]
−2 −6 −4


3 0 1 0 1 3
𝑑𝑒𝑡(𝐴𝑇) = 3 | | − (−5) | | + (−1) | |
−6 −4 −2 −4 −2 −6

= 3(−12 − 0) + 5(−4 − 0) − 1(−6 + 6)

= −56

, 𝑊𝑒 𝑐𝑎𝑛 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡: 𝑑𝑒𝑡(−𝐴) = −𝑑𝑒𝑡(𝐴𝑇)




QUESTION 3
𝐸𝑓𝑓𝑒𝑐𝑡𝑠 𝑜𝑓 𝑟𝑜𝑤 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑛 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑠:


𝑅𝑜𝑤 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐸𝑓𝑓𝑒𝑐𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡

𝑅𝑖↔𝑅𝑗 𝑑𝑒𝑡(𝐵) = −𝑑𝑒𝑡(𝐴)
𝐴→−−−→𝐵
𝑐𝑅𝑖→𝑅𝑖 𝑑𝑒𝑡(𝐵) = 𝑐 ∙ 𝑑𝑒𝑡(𝐴)
𝐴→−−−→𝐵
𝑎𝑅𝑖 +𝑏𝑅𝑖 →𝑅𝑖 𝑜𝑟 𝑗 𝑑𝑒𝑡(𝐵) = −𝑑𝑒𝑡(𝐴)
𝐴 −→ 𝐵
→−−−−−−−−−

3.1).


1 0 0 0
0 1 0 0
𝐿𝑒𝑡: 𝐴 = [ ]
0 0 1 0
0 0 0 −2
1 0 0 0 1 0 0 0
0 −2𝑅4→𝑅4 0 1 0 0 ]=𝐴
𝐼4 = [0 1 0 ]→−−−−−→[
0 0 1 0 0 0 1 0
0 0 0 1 0 0 0 −2

𝑑𝑒𝑡(−𝐴) = −2𝑑𝑒𝑡(𝐼4) ∴ 𝑑𝑒𝑡(𝐼𝑛) = 1

𝑑𝑒𝑡(−𝐴) = −2(1)

𝑑𝑒𝑡(−𝐴) = −2



3.2).


1 0 0 0
0 1 0 0]
𝐿𝑒𝑡 ∶ 𝐴 = [ 0 1
0 0
0 0 1⁄ 0
4