3-1) Let ẞ = {b_1, b_2} and C = {C_1, C_2} be bases for R². b₁=[[7],[3]], b_2 = [[-3],[-1]], C₁ = [[1],[-5]], C2 = [[-2],[2]], and x =[[4],[6]]

a) Find the change-of-coordinate matrix from ẞ to the standard basis, and obtain the coordinate vector [x]_ẞ of x relative to β.

b) Find the change-of-coordinate matrix from C to the standard basis, and obtain the coordinate vector [x]_c of x relative to C.

c) Find the change-of-coordinate matrix from ẞ to C and the change-of-coordinate matrix from C to β.

d) Transform [x]_c to [x]_β by using the change-of-coordinate matrix you found in this problem.



solve it step by step

Let's solve this problem step by step.

### Part (a) - Change-of-Coordinate Matrix from \( \beta \) to the Standard Basis, and Coordinate Vector \( [x]_{\beta} \) of \( x \) Relative to \( \beta \)

#### 1. Find the Change-of-Coordinate Matrix from \( \beta \) to the Standard Basis

Given:
- \( \beta = \{ b_1, b_2 \} \)
- \( b_1 = \begin{bmatrix} 7 \\ 3 \end{bmatrix} \), \( b_2 = \begin{bmatrix} -3 \\ -1 \end{bmatrix} \)

The change-of-coordinate matrix \( P_{\beta \to \text{std}} \) from \( \beta \) to the standard basis is formed by placing the basis vectors \( b_1 \) and \( b_2 \) as columns in a matrix:

\[
P_{\beta \to \text{std}} = \begin{bmatrix} b_1 & b_2 \end{bmatrix} = \begin{bmatrix} 7 & -3 \\ 3 & -1 \end{bmatrix}
\]

#### 2. Find the Coordinate Vector \( [x]_{\beta} \)

Given:
- \( x = \begin{bmatrix} 4 \\ 6 \end{bmatrix} \)

To find \( [x]_{\beta} \), we need to solve the equation:

\[
P_{\beta \to \text{std}} \cdot [x]_{\beta} = x
\]

This translates to:

\[
\begin{bmatrix} 7 & -3 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}
\]

This is a system of linear equations:
\[
\begin{aligned}
7c_1 - 3c_2 &= 4 \\
3c_1 - c_2 &= 6
\end{aligned}
\]

Let's solve this system using the method of substitution or elimination.

Multiplying the second equation by 3:

\[
9c_1 - 3c_2 = 18
\]

Subtracting the first equation from this:

\[
(9c_1 - 3c_2) - (7c_1 - 3c_2) = 18 - 4
\]

\[
2c_1 = 14 \implies c_1 = 7
\]

Now, substitute \( c_1 = 7 \) into the second equation:

\[
3(7) - c_2 = 6 \implies 21 - c_2 = 6 \implies c_2 = 15
\]

So, \( [x]_{\beta} = \begin{bmatrix} 7 \\ 15 \end{bmatrix} \).

### Part (b) - Change-of-Coordinate Matrix from \( C \) to the Standard Basis, and Coordinate Vector \( [x]_C \) of \( x \) Relative to \( C \)

#### 1. Find the Change-of-Coordinate Matrix from \( C \) to the Standard Basis

Given:
- \( C = \{ C_1, C_2 \} \)
- \( C_1 = \begin{bmatrix} 1 \\ -5 \end{bmatrix} \), \( C_2 = \begin{bmatrix} -2 \\ 2 \end{bmatrix} \)

The change-of-coordinate matrix \( P_{C \to \text{std}} \) from \( C \) to the standard basis is:

\[
P_{C \to \text{std}} = \begin{bmatrix} C_1 & C_2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -5 & 2 \end{bmatrix}
\]

#### 2. Find the Coordinate Vector \( [x]_C \)

We need to solve:

\[
P_{C \to \text{std}} \cdot [x]_C = x
\]

This translates to:

\[
\begin{bmatrix} 1 & -2 \\ -5 & 2 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \end{bmatrix}
\]

This is a system of linear equations:

\[
\begin{aligned}
d_1 - 2d_2 &= 4 \\
-5d_1 + 2d_2 &= 6
\end{aligned}
\]

Let's solve this system.

Multiply the first equation by 2:

\[
2d_1 - 4d_2 = 8
\]

Add this to the second equation:

\[
(-5d_1 + 2d_2) + (2d_1 - 4d_2) = 6 + 8
\]

\[
-3d_1 - 2d_2 = 14 \implies d_1 = -14
\]

Substitute \( d_1 = -14 \) into the first equation:

\[
-14 - 2d_2 = 4 \implies -2d_2 = 18 \implies d_2 = -9
\]

So, \( [x]_C = \begin{bmatrix} -14 \\ -9 \end{bmatrix} \).

### Part (c) - Change-of-Coordinate Matrices from \( \beta \) to \( C \) and from \( C \) to \( \beta \)

#### 1. Change-of-Coordinate Matrix from \( \beta \) to \( C \)

The change-of-coordinate matrix from \( \beta \) to \( C \) is given by:

\[
P_{\beta \to C} = P_{C \to \text{std}}^{-1} \cdot P_{\beta \to \text{std}}
\]

Let's compute this.

- We already have \( P_{\beta \to \text{std}} = \begin{bmatrix} 7 & -3 \\ 3 & -1 \end{bmatrix} \)
- \( P_{C \to \text{std}} = \begin{bmatrix} 1 & -2 \\ -5 & 2 \end{bmatrix} \)

First, compute the inverse of \( P_{C \to \text{std}} \):

\[
P_{C \to \text{std}}^{-1} = \frac{1}{(1)(2) - (-5)(-2)} \begin{bmatrix} 2 & 2 \\ 5 & 1 \end{bmatrix} = \frac{1}{2 - 10} \begin{bmatrix} 2 & 2 \\ 5 & 1 \end{bmatrix} = \frac{1}{-8} \begin{bmatrix} 2 & 2 \\ 5 & 1 \end{bmatrix}
\]

\[
P_{C \to \text{std}}^{-1} = \begin{bmatrix} -\frac{1}{4} & -\frac{1}{4} \\ -\frac{5}{8} & -\frac{1}{8} \end{bmatrix}
\]

Now, multiply by \( P_{\beta \to \text{std}} \):

\[
P_{\beta \to C} = \begin{bmatrix} -\frac{1}{4} & -\frac{1}{4} \\

Learn how to solve problems involving change of coordinate matrices in linear algebra step-by-step. This problem covers finding change-of-coordinate matrices from given bases to the standard basis, and transforming coordinate vectors relative to different bases. Suitable for college-level students studying linear algebra.

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