91Ó°ÊÓ

Determine whether or not the given scalar \(k\) is an eigenvalue of the matrix \(A .\) $$(a) A=\left[ \begin{array}{ll}{1} & {2} \\ {2} & {1}\end{array}\right] \quad k=3$$ $$(b)A=\left[ \begin{array}{lll}{0} & {2} & {1} \\ {2} & {1} & {0} \\ {0} & {2} & {1}\end{array}\right] \quad k=0$$ $$\begin{array}{l}{\text { (c) } A=\left[ \begin{array}{cc}{5} & {2} \\ {0} & {1}\end{array}\right] \quad k=2} \\ {\text { (d) } A=\left[ \begin{array}{rr}{1} & {-1} \\ {1} & {1}\end{array}\right] \quad k=1-i}\end{array}$$ $$(e)A=\left[ \begin{array}{lll}{1} & {2} & {1} \\ {0} & {2} & {0} \\ {2} & {1} & {0}\end{array}\right] \quad k=0$$ $$(\mathrm{f}) A=\left[ \begin{array}{ll}{1} & {a} \\ {1} & {1}\end{array}\right] \quad k=1+\sqrt{a}$$

Step by step solution

01

Understanding Eigenvalue Condition

To determine if a scalar \( k \) is an eigenvalue of a matrix \( A \), we need to evaluate the determinant of \( A - kI \), where \( I \) is the identity matrix of the same size as \( A \). If \( \det(A - kI) = 0 \), \( k \) is an eigenvalue.

02

Step 2a: Check Eigenvalue for Matrix A1

Given the matrix \( A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \) and \( k = 3 \), we compute \( A - 3I = \begin{bmatrix} 1 - 3 & 2 \ 2 & 1 - 3 \end{bmatrix} = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix} \). The determinant is \( (-2)(-2) - (2)(2) = 4 - 4 = 0 \). Thus, \( k = 3 \) is an eigenvalue of \( A \).

03

Step 2b: Check Eigenvalue for Matrix A2

Given the matrix \( A = \begin{bmatrix} 0 & 2 & 1 \ 2 & 1 & 0 \ 0 & 2 & 1 \end{bmatrix} \) and \( k = 0 \), we compute \( A - 0I = A \). The determinant can be evaluated using cofactor expansion or row reduction and results in \( \det(A) = -6 \). Since \( \det(A) eq 0 \), \( k = 0 \) is not an eigenvalue.

04

Step 2c: Check Eigenvalue for Matrix A3

Given \( A = \begin{bmatrix} 5 & 2 \ 0 & 1 \end{bmatrix} \) and \( k = 2 \), compute \( A - 2I = \begin{bmatrix} 5 - 2 & 2 \ 0 & 1 - 2 \end{bmatrix} = \begin{bmatrix} 3 & 2 \ 0 & -1 \end{bmatrix} \). The determinant is \( (3)(-1) - (2)(0) = -3 \). Since \( \det(A - 2I) eq 0 \), \( k = 2 \) is not an eigenvalue.

05

Step 2d: Check Eigenvalue for Matrix A4

For matrix \( A = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} \) and \( k = 1-i \), compute \( A - (1-i)I = \begin{bmatrix} 1-(1-i) & -1 \ 1 & 1-(1-i) \end{bmatrix} = \begin{bmatrix} i & -1 \ 1 & i \end{bmatrix} \). The determinant is \( i(i) - (-1)(1) = -1 + 1 = 0 \). Thus, \( k = 1-i \) is an eigenvalue.

06

Step 2e: Check Eigenvalue for Matrix A5

For matrix \( A = \begin{bmatrix} 1 & 2 & 1 \ 0 & 2 & 0 \ 2 & 1 & 0 \end{bmatrix} \) and \( k = 0 \), compute \( A - 0I = A \). The determinant is computed as \( -6 + 2 = -4 \). Since \( \det(A) eq 0 \), \( k = 0 \) is not an eigenvalue.

07

Step 2f: Check Eigenvalue for Matrix A6

For matrix \( A = \begin{bmatrix} 1 & a \ 1 & 1 \end{bmatrix} \) and \( k = 1+\sqrt{a} \), compute \( A - (1+\sqrt{a})I = \begin{bmatrix} 1-(1+\sqrt{a}) & a \ 1 & 1-(1+\sqrt{a}) \end{bmatrix} = \begin{bmatrix} -\sqrt{a} & a \ 1 & -\sqrt{a} \end{bmatrix} \). The determinant is \( (-\sqrt{a})(-\sqrt{a}) - (a)(1) = a - a = 0 \). Hence, \( k = 1+\sqrt{a} \) is an eigenvalue.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It provides a framework for understanding and solving equations involving linear relationships.
Linear algebra is extremely important in various fields such as physics, computer science, and engineering. This is because it simplifies complex systems to more manageable parts.
In the context of eigenvalues and eigenvectors, linear algebra assists in identifying specific scalars and vectors that describe important properties of transformations represented by matrices. Let's delve deeper into the concept of matrices.

Matrices

Matrices are fundamental to linear algebra and are used to organize and manage linear equations and transformations.
A matrix is a rectangular array of numbers arranged in rows and columns. It can represent a multitude of data forms, such as systems of linear equations or transformations in space.
For example, matrices help perform linear transformations, coordinate changes, and manage equations. They come in different types, such as square matrices (same number of rows and columns) which are critical in determining eigenvalues and eigenvectors. Understanding and manipulating matrices are crucial in many technical fields, including computer graphics and optimization problems.

Determinants

The determinant is a special number that can be calculated from a square matrix.
It plays a pivotal role in matrix algebra as it determines a matrix's properties, such as invertibility and eigenvalues.
In the context of eigenvalues, the determinant helps us establish the eigenvalue condition. We consider the determinant of the matrix \(A - kI\) (where \(I\) is the identity matrix) to determine if a scalar \(k\) is an eigenvalue of the matrix \(A\). The eigenvalue condition states that if \(\det(A - kI) = 0\), then \(k\) is indeed an eigenvalue. Calculating determinants for larger matrices can be complex, requiring cofactor expansion or other methods to simplify computations.

Eigenvectors

An eigenvector is a non-zero vector that changes at most by a scalar factor when a linear transformation is applied to it.
Given a matrix \(A\) and its eigenvalue \(k\), the corresponding eigenvectors \(\mathbf{v}\) satisfy the equation:\[A \mathbf{v} = k \mathbf{v}.\] Finding eigenvectors allows further insight into the structure of a matrix and its effects in transformation.
These vectors are essential in areas such as quantum mechanics, vibration analysis, and systems dynamics.
While each eigenvalue has several eigenvectors (forming an eigenspace), understanding these vectors help untangle complex transformations into simpler, scalable components that retain the essence of the original operation.