Math Problems With Solution 2022

Added on 2022-09-28

6 Pages1211 Words19 Views

Surname 1
Student Name
Instructor Name
Course
15, August 2019
Assignment 1 for MATH203
Q1.
a.
Solution
The transpose AT of 𝑚 𝑥 𝑛 matrix A is 𝑚 𝑥 𝑛 matrix
It is gotten from A by alternating both rows and columns.
For symmetric matrix then
AT = A.
The set of all symmetric matrix in Mn(R) is a subspace of Mn(R).
Verifying for M2(R) is a vector space
𝐴 = (𝑎1 𝑎2
𝑎3 𝑎4) , 𝐵 = (𝑏1 𝑏2
𝑏3 𝑏4
)
The Sum of a 2 𝑥 2 matrices is hence a 2 𝑥 2 matrix
𝐴 + 𝐵 → [𝑎1 + 𝑏1 𝑎2 + 𝑏2
𝑎3 + 𝑏3 𝑎4 + 𝑏4
]
But,
[𝑏1 + 𝑎1 𝑏2 + 𝑎2
𝑏3 + 𝑎3 𝑏4 + 𝑎4
] = 𝐵 + 𝐴
A 2 𝑥 2 matrix multiplied by a scalar also results in a 2 𝑥 2 matrix
In three 2 𝑥 2 matrix,
𝐴 = (𝑎1 𝑎2
𝑎3 𝑎4) , 𝐵 = (𝑏1 𝑏2
𝑏3 𝑏4
) , 𝐶 = (𝑐1 𝑐2
𝑐3 𝑐4)
We have
(𝐴 + 𝐵) + 𝐶 = 𝐴 + (𝐵 + 𝐶)

Surname 2
The additive inverse of 𝐴 = (𝑎 𝑏
𝑐 𝑑) 𝑖𝑠 − 𝐴 = (−𝑎 −𝑏
−𝑐 −𝑑)
If A is a matrix then 1A = A
Given matrix 𝐴 = (𝑎 𝑏
𝑐 𝑑) and scalars t and h, we have
(𝑡 + ℎ)𝐴 = ((𝑡 + ℎ)𝑎 (𝑡 + ℎ)𝑏
(𝑡 + ℎ)𝑐 (𝑡 + ℎ)𝑑)
= (𝑡𝑎 + ℎ𝑎 𝑡𝑏 + ℎ𝑏
𝑟𝑐 + ℎ𝑐 𝑡𝑑 + ℎ𝑑)
= 𝑡𝐴 + ℎ𝐴
b.
Solution
If A, B are upper-triangular matrices
𝐴𝑖𝑗 𝑎𝑛𝑑 𝐵𝑖𝑗 𝑎𝑟𝑒 0 𝑤ℎ𝑒𝑛𝑒𝑣𝑒𝑟 𝑖 > 𝑗
This means (𝐴 + 𝐵)𝑖𝑗 = 𝐴𝑖𝑗 + 𝐵𝑖𝑗 𝑖𝑠 0 𝑤ℎ𝑒𝑛 𝑖 > 𝑗
Hence, (𝐴 + 𝐵) is the upper-triangular.
Hence if c is a constant scalar
Then we have
(𝑐𝐴)𝑖𝑗 = 𝑐(𝐴𝑖𝑗) = 0 𝑤ℎ𝑒𝑛 𝑖 > 𝑗
Hence (𝑐𝐴) is upper-triangular.
Therefore, 0 matrix is a upper-triangular.
Hence the upper-triangular matrices will form a subspace of Mn(R).

Surname 3
Q2.
Solution
𝑃2(𝑅) 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠 𝑜𝑠 𝑃3(𝑅)
Suppose V → vector space.
U is a subspace of V
If and only if it satisfies the three properties:
1... U = nonempty subset of V
2... x + y ∈ U for all x, y ∈ U ...under addition
3... tx ∈ U for all x ∈ U and t ∈ R ... under a scalar multiplication
The elements of 𝑅2 have exactly 2 entries, and the elements of 𝑅3 have three entries.
Therefore 𝑅2 is not a subspace of 𝑅3
And the inverse is also true: 𝑅3 is not a subset of 𝑅2.
Similarly, M(2, 2) is not a subspace of M(2, 3), because M(2, 2) is not a subset of M(2, 3).
Q 3.
a.
Solution
𝐹 = {𝑓 ∶ [−1, 1] → 𝑅
𝐸[−1, 1] = 𝑓 − (𝑥) → 𝑓(𝑥)}
𝑂 = 𝑓(−𝑥) = −𝑓(𝑥)
To find the ℝ 𝑖𝑛 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟,
𝐸 = ∫ −𝑥
1
−1
= 𝑓(𝑥) 𝑂 = ∫ −𝑥
1
−1
= −𝑓(𝑥)
Hence, since
𝐸 = ∫ −𝑥
1
−1
= 𝑓(𝑥)

End of preview

Want to access all the pages? Upload your documents or become a member.