MAT2611 SEMESTER 1 ASSIGNMENT 4 2022
Problem 11
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 𝑡)} 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑀𝑒𝑡ℎ𝑜𝑑 1
𝑂𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑘1 (3, 1, 4) + 𝑘2 (2, −3, 5) + 𝑘3 (5, 9, 𝑡) = (0, 0, 0)
𝑇ℎ𝑒 𝑘𝑖 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑏𝑒 𝑎𝑙𝑙 𝑒𝑞𝑢𝑎𝑙 𝑧𝑒𝑟𝑜
(𝑘1 , 𝑘2 , 𝑘3 ) ≠ (0, 0, 0)
(3𝑘1 , 1𝑘1 , 4𝑘1 ) + (2𝑘2 , −3𝑘2 , 5𝑘2 ) + (5𝑘3 , 9𝑘3 , 𝑡𝑘3 ) = (0, 0, 0)
(3𝑘1 + 2𝑘2 + 5𝑘3 , 1𝑘1 − 3𝑘2 + 9𝑘3 , 4𝑘1 + 5𝑘2 + 𝑡𝑘3 ) = (0, 0, 0)
3𝑘1 + 2𝑘2 + 5𝑘3 = 0
1𝑘1 − 3𝑘2 + 9𝑘3 = 0
4𝑘1 + 5𝑘2 + 𝑡𝑘3 = 0
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑚𝑢𝑠𝑡 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑢𝑛𝑖𝑞𝑢𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (0, 0, 0).
𝐼𝑡 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
3 2 5 0
[1 −3 9| 0]
4 5 𝑡 0
1
𝑅2 → 𝑅2 − 𝑅1
3
4
𝑅3 → 𝑅3 − 𝑅1
3
3 2 5 0
1 1 1 1
1 − (3) −3 − (2) 9 − (5)| 0 − (0)
3 3 3 | 3
4 4 4 4
4 − (3) 5 − (2) 𝑡 − (5) 0 − (0)
[ 3 3 3 3 ]
3 2 5
11 22 0
0 − |
3 3 |0
7 20 0
[0 3
𝑡−
3 ]
, 3 2 5
11 22 0
0 − |
3 3 |0
7 20 0
[0 3
𝑡−
3 ]
𝑅2 → 3𝑅2
𝑅3 → 3𝑅3
3 2 5
11 22 0
3(0) 3 (− ) 3( ) | ( )
3 3 3 0
|
7 20 3(0)
[3(0) 3( )
3
3 (𝑡 − )
3 ]
3 2 5 0
[0 −11 22 | 0]
0 7 3𝑡 − 20 0
7
𝑅3 → 𝑅3 + 𝑅
11 2
3 2 5 0
0 −11 22 0
[ 7 7 | 7 ]
0 7 + (−11) 3𝑡 − 20 + (22) 0 + (0)
11 11 11
3 2 5 0
[0 −11 22 | 0]
0 0 3𝑡 − 6 0
𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑓:
3𝑡 − 6 = 0
3𝑡 = 6
6
𝑡=
3
𝑡=2
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 2)} 𝑖𝑠 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑀𝑒𝑡ℎ𝑜𝑑 2
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 𝑡)} 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 (5, 9, 𝑡) 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 (3, 1, 4) 𝑎𝑛𝑑 (2, −3, 5)
(5, 9, 𝑡) = 𝑎1 (3, 1, 4) + 𝑎2 (2, −3, 5) 𝑤ℎ𝑒𝑟𝑒 𝑎1 𝑎𝑛𝑑 𝑎2 ∈ ℝ
(5, 9, 𝑡) = (3𝑎1 , 1𝑎1 , 4𝑎1 ) + (2𝑎2 , −3𝑎2 , 5𝑎2 )
(5, 9, 𝑡) = (3𝑎1 + 2𝑎2 , 1𝑎1 − 3𝑎2 , 4𝑎1 + 5𝑎2 )
Problem 11
