FINAL TUTORIAL MTH3201 LINEAR ALGEBRA

1. Linear combination or not

(a) 𝛵: ℛ3 → ℛ2; 𝛵 𝑥, 𝑦, 𝑧 = (𝑥 − 𝑦, 𝑦 − 𝑧)

𝐿𝑒𝑡 𝑢 = 𝑥1, 𝑦1, 𝑧1 𝑎𝑛𝑑 𝑣 = 𝑥2, 𝑦2, 𝑧2

(i) 𝑢 + 𝑣 = 𝑥1 + 𝑥2, 𝑦1 + 𝑦2, 𝑧1 + 𝑧2

Condition given, please follow

𝛵 𝑢 + 𝑣 = 𝛵 𝑥1 + 𝑥2, 𝑦1 + 𝑦2, 𝑧1 + 𝑧2

By follow the condition,

= 𝑥1 + 𝑥2 − 𝑦1 − 𝑦2, 𝑦1 + 𝑦2 − 𝑧1 − 𝑧2

= 𝑥1 − 𝑦1 + 𝑥2 − 𝑦2, 𝑦1 − 𝑧1 + 𝑦2 − 𝑧2

= 𝑥1 − 𝑦1, 𝑦1 − 𝑧1) + (𝑥2 − 𝑦2, 𝑦2 − 𝑧2

= 𝛵(𝑢 ) + 𝛵(𝑣 ) (ii) 𝐼𝑓 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑠𝑐𝑎𝑙𝑎𝑟, 𝑘 ∈ ℜ,

𝑘𝑢 = 𝑘𝑥1, 𝑘𝑦1, 𝑘𝑧1

𝛵 (𝑘𝑢 ) = 𝛵 𝑘𝑥1, 𝑘𝑦1, 𝑘𝑧1

= 𝑘𝑥1 − 𝑘𝑦1, 𝑘𝑦1 − 𝑘𝑧1

= 𝑘 𝑥1 − 𝑦1, 𝑦1 − 𝑧1

= 𝑘𝛵(𝑢 )

∴ 𝛵 is linear combination

𝑆𝑖𝑛𝑐𝑒 𝛵 𝑢 + 𝑣 = 𝛵 𝑢 + 𝛵 𝑣 ,

1. Linear combination or not

(d) 𝛵: ℛ2 → ℛ; 𝛵 𝑥, 𝑦 =𝑥 𝑦

𝑥 + 𝑦 𝑥 − 𝑦 𝐿𝑒𝑡 𝑢 = 𝑥1, 𝑦1 𝑎𝑛𝑑 𝑣 = 𝑥2, 𝑦2

(i) 𝑢 + 𝑣 = 𝑥1 + 𝑥2, 𝑦1 + 𝑦2 ,

Condition given, please follow

𝛵 𝑢 + 𝑣 = 𝛵 𝑥1 + 𝑥2, 𝑦1 + 𝑦2

By follow the condition, 𝛵 𝑢 + 𝑣 =𝑥1 + 𝑥2 𝑦1 + 𝑦2

𝑥1 + 𝑥2 + 𝑦1 + 𝑦2 𝑥1 + 𝑥2 − 𝑦1 − 𝑦2

∗∗ 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝛵 𝑢 + 𝛵 𝑣 ? ? ?

= 𝑥1 + 𝑥2 𝑥1 + 𝑥2 − 𝑦1 − 𝑦2 − 𝑦1 + 𝑦2 𝑥1 + 𝑥2 + 𝑦1 + 𝑦2

= 𝑥12 + 𝑥1𝑥2 − 𝑥1𝑦1 − 𝑥1𝑦2 + 𝑥1𝑥2 + 𝑥2

2 − 𝑥2𝑦1 − 𝑥2𝑦2

− 𝑦1𝑥1 + 𝑦1𝑥2 + 𝑦12 + 𝑦1𝑦2 + 𝑦2𝑥1 + 𝑦2𝑥2 + 𝑦2𝑦1 + 𝑦2

2

= 𝑥12 + 𝑥2

2 − 𝑦12 − 𝑦2

2 + 2𝑥1𝑥2 − 2𝑦1𝑥2 − 2𝑥1𝑦1 − 2𝑥1𝑦2 − 2𝑥2𝑦1 − 2𝑥2𝑦2

𝛵 𝑢 + 𝛵 𝑣 = 𝛵 𝑥1, 𝑦1 + 𝛵 𝑥2, 𝑦2

=𝑥1 𝑦1

𝑥1 + 𝑦1 𝑥1 − 𝑦1+

𝑥2 𝑦2

𝑥2 + 𝑦2 𝑥2 − 𝑦2

= 𝑥12 + 𝑥2

2 − 𝑦12 − 𝑦2

2 − 2𝑥1𝑦1 − 2𝑥2𝑦2

∴ 𝛵 is not linear combination

compare

𝑆𝑖𝑛𝑐𝑒 𝛵 𝑢 + 𝑣 ≠ 𝛵 𝑢 + 𝛵 𝑣 ,

2(a) 𝛵2 ∙ 𝛵1 𝑝 𝑥 = 𝛵2 𝛵1(𝑝 𝑥 )

= 𝛵2 𝑝(𝑥 − 1)

= 𝑝(𝑥 − 1 + 2)

= 𝑝(𝑥 + 1)

(b) 𝛵1 ∙ 𝛵2 𝑝 𝑥 = 𝛵1 𝛵2(𝑝 𝑥 )

