Math Problem Statement

The answer: 1) 𝑒1 βƒ— βƒ— βƒ—βƒ— = ( 3 βˆ’5) , 𝑒2 βƒ— βƒ— βƒ—βƒ— = (4 0) The question: Find the basic vectors of the basis E

βƒ—βƒ—

,

If:

  1. a

βƒ—

= (22

βˆ’10), b

βƒ—βƒ—

= (21

5 ) and

A βƒ— E βƒ— βƒ— = (2 4), b βƒ—βƒ— E βƒ— βƒ— = (-1 6 ) coordinate vectors in the basis E

βƒ—βƒ—

.

Solution

To find the basic vectors of the basis EE from the information given, we’ll proceed as follows:

Step 1: Analyze Given Basis Vectors

We are given: e1=(3βˆ’5),e2=(40)\mathbf{e}_1 = \begin{pmatrix} 3 \\ -5 \end{pmatrix}, \quad \mathbf{e}_2 = \begin{pmatrix} 4 \\ 0 \end{pmatrix} These vectors form the basis EE.

Step 2: Write Down Coordinate Transformations

The question also provides coordinates for two vectors a\mathbf{a} and b\mathbf{b} in terms of both the standard basis and basis EE:

  1. Vector a\mathbf{a} in the standard basis is a=(22βˆ’10)\mathbf{a} = \begin{pmatrix} 22 \\ -10 \end{pmatrix}.
  2. The coordinates of a\mathbf{a} in basis EE are given as aE=(24)\mathbf{a}_E = \begin{pmatrix} 2 \\ 4 \end{pmatrix}.
  3. Vector b\mathbf{b} in the standard basis is b=(215)\mathbf{b} = \begin{pmatrix} 21 \\ 5 \end{pmatrix}.
  4. The coordinates of b\mathbf{b} in basis EE are given as bE=(βˆ’16)\mathbf{b}_E = \begin{pmatrix} -1 \\ 6 \end{pmatrix}.

Step 3: Verify Basis Vectors of EE

To confirm that e1\mathbf{e}_1 and e2\mathbf{e}_2 form the basis EE, we can check if a\mathbf{a} and b\mathbf{b} can be expressed as linear combinations of e1\mathbf{e}_1 and e2\mathbf{e}_2 using the provided coordinates in basis EE.

For a\mathbf{a}:

a=2β‹…e1+4β‹…e2=2(3βˆ’5)+4(40)\mathbf{a} = 2 \cdot \mathbf{e}_1 + 4 \cdot \mathbf{e}_2 = 2 \begin{pmatrix} 3 \\ -5 \end{pmatrix} + 4 \begin{pmatrix} 4 \\ 0 \end{pmatrix} Calculating this: a=(6βˆ’10)+(160)=(22βˆ’10)\mathbf{a} = \begin{pmatrix} 6 \\ -10 \end{pmatrix} + \begin{pmatrix} 16 \\ 0 \end{pmatrix} = \begin{pmatrix} 22 \\ -10 \end{pmatrix} This matches the given a\mathbf{a}.

For b\mathbf{b}:

b=βˆ’1β‹…e1+6β‹…e2=βˆ’1(3βˆ’5)+6(40)\mathbf{b} = -1 \cdot \mathbf{e}_1 + 6 \cdot \mathbf{e}_2 = -1 \begin{pmatrix} 3 \\ -5 \end{pmatrix} + 6 \begin{pmatrix} 4 \\ 0 \end{pmatrix} Calculating this: b=(βˆ’35)+(240)=(215)\mathbf{b} = \begin{pmatrix} -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 24 \\ 0 \end{pmatrix} = \begin{pmatrix} 21 \\ 5 \end{pmatrix} This matches the given b\mathbf{b}.

Conclusion

The basis EE is confirmed to be: e1=(3βˆ’5),e2=(40)\mathbf{e}_1 = \begin{pmatrix} 3 \\ -5 \end{pmatrix}, \quad \mathbf{e}_2 = \begin{pmatrix} 4 \\ 0 \end{pmatrix}

Let me know if you'd like additional details or have questions.


Further Questions:

  1. How can we find the inverse of a coordinate transformation?
  2. What are the properties of a vector space basis?
  3. How can we check if a set of vectors forms a basis?
  4. What is the significance of linear independence in a basis?
  5. How do coordinate transformations affect vector norms?

Tip: Always verify that your calculated coordinates match the given ones when dealing with different bases to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Coordinate Transformations
Basis and Dimension

Formulas

Vector representation in terms of basis: v = a1 * e1 + a2 * e2
Coordinate transformation formula for vector spaces

Theorems

Linear Independence
Basis Definition

Suitable Grade Level

College Level or Advanced High School