2. FINAL TUTORIAL
MTH3201 LINEAR ALGEBRA

3. 1. Linear combination or not
(a) 𝛵: ℛ 3 → ℛ 2 ; 𝛵 𝑥, 𝑦, 𝑧 = (𝑥 − 𝑦, 𝑦 − 𝑧)
Condition given, please follow
𝐿𝑒𝑡 𝑢 = 𝑥1 , 𝑦1 , 𝑧1 𝑎𝑛𝑑 𝑣 = 𝑥2 , 𝑦2 , 𝑧2
(i) 𝑢 + 𝑣 = 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝑧1 + 𝑧2
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 𝑆𝑖𝑛𝑐𝑒 𝛵 𝑢 + 𝑣 = 𝛵 𝑢 + 𝛵 𝑣 ,
𝛵 (𝑘𝑢) = 𝛵 𝑘𝑥1 , 𝑘𝑦1 , 𝑘𝑧1 ∴ 𝛵 is linear combination
= 𝑘𝑥1 − 𝑘𝑦1 , 𝑘𝑦1 − 𝑘𝑧1
= 𝑘 𝑥1 − 𝑦1 , 𝑦1 − 𝑧1
= 𝑘𝛵(𝑢 )

4. 1. Linear combination or not
𝑥 𝑦
(d) 𝛵: ℛ 2 → ℛ; 𝛵 𝑥, 𝑦 = 𝐿𝑒𝑡 𝑢 = 𝑥1 , 𝑦1 𝑎𝑛𝑑 𝑣 = 𝑥2 , 𝑦2
𝑥+ 𝑦 𝑥− 𝑦
Condition given, please follow
(i) 𝑢 + 𝑣 = 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 𝛵 𝑢 + 𝑣 = 𝛵 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2
𝑥 + 𝑥 𝑦1 + 𝑦2
By follow the condition,
𝛵 𝑢 + 𝑣 = 𝑥 + 𝑥1 + 𝑦2 + 𝑦 𝑥1 + 𝑥2 − 𝑦1 − 𝑦2
1 2 1 2
= 𝑥1 + 𝑥2 𝑥1 + 𝑥2 − 𝑦1 − 𝑦2 − 𝑦1 + 𝑦2 𝑥1 + 𝑥2 + 𝑦1 + 𝑦2
= 𝑥1 2 + 𝑥1 𝑥2 − 𝑥1 𝑦1 − 𝑥1 𝑦2 + 𝑥1 𝑥2 + 𝑥2 2 − 𝑥2 𝑦1 − 𝑥2 𝑦2
− 𝑦1 𝑥1 + 𝑦1 𝑥2 + 𝑦1 2 + 𝑦1 𝑦2 + 𝑦2 𝑥1 + 𝑦2 𝑥2 + 𝑦2 𝑦1 + 𝑦2 2
= 𝑥1 2 + 𝑥2 2 − 𝑦1 2 − 𝑦2 2 + 2𝑥1 𝑥2 − 2𝑦1 𝑥2 − 2𝑥1 𝑦1 − 2𝑥1 𝑦2 − 2𝑥2 𝑦1 − 2𝑥2 𝑦2
∗∗ 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝛵 𝑢 + 𝛵 𝑣 ? ? ?
𝛵 𝑢 + 𝛵 𝑣 = 𝛵 𝑥1 , 𝑦1 + 𝛵 𝑥2 , 𝑦2 compare
𝑥1 𝑦1 𝑥2 𝑦2
= 𝑥 + 𝑦 𝑥1 − 𝑦1 + 𝑥 + 𝑦 𝑥2 − 𝑦2
1 1 2 2
= 𝑥1 2 + 𝑥2 2 − 𝑦1 2 − 𝑦2 2 − 2𝑥1 𝑦1 − 2𝑥2 𝑦2
𝑆𝑖𝑛𝑐𝑒 𝛵 𝑢 + 𝑣 ≠ 𝛵 𝑢 + 𝛵 𝑣 , ∴ 𝛵 is not linear combination

5. 2(a) 𝛵2 ∙ 𝛵1 𝑝 𝑥 = 𝛵2 𝛵1 (𝑝 𝑥 ) 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑛: 𝛵1 𝑝(𝑥) = 𝑝 𝑥 − 1 ,
= 𝛵2 𝑝(𝑥 − 1) 𝛵2 𝑝 𝑥 = 𝑝 𝑥 + 2
= 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)

6. 3(a) 𝑎 𝑏 𝑎 𝑏 𝑎 𝑐
𝛵1 ∙ 𝛵2 = 𝛵1 𝛵2 = 𝛵1 = 𝑎 − 𝑐 + 4𝑏 − 𝑑
𝑐 𝑑 𝑐 𝑑 𝑏 𝑑
𝑎 𝑏
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑛: 𝛵1 = 𝑎 − 𝑏 + 4𝑐 − 𝑑,
𝑐 𝑑
𝑎 𝑏 𝑎 𝑐
𝛵2 =
𝑐 𝑑 𝑏 𝑑
𝑎 𝑏 𝑎 𝑏
(b) 𝛵2 ∙ 𝛵1 = 𝛵2 𝛵1 = 𝛵2 (𝑎 − 𝑏 + 4𝑐 − 𝑑)
𝑐 𝑑 𝑐 𝑑
∴ 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡, 𝑖𝑚𝑎𝑔𝑒 𝑇1 𝑛𝑜𝑡 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑑𝑜𝑚𝑎𝑖𝑛 𝑇2