= 𝛵1 𝑝(𝑥 + 2)

= 𝑝(𝑥 + 2 − 1)

= 𝑝(𝑥 + 1)

𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑛: 𝛵1 𝑝(𝑥) = 𝑝 𝑥 − 1 , 𝛵2 𝑝 𝑥 = 𝑝 𝑥 + 2

3(a) 𝛵1 ∙ 𝛵2𝑎 𝑏𝑐 𝑑

= 𝛵1 𝛵2𝑎 𝑏𝑐 𝑑

= 𝛵1𝑎 𝑐𝑏 𝑑

= 𝑎 − 𝑐 + 4𝑏 − 𝑑

(b)

𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑛: 𝛵1𝑎 𝑏𝑐 𝑑

= 𝑎 − 𝑏 + 4𝑐 − 𝑑,

𝛵2𝑎 𝑏𝑐 𝑑

=𝑎 𝑐𝑏 𝑑

𝛵2 ∙ 𝛵1𝑎 𝑏𝑐 𝑑

= 𝛵2 𝛵1𝑎 𝑏𝑐 𝑑

= 𝛵2(𝑎 − 𝑏 + 4𝑐 − 𝑑)

∴ 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡, 𝑖𝑚𝑎𝑔𝑒 𝑇1 𝑛𝑜𝑡 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑑𝑜𝑚𝑎𝑖𝑛 𝑇2

4(a)

𝛵 𝑝 𝑥 = 𝑥2𝑝(𝑥)

𝑥2𝑝 𝑥 = 𝑥2 + 𝑥 𝑝 𝑥 = 1 + 1/𝑥

𝑝 𝑥 is not in domain of 𝑝2

∴ x2 +x is not in range(T)

𝑥2𝑝 𝑥 = 𝑥 + 1 𝑝 𝑥 = 1/𝑥 + 1/𝑥2

𝑝 𝑥 is not in domain of 𝑝2

∴ x + 1 is not in range(T)

𝑥2𝑝 𝑥 = 3 − 𝑥2

𝑝 𝑥 =3

𝑥2 − 1

𝑝 𝑥 is not in domain of 𝑝2

∴ 3 − x2 is not in range(T)

(i)

(ii)

(iii)

4(b)

𝛵 𝑝 𝑥 = 𝑥2𝑝(𝑥)

𝛵 𝑥2 = 𝑥2 ∙ 𝑥2= 𝑥4 𝑥4 ≠ 0

∴not in Kernel(T) (i)

(ii)

(iii)

𝛵 0 = 𝑥2 ∙ 0 = 0 ∴ in Kernel(T)

𝛵 𝑥 + 1 = 𝑥2 𝑥 + 1 = 𝑥3 + 𝑥2 ≠ 0 ∴not in Kernel(T)

5(a) 𝐴𝑥 = 0

4 5 7−6 1 −13 6 4

000

𝑟1/4

3𝑟1 + 2𝑟2

𝑟2 + 3𝑟3

1 5/4 7/40 17 190 13 7

000

𝑟2/17

13𝑟2 − 17𝑟3 1 5/4 7/40 1 19/170 0 128

000

𝑟3/128

𝑟1 −5

4𝑟2

1 0 6/170 1 19/170 0 1

000

𝑟1 −6

17𝑟3

𝑟2 −19

17𝑟3 1 0 0

0 1 00 0 1

000

∴ Since 𝑥1 = 0, 𝑥2 = 0, 𝑥3 = 0. There is no basis for Kernel (T)

∴ Basis for image (T)=4

−63

,516

,7

−14

5(b)

1 −1 35 6 −47 4 2

000

5𝑟1 − 𝑟2

7𝑟1 − 𝑟3

1 −1 30 −11 190 −11 19

000

𝐴𝑥 = 0

𝑟2/-11

𝑟2 − 𝑟3 1 −1 30 1 −19/110 0 0

000

𝑟1 + 𝑟2 1 0 14/110 1 −19/110 0 0

000

𝐿𝑒𝑡 𝑥3 = 𝑡 ,

𝑥2 =19

11𝑡, 𝑥1 = −

14

11𝑡

𝑥 =

−14

11𝑡

19

11𝑡

𝑡

=𝑡

11

−141911

∴ Basis for range (T)=157

,−164

∴ Basis for kernel (T)=−141911

6. 𝑇 𝑥, 𝑦, 𝑧 = (0,0,0)

2 4 −61 −2 15 −2 −3

000

𝑟1 − 𝑟2

5𝑟2 − 𝑟3

1 2 −30 8 −80 −8 8

000

𝑟2 − 2𝑟2 𝐿𝑒𝑡 𝑥3 = 𝑡 ,

𝑥2 = 𝑡, 𝑥1 = 𝑡

𝑥 =𝑡 𝑡𝑡

= 𝑡111

∴ Basis for range (T)=215

,4

−2−2

∴ Basis for kernel (T)=111

(2𝑥 + 4𝑦 − 6𝑧, 𝑥 − 2𝑦 + 𝑧, 5𝑥 − 2𝑦 − 3𝑧) = (0,0,0) 2𝑥 + 4𝑦 − 6𝑧 = 0

𝑥 − 2𝑦 + 𝑧 = 0 5𝑥 − 2𝑦 − 3𝑧 = 0

𝑟1/2

𝑟2/8

𝑟3 + 𝑟2 1 2 −30 1 −10 0 0

000

1 0 −10 1 −10 0 0

000

TAMAT