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 𝑡)} 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑀𝑒𝑡ℎ𝑜𝑑 1
𝑂𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑘1 (3, 1, 4) + 𝑘2 (2, −3, 5) + 𝑘3 (5, 9, 𝑡) = (0, 0, 0)
𝑇ℎ𝑒 𝑘𝑖 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑏𝑒 𝑎𝑙𝑙 𝑒𝑞𝑢𝑎𝑙 𝑧𝑒𝑟𝑜
(𝑘1 , 𝑘2 , 𝑘3 ) ≠ (0, 0, 0)
(3𝑘1 , 1𝑘1 , 4𝑘1 ) + (2𝑘2 , −3𝑘2 , 5𝑘2 ) + (5𝑘3 , 9𝑘3 , 𝑡𝑘3 ) = (0, 0, 0)
(3𝑘1 + 2𝑘2 + 5𝑘3 , 1𝑘1 − 3𝑘2 + 9𝑘3 , 4𝑘1 + 5𝑘2 + 𝑡𝑘3 ) = (0, 0, 0)
3𝑘1 + 2𝑘2 + 5𝑘3 = 0
1𝑘1 − 3𝑘2 + 9𝑘3 = 0
4𝑘1 + 5𝑘2 + 𝑡𝑘3 = 0
𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑚𝑢𝑠𝑡 𝑛𝑜𝑡 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑢𝑛𝑖𝑞𝑢𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (0, 0, 0).
𝐼𝑡 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
3 2 5 0
[1 −3 9| 0]
4 5 𝑡 0
1
𝑅2 → 𝑅2 − 𝑅1
3
4
𝑅3 → 𝑅3 − 𝑅1
3
3 2 5 0
1 1 1 1
1 − (3) −3 − (2) 9 − (5)| 0 − (0)
3 3 3 | 3
4 4 4 4
4 − (3) 5 − (2) 𝑡 − (5) 0 − (0)
[ 3 3 3 3 ]
3 2 5
11 22 0
0 − |
3 3 |0
7 20 0
[0 3
𝑡−
3 ]
, 3 2 5
11 22 0
0 − |
3 3 |0
7 20 0
[0 3
𝑡−
3 ]
𝑅2 → 3𝑅2
𝑅3 → 3𝑅3
3 2 5
11 22 0
3(0) 3 (− ) 3( ) | ( )
3 3 3 0
|
7 20 3(0)
[3(0) 3( )
3
3 (𝑡 − )
3 ]
3 2 5 0
[0 −11 22 | 0]
0 7 3𝑡 − 20 0
7
𝑅3 → 𝑅3 + 𝑅
11 2
3 2 5 0
0 −11 22 0
[ 7 7 | 7 ]
0 7 + (−11) 3𝑡 − 20 + (22) 0 + (0)
11 11 11
3 2 5 0
[0 −11 22 | 0]
0 0 3𝑡 − 6 0
𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑓:
3𝑡 − 6 = 0
3𝑡 = 6
6
𝑡=
3
𝑡=2
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 2)} 𝑖𝑠 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑀𝑒𝑡ℎ𝑜𝑑 2
𝑇ℎ𝑒 𝑠𝑒𝑡 {(3, 1, 4), (2, −3, 5), (5, 9, 𝑡)} 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑛𝑜𝑡 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.
𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 (5, 9, 𝑡) 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 (3, 1, 4) 𝑎𝑛𝑑 (2, −3, 5)
(5, 9, 𝑡) = 𝑎1 (3, 1, 4) + 𝑎2 (2, −3, 5) 𝑤ℎ𝑒𝑟𝑒 𝑎1 𝑎𝑛𝑑 𝑎2 ∈ ℝ
(5, 9, 𝑡) = (3𝑎1 , 1𝑎1 , 4𝑎1 ) + (2𝑎2 , −3𝑎2 , 5𝑎2 )
(5, 9, 𝑡) = (3𝑎1 + 2𝑎2 , 1𝑎1 − 3𝑎2 , 4𝑎1 + 5𝑎2 )
