Mat1503 Assignment 2 2022

MAT1503
ASSIGNMENT 2
SEMESTER 1
2022

,Solution:


(i)

𝐵𝐷 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.



(ii)

𝐵𝐷 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑



(iii)

𝐷𝐶 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑




Solution:
3𝑥 = 3
𝑥=1



𝑦−𝑥 =1
𝑦−1=
𝑦=2

, 𝑧 7
𝑡+ = 1
2 2
𝑡−𝑧 =3 2

𝐹𝑟𝑜𝑚 2 𝑡 = 3 + 𝑧 𝑡ℎ𝑒𝑛 𝑠𝑢𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑖𝑛𝑡𝑜 1


𝑧 7
𝑡+ = 1
2 2
𝑧 7
3+𝑧+ =
2 2
𝑧 7
𝑧+ = −3
2 2
3𝑧 1
=
2 2
1
𝑧=
3


𝑡 =3+𝑧
1
𝑡 =3+
3
10
𝑡=
3




Solution:


3 −1
𝐴=[ ]
0 −2
3 −1 3 −1 9 −2
𝐴2 = [ ][ ]=[ ]
0 −2 0 −2 0 4
1 0
𝑃(𝐴) = 𝐴2 − 𝐴 − 6𝐼 𝑤ℎ𝑒𝑟𝑒 𝐼 = [ ]
0 1
9 −2 3 −1 1 0
𝑃(𝐴) = [ ]−[ ]− 6[ ]
0 4 0 −2 0 1

, 6 −1 6 0
𝑃(𝐴) = [ ]−[ ]
0 6 0 6
0 −1
𝑃(𝐴) = [ ]
0 0




Solution:


3 0 2 −5 1 1 1 1 1
𝐴 = [ 4 −6 3] , 𝐵 = [ 0 3 0] 𝑎𝑛𝑑 𝐶 = [2 3 −1]
−2 1 8 7 6 2 3 −5 −7




(i)

𝐴 + (𝐵 − 𝐶) = (𝐴 + 𝐵) − 𝐶
3 0 2 −5 1 1 1 1 1 3 0 2 −5 1 1 1 1 1
[4 −6 3] + ([ 0 3 0] − [ 2 3 −1]) = ([ 4 −6 3] + [ 0 3 0]) − [2 3 −1]
−2 1 8 7 6 2 3 −5 −7 −2 1 8 7 6 2 3 −5 −7
3 0 2 −6 0 0 −2 1 3 1 1 1
[4 −6 3] + [−2 0 1] = [ 4 −3 3 ] − [2 3 −1]
−2 1 8 4 11 9 5 7 10 3 −5 −7
−3 0 2 −3 0 2
[2 −6 4]=[ 2 −6 4 ] (𝑇𝑅𝑈𝐸)
2 12 17 2 12 17


𝐴(𝐵𝐶) = (𝐴𝐵)𝐶
3 0 2 −5 1 1 1 1 1 3 0 2 −5 1 1 1 1 1
[4 −6 3] ([ 0 3 0] [ 2 3 −1]) = ([ 4 −6 3] [ 0 3 0]) [2 3 −1]
−2 1 8 7 6 2 3 −5 −7 −2 1 8 7 6 2 3 −5 −7
3 0 2 0 −7 −13 −1 15 7 1 1 1
[4 −6 3] [ 6 9 −3 ] = [ 1 4 10] [2 3 −1]
−2 1 8 25 15 −12 66 49 14 3 −5 −7