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Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{cc}{-3} & {3 / 4} \\ {-5} & {1}\end{array}\right) $$

Short Answer

Expert verified
Matrix A: $$ A=\left(\begin{array}{cc}{-3} & {3 / 4} \\\ {-5} & {1}\end{array}\right) $$ Answer: The eigenvalues and eigenvectors of the given matrix A are as follows: - Eigenvalue 位鈧 = -0.5 鈫 Eigenvector \(\begin{pmatrix}\frac{3}{2}\\1\end{pmatrix}\) - Eigenvalue 位鈧 = -1.5 鈫 Eigenvector \(\begin{pmatrix}-3\\5\end{pmatrix}\)

Step by step solution

01

Calculate the Characteristic Equation

We find the determinant of the matrix (A - 位滨) to get the characteristic equation: $$ \text{det}(A-\lambda I)=\left|\begin{array}{cc}{-3-\lambda} & {3 / 4} \\\ {-5} & {1-\lambda}\end{array}\right| $$ Now, calculate the determinant: $$ (-3-\lambda)(1-\lambda) - \left(-\frac{15}{4}\right) = 0 $$
02

Find Eigenvalues

We have the following equation: $$ (-3-\lambda)(1-\lambda) + \frac{15}{4} = 0 $$ Or, $$ \lambda^{2}+2\lambda+\frac{3}{4}= 0 $$ Solving for 位, we find two eigenvalues: 位鈧 = -0.5 and 位鈧 = -1.5.
03

Find Eigenvectors

Now, we will find the eigenvectors corresponding to each eigenvalue by solving the following system of linear equations: For 位鈧 = -0.5: $$ (A - (-0.5)I) \mathbf{v}=\begin{pmatrix}-2.5 & 3/4\\ -5 & 1.5\end{pmatrix}\mathbf{v}=0 $$ We can solve this system of linear equations to find the eigenvector: $$ \mathbf{v}=\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}= \begin{pmatrix}\frac{3}{2}\\1\end{pmatrix} $$ For 位鈧 = -1.5: $$ (A - (-1.5)I) \mathbf{v}=\begin{pmatrix}-1.5 & 3/4\\ -5 & 2.5\end{pmatrix}\mathbf{v}=0 $$ We can solve this system of linear equations to find the eigenvector: $$ \mathbf{v}=\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}= \begin{pmatrix}-3\\5\end{pmatrix} $$ So we have found the eigenvalues and their corresponding eigenvectors: - Eigenvalue 位鈧 = -0.5 鈫 Eigenvector \(\begin{pmatrix}\frac{3}{2}\\1\end{pmatrix}\) - Eigenvalue 位鈧 = -1.5 鈫 Eigenvector \(\begin{pmatrix}-3\\5\end{pmatrix}\)

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Key Concepts

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

Characteristic Equation
The characteristic equation is a fundamental concept in linear algebra, particularly when dealing with matrices and eigenvalues. It is an algebraic equation arising from the search for eigenvalues of a matrix. Let's understand this with a clear example from our exercise.

To find the characteristic equation, we consider a square matrix A and subtract 位滨 (where is a scalar and I is the identity matrix of the same size as A). The determinant of this resulting matrix is set to zero:
\[\text{det}(A-\lambda I) = 0 \]
This equation allows us to solve for , which are the eigenvalues. In our exercise, the provided matrix is used to form an equation leading to a quadratic polynomial. The roots of the polynomial, 位鈧 = -0.5 and 位鈧 = -1.5, are the eigenvalues. Understanding the connection between matrices and their characteristic equation is key in linear algebra. It fundamentally represents the conditions under which our system (matrix) has non-trivial solutions that scale by factors of .
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It provides a way to concisely represent and operate on sets of linear equations. For example, in our exercise, we use the linear algebra concept of eigenvalues and eigenvectors that represent special solutions to matrix equations.

Eigenvalues () are scalars that, when applied to an eigenvector (v), maintain its direction but may alter its magnitude. The eigenvector remains parallel to its original direction even after the transformation by the matrix is applied. Finding eigenvalues and eigenvectors can reveal fundamental properties about the linear transformation that the matrix represents, like scaling, rotating, or projecting. By mastering the language and operations of linear algebra, we can solve various problems in engineering, physics, computer science, and more.
Systems of Linear Equations
A system of linear equations consists of multiple linear equations that we wish to solve simultaneously. Our aim is to find values for the variables that satisfy all equations in the system. In the context of our exercise, once we have the eigenvalues, we need to find the eigenvectors by solving systems of linear equations of the form:
\[(A - \lambda I) \mathbf{v} = 0 \]
This system arises from the definition of an eigenvector v, which must satisfy the equation when multiplied by the matrix transformed by subtracting