MAT1503 Assignment 2 2021

MAT1503 ASSIGNMENT 2 2021



Question 1



𝑇ℎ𝑒 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑎 𝑚𝑎𝑡𝑟𝑖𝑥 𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑠 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑐𝑜𝑙𝑢𝑚𝑛𝑠.
𝐼𝑓 𝑎 𝑚𝑎𝑡𝑟𝑖𝑥 ℎ𝑎𝑠 𝑛 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑘 𝑐𝑜𝑙𝑢𝑚𝑛𝑠, 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.


−2 3 1 5
𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝐻 = [−4 8 0 6 ] . 𝑇ℎ𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝐻 ℎ𝑎𝑠 3 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 4 𝑐𝑜𝑙𝑢𝑚𝑛𝑠.
7 1 9 −1
𝑆𝑜, 𝐻 𝑖𝑠 𝑎 3 × 4 𝑚𝑎𝑡𝑟𝑖𝑥.



𝐺𝑖𝑣𝑒𝑛 𝑡𝑤𝑜 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝐸 𝑎𝑛𝑑 𝐹.
𝑊𝑒 𝑚𝑎𝑦 𝑜𝑛𝑙𝑦 𝑝𝑒𝑟𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝐸𝐹 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑖𝑛 𝐸 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠
𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑤𝑠 𝑖𝑛 𝐹. 𝑇ℎ𝑎𝑡 𝑖𝑠, 𝐸 𝑖𝑠 𝑎𝑛 𝑛 × 𝑚 𝑚𝑎𝑡𝑟𝑖𝑥 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑛 𝑚 × 𝑝 𝑚𝑎𝑡𝑟𝑖𝑥. 𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔
𝑚𝑎𝑡𝑟𝑖𝑥 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎𝑛 𝑛 × 𝑝 𝑚𝑎𝑡𝑟𝑖𝑥.



𝑊𝑒 𝑚𝑎𝑦 𝑜𝑛𝑙𝑦 𝑝𝑒𝑟𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝐹𝐸 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑖𝑛 𝐹 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠
𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑤𝑠 𝑖𝑛 𝐸. 𝑇ℎ𝑎𝑡 𝑖𝑠, 𝐹 𝑖𝑠 𝑎𝑛 𝑠 × 𝑡 𝑚𝑎𝑡𝑟𝑖𝑥 𝑎𝑛𝑑 𝐸 𝑖𝑠 𝑎 𝑡 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥. 𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔
𝑚𝑎𝑡𝑟𝑖𝑥 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎𝑛 𝑠 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.



𝑊𝑒 𝑚𝑎𝑦 𝑜𝑛𝑙𝑦 𝑎𝑑𝑑 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝑖𝑓 𝑡ℎ𝑒𝑦 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓
𝑐𝑜𝑙𝑢𝑚𝑛𝑠.
𝑇ℎ𝑎𝑡 𝑖𝑠, 𝑤𝑒 𝑚𝑎𝑦 𝑝𝑒𝑟𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝐸 + 𝐹 𝑖𝑓 𝐸 𝑖𝑠 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.
𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑚𝑎𝑡𝑟𝑖𝑥 𝑤𝑖𝑙𝑙 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.



𝑊𝑒 𝑚𝑎𝑦 𝑜𝑛𝑙𝑦 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝑖𝑓 𝑡ℎ𝑒𝑦 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑤𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓
𝑐𝑜𝑙𝑢𝑚𝑛𝑠.
𝑇ℎ𝑎𝑡 𝑖𝑠, 𝑤𝑒 𝑚𝑎𝑦 𝑝𝑒𝑟𝑓𝑜𝑟𝑚 𝑡ℎ𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝐸 − 𝐹 𝑖𝑓 𝐸 𝑖𝑠 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.
𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑚𝑎𝑡𝑟𝑖𝑥 𝑤𝑖𝑙𝑙 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑎𝑛 𝑛 × 𝑘 𝑚𝑎𝑡𝑟𝑖𝑥.

,𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝐴 𝑖𝑠 5 × 2, 𝐵 𝑖𝑠 4 × 2, 𝐶 𝑖𝑠 4 × 5 𝑎𝑛𝑑 𝐷 𝑖𝑠 4 × 5.
(i) 𝐷𝐶

𝐷 𝑖𝑠 4 × 5 𝑏𝑢𝑡 𝐶 𝑖𝑠 4 × 5. 𝑇ℎ𝑒 𝑖𝑛𝑛𝑒𝑟 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 (5 𝑎𝑛𝑑 4) 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙.
𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝐷𝐶 𝑖𝑠 𝑁𝑂𝑇 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.

(ii) −𝐶𝐴 + 𝐵

−𝐶 𝑖𝑠 4 × 5 𝑎𝑛𝑑 𝐴 𝑖𝑠 5 × 2. 𝑇ℎ𝑒 𝑖𝑛𝑛𝑒𝑟 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 (5 𝑎𝑛𝑑 5) 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙.
𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − 𝐶𝐴 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑. −𝐶𝐴 𝑤𝑖𝑙𝑙 𝑎 4 × 2 𝑚𝑎𝑡𝑟𝑖𝑥.

