FInal Examination

Question 1

Explain the solution

Answer 1

Since these two vectors are linearly independent, we can find the perpendicular basis by simply setting them to row vectors and solving for the new basis.

|1 2 3 4 | —-> C [1 -2 1 0]
|5 6 7 8 | **D [2 -3 0 1] **

Question 2

Explain the solution

Answer 2

Find the projection onto each vector using the (dot product)/magnitudes * unit vector of each vector.

Then add up all the new projections.

Question 3

Explain the Solution

Answer 3

₁₂Start by finding the Unit Vector for 𝓥 using:

𝓾₁ = [1 / ||𝓥₁||] * [𝓥₁]

Then we will find the perpendicular 𝓥₂ using the equation: 𝓥₂⊥ = 𝓥₂ - (𝓾₁ ⋅ 𝓥₂)𝓾₁

then we find 𝓾₂ using

𝓾₂ = [1 / ||𝓥₂⊥||] * [𝓥₂⊥]

Question 4

What is equation to find 𝓥₃⊥ and 𝓾₃ using Gram–Schmidt process ?

Answer 4

𝓥₃⊥ = 𝓥₃ - (𝓾₁ ⋅ 𝓥₃)𝓾₁ - (𝓾₂ ⋅ 𝓥₃)𝓾₂

then: 𝓾₃ = [1/||𝓥₃⊥||] * 𝓥₃⊥

Question 5

Go about solving this problem ?

Answer 5

Start by setting this matrix equal to 0 and solving for the Kernel.

Create a matrix out of the Kernel [-t -r r t] then set to two kernels to their own linearly independent vectors

then solve using the Gram–Schmidt process for the Orthonormal Basis

Question 6

With regard to Determinants what condition is a matrix invertible ?

Answer 6

when det(A) = ad - bc ≠ 0

Question 6

What can we conclude about the relationship between det(A) and its transposed matrix det(A^t)

Answer 6

det(A) = det(A^t) so long as its a square matrix

Question 6

How to find Determinants of 4x4 matrix ?

Answer 6

set the first row as the coefficient and follow the crossout method in order to set a few rows sets of 3x3 alternating in positivity. then solve each 3x3 using the same method as before.

Question 6

Explain the Solutions to these

Answer 6

11. Since there is a scalar being applied to a row we must multiply 8 * -9 = -72
12. Since a row was switched it must be (-1)¹8 = -8
13. Since there is 2 row switched it must be (-1)²8 = 8
14. The operation of adding another row times a scalar doesnt change the det(A) so 8

Question 7

How to find Basis of the Subspace of R^3 defined by equation:

2x₁ + 3x₂ + x₃ = 0 ?

Answer 7

adjust the equation:

x₁ = - (3/2)x₂ - (1/2)x₃
then set x₂ = t; x₃ = s

we now have [x₁, x₂, x₃] = [- 3/2t -1/2s, t, s]

this is now the two basis t[-3, 2, 0] & s[-1, 0, 2]

Question 8

How can i tell if a Vector x is a part of Span V of vectors 𝓥₁ …… 𝓥ₘ

Answer 8

We can set all the vectors into an augmented matrix equal to vector x and determine whether there a coordinates which can be solved for.

Question 9

Solve

Answer 9

Convert from 2x2 format to R^4. line the vectors into columns. then solve for linear independence.

Question 10

Solve to Basis

|1 1|
|0 1|

Answer 10

SImply set a Basis with arbitrary letters
|a b|
|0 c| Then solve a dot product between the two matrix’s

|1 2| |a b| = |a b+c |
|0 3| |0 c| |0 3c | Then align vertically for our matrix and solve for RREF

Question 11

Explain Solution

Answer 11

Set M at each Basis
Align the new Matrix Vertically
Then solve RREF

Question 12

Explain Solution

Answer 12

Use Diagram:

x + iy → x - iy
| |
|x| |x |
|y| |-y| Then convert to matrix and solve RREF

Question 13

Find all the 2x2 Matrix such that e₁ = |1|
|0|
is an EigenVector with EigenValue of 5

Answer 13

Start by setting an arbitrary 2x2 matrix times e₁ = 5 e₁

|a b||1| = 5|1|
|c d||0|……..|0|

Then solve for the values of a b c d that would make the 2x2 which would satisfy the statement.

Question 14

Find an EigenBasis for a Rotation 180

Answer 14

All we are doing is flipping the axis of a basic Identity Vector:

|-1 0|x
|0 -1|y

Question 15

equation for EigenValues

Answer 15

We can simply use the determamite, set it to 0 and solve for a value of λ that makes the det = 0

det(A - λＩₙ) = 0

Question 16

Answer all

Answer 16

1. Yes we can solve A³ down to λ³𝓥 through extrapolation

2. Yes it is simply 1/λ as we can basic algebra to solve for this

3. Yes it is simply λ + 2

4. Yes it is simply 7λ

Question 16

How to solve for the EigenBasis ?

Answer 16

Start by setting the det(A - λＩₙ) = 0

det |7 8| - |λ 0| = 0
……..|0 9| |0 λ|

|7-λ 8 | → (7-λ)(9-λ) = 0
|0 9-λ|…..our EigenValues are 7 and 9

now we will solve for the Kernel with these new values:

Ker…|7 8| - |7 0|
…………..|0 9|……….|0 7|

|0 8|x = 0 → y = 0…….our first kernel = |1|
|0 2|y………………………………………………………|0|