please box answers in correct order so I can learn clearly thank you. For \(\lambda = -1\), we found the eigenspace consists of solutions to the following:\[\begin{bmatrix}2 & 0 & 2 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{bmatrix}\]Writing this as a system of equations gives \(2x_1 + 2x_3 = 0\), or \(x_1 = \left(\rule{20px}{0.5pt}\right)x_3\). Let \(x_3 = s\) be the free variable in the equation. Also, note that \(x_2 = t\) is a free variable as it does not appear in the equations.Therefore, the set of solutions is given by the vector \(\begin{bmatrix} \left(\rule{20px}{0.5pt}\right)s \\ t \\ s \end{bmatrix}\). Thus, the eigenvectors are of the form\[s \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} + t \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.\]Find the spanning set for the eigenspace for \(\lambda = -1\).\[\text{span}\left( \begin{bmatrix} \rule{20px}{0.5pt} \\ \rule{20px}{0.5pt} \\ \rule{20px}{0.5pt} \end{bmatrix} \right)\] Consider the following.\[ A = \begin{bmatrix} 1 & 0 & 2 \\ 3 & -1 & 3 \\ 2 & 0 & 1 \end{bmatrix} \](a) Compute the characteristic polynomial of \( A \).(b) Compute the eigenvalues and bases of the corresponding eigenspaces of \( A \).(c) Compute the algebraic and geometric multiplicity of each eigenvalue.

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For 2 = -1, we found the eigenspace consists of solutions to the following. 20 2 0 0 0 0 0 0 0 0 0 Writing this as a system of equations gives 2x, + 2x3 = 0, or x, = ] )*3• Let x3 = s be the free variable in the equation. Also, note that x, = t is a free variable as it does not appear in the equations. Therefore, the set of solutions is given by the vector . Thus, the eigenvectors are of the form Find the spanning set for the eigenspace for 1 = -1. span

For 2 = -1, we found the eigenspace consists of solutions to the following. 20 2 0 0 0 0 0 0 0 0 0 Writing this as a system of equations gives 2x, + 2x3 = 0, or x, = ] )*3• Let x3 = s be the free variable in the equation. Also, note that x, = t is a free variable as it does not appear in the equations. Therefore, the set of solutions is given by the vector . Thus, the eigenvectors are of the form Find the spanning set for the eigenspace for 1 = -1. span

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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