7. 4(a) 𝑥2 𝑝 𝑥 = 𝑥2 + 𝑥
(i) 𝑝 𝑥 = 1 + 1/𝑥
𝑝 𝑥 is not in domain of 𝑝2
∴ x 2 +x is not in range(T)
𝛵 𝑝 𝑥 = 𝑥 2 𝑝(𝑥)
(ii)
𝑥2 𝑝 𝑥 = 𝑥 + 1
(iii) 𝑝 𝑥 = 1/𝑥 + 1/𝑥 2
𝑝 𝑥 is not in domain of 𝑝2
∴ x + 1 is not in range(T)
𝑥2 𝑝 𝑥 = 3 − 𝑥2
3
𝑝 𝑥 = 2−1
𝑥
𝑝 𝑥 is not in domain of 𝑝2
∴ 3 − x 2 is not in range(T)

8. 4(b) 𝛵 𝑥 2 = 𝑥 2 ∙ 𝑥 2= 𝑥 4
(i) 𝑥4 ≠ 0
∴not in Kernel(T)
𝛵 𝑝 𝑥 = 𝑥 2 𝑝(𝑥)
(ii)
𝛵 0 = 𝑥2 ∙ 0 = 0
(iii) ∴ in Kernel(T)
𝛵 𝑥 + 1 = 𝑥2 𝑥 + 1 = 𝑥3 + 𝑥2 ≠ 0
∴not in Kernel(T)

9. 5(a) 𝐴𝑥 = 0
𝑟1 /4
4 5 7 0 3𝑟1 + 2𝑟2 1 5/4 7/4 0 13𝑟2 − 17𝑟3 1 5/4 7/4 0
−6 1 −1 0 0 17 19 0 0 1 19/17 0
𝑟2 + 3𝑟3 𝑟2 /17
3 6 4 0 0 13 7 0 0 0 128 0
19
𝑟3 /128 1 0 6/17 0 𝑟2 − 𝑟 1 0 0 0
17 3
0 1 19/17 0 0 1 0 0
5 6
𝑟1 − 𝑟 0 0 1 0 𝑟1 − 𝑟 0 0 1 0
4 2 17 3
∴ Since 𝑥1 = 0, 𝑥2 = 0, 𝑥3 = 0. There is no basis for Kernel (T)
4 5 7
∴ Basis for image (T)= −6 , 1 , −1
3 6 4

10. 5(b) 𝐴𝑥 = 0
1 −1 3 0 5𝑟1 − 𝑟2 1 −1 3 0 𝑟2 − 𝑟3 1 −1 3 0
5 6 −4 0 0 −11 19 0 0 1 −19/11 0
7 4 2 0 7𝑟1 − 𝑟3 0 −11 19 0 𝑟2 /-11 0 0 0 0
19 14
𝑟1 + 𝑟2 1 0 14/11 0 𝐿𝑒𝑡 𝑥3 = 𝑡 , 𝑥2 = 𝑡, 𝑥1 = − 𝑡
11 11
0 1 −19/11 0
0 14
0 0 0 − 𝑡
11 𝑡 −14
𝑥 = 19 = 19
𝑡 11
11 11
𝑡
1 −1
∴ Basis for range (T)= 5 , 6
7 4
−14
∴ Basis for kernel (T)= 19
11

11. 6. 𝑇 𝑥, 𝑦, 𝑧 = (0,0,0) 2𝑥 + 4𝑦 − 6𝑧 = 0
(2𝑥 + 4𝑦 − 6𝑧, 𝑥 − 2𝑦 + 𝑧, 5𝑥 − 2𝑦 − 3𝑧) = (0,0,0) 𝑥 − 2𝑦 + 𝑧 = 0
5𝑥 − 2𝑦 − 3𝑧 = 0
𝑟1 /2
2 4 −6 0 𝑟1 − 𝑟2 1 2 −3 0 𝑟3 + 𝑟2 1 2 −3 0
1 −2 1 0 0 8 −8 0 0 1 −1 0
5𝑟2 − 𝑟3 𝑟2 /8 0 0 0 0
5 −2 −3 0 0 −8 8 0
𝑟2 − 2𝑟2 1 0 −1 0 𝐿𝑒𝑡 𝑥3 = 𝑡 , 𝑥2 = 𝑡, 𝑥1 = 𝑡
0 1 −1 0
𝑡 1
0 0 0 0
𝑥= 𝑡 = 𝑡 1
𝑡 1
2 4
∴ Basis for range (T)= 1 , −2
5 −2
1
∴ Basis for kernel (T)= 1
1

12. TAMAT