In Exercises 19 through 24, find the matrix B of the lin- ear transformation T(x) = Ax with respect to the basis B = (01, 02). For practice, solve each problem in three ways: (a) Use the formula B = S-'AS, (b) use a commu- tative diagram (as in Examples 3 and 4), and (c) construct B "column by column."
In Exercises 19 through 24, find the matrix B of the lin- ear transformation T(x) = Ax with respect to the basis B = (01, 02). For practice, solve each problem in three ways: (a) Use the formula B = S-'AS, (b) use a commu- tative diagram (as in Examples 3 and 4), and (c) construct B "column by column."
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please answer part b) and c) in particular from section 3.4 of "
![In Exercises 19 through 24, find the matrix \( B \) of the linear transformation \( T(\vec{x}) = A\vec{x} \) with respect to the basis \( \mathfrak{B} = (\vec{v}_1, \vec{v}_2) \). For practice, solve each problem in three ways: (a) Use the formula \( B = S^{-1}AS \), (b) use a commutative diagram (as in Examples 3 and 4), and (c) construct \( B \) "column by column."
**20.**
\[ A = \begin{bmatrix} -3 & 4 \\ 4 & 3 \end{bmatrix} \]
\[ \vec{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
\[ \vec{v}_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b1372ae-e668-4971-89ee-29da7ac7c466%2F026e7819-02a9-4758-b522-d3b0bbc7292e%2Fhfsj71_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In Exercises 19 through 24, find the matrix \( B \) of the linear transformation \( T(\vec{x}) = A\vec{x} \) with respect to the basis \( \mathfrak{B} = (\vec{v}_1, \vec{v}_2) \). For practice, solve each problem in three ways: (a) Use the formula \( B = S^{-1}AS \), (b) use a commutative diagram (as in Examples 3 and 4), and (c) construct \( B \) "column by column."
**20.**
\[ A = \begin{bmatrix} -3 & 4 \\ 4 & 3 \end{bmatrix} \]
\[ \vec{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
\[ \vec{v}_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]
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