−𝐶𝐴 𝑖𝑠 4 × 2 𝑎𝑛𝑑 𝐵 𝑖𝑠 4 × 2. 𝑇ℎ𝑒 𝑠𝑢𝑚 − 𝐶𝐴 + 𝐵 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
−𝐶𝐴 + 𝐵 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎 4 × 2 𝑚𝑎𝑡𝑟𝑖𝑥.

(iii) 𝐶𝐷 − 𝐷

𝐶 𝑖𝑠 4 × 5 𝑏𝑢𝑡 𝐷 𝑖𝑠 4 × 5. 𝑇ℎ𝑒 𝑖𝑛𝑛𝑒𝑟 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 (5 𝑎𝑛𝑑 4) 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙.
𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝐶𝐷 𝑖𝑠 𝑁𝑂𝑇 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.
𝑆𝑖𝑛𝑐𝑒 𝐶𝐷 𝑖𝑛 𝑁𝑂𝑇 𝑑𝑒𝑓𝑖𝑛𝑒𝑑, 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝐶𝐷 − 𝐷 𝑖𝑠 𝑁𝑂𝑇 𝑑𝑒𝑓𝑖𝑛𝑒𝑑.



Question 2


3𝑥 𝑦−𝑥 3 1
[ 1 ] = [7 ]
𝑡+ 𝑧 𝑡−𝑧 3
2 2


𝑊ℎ𝑒𝑛 𝑡𝑤𝑜 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙, 𝑖𝑡 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑡𝑟𝑖𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙.



3𝑥 = 3 … . . (1)
𝑦−𝑥 =1 … … (2)
1 7
𝑡+ 𝑧= … … (3)
2 2
𝑡−𝑧 =3 … … (4)



𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 (1), 3𝑥 = 3
3𝑥 3
=
3 3
𝑥=1
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑛 𝑒𝑞𝑛 (2):
𝑦−𝑥 =1
𝑦 − (1) = 1

,𝑦−1=1
𝑦 =1+1
𝑦=2


1 7
𝑡+ 𝑧= … … (3)
2 2
𝑡−𝑧 =3 … … (4)
𝑒𝑞𝑛 (3) − 𝑒𝑞𝑛 (4) 𝑔𝑖𝑣𝑒𝑠:
1 7
𝑡 − 𝑡 + 𝑧 − (−𝑧) = − 3
2 2
1 7
𝑧+𝑧 = −3
2 2
1 7
( 𝑧 + 𝑧) × 2 = ( − 3) × 2
2 2
1 7
𝑧×2+𝑧×2= ×2−3×2
2 2
𝑧 + 2𝑧 = 7 − 6
3𝑧 = 1
3𝑧 1
=
3 3
1
𝑧=
3
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑧 𝑖𝑛 𝑒𝑞𝑛 (4)
𝑡−𝑧 =3
1
𝑡−( )=3
3
1
𝑡− =3
3
1
𝑡 =3+
3
3 1
𝑡= +
1 3
3×3 1
𝑡= +
1×3 3
9 1
𝑡= +
3 3
9+1
𝑡=
3
10
𝑡=
3

, Question 3



𝑃(𝑥) = 𝑥 2 − 𝑥 − 6
3 −1
𝐴=[ ]
0 −2


𝑃(𝐴) = 𝐴2 − 𝐴 − 6𝐼2



3 −1 2 3 −1 1 0
𝑃(𝐴) = [ ] −[ ]− 6[ ]
0 −2 0 −2 0 1
3 −1 3 −1 3 −1 6×1 6×0
𝑃(𝐴) = [ ][ ]−[ ]−[ ]
0 −2 0 −2 0 −2 6×0 6×1
(3 , −1) • (3 , 0) (3 , −1) • (−1 , −2) 3 −1 6 0
𝑃(𝐴) = [ ]−[ ]−[ ]
(0 , −2) • (3 , 0) (0 , −2) • (−1 , −2) 0 −2 0 6
3 × 3 + (−1 × 0) 3 × −1 + (−1 × −2) 3 −1 6 0
𝑃(𝐴) = [ ]−[ ]−[ ]
0 × 3 + (−2 × 0) 0 × −1 + (−2 × −2) 0 −2 0 6
9 + 0 −3 + 2 3 −1 6 0
𝑃(𝐴) = [ ]−[ ]−[ ]
0+0 0+4 0 −2 0 6
9 −1 3 −1 6 0
𝑃(𝐴) = [ ]−[ ]−[ ]
0 4 0 −2 0 6
9−3−6 −1 − (−1) − 0
𝑃(𝐴) = [ ]
0−0−0 4 − (−2) − 6
9 − 3 − 6 −1 + 1 − 0
𝑃(𝐴) = [ ]
0−0−0 4+2−6
6−6 0−0
𝑃(𝐴) = [ ]
0−0 6−6
0 0
𝑃(𝐴) = [ ]
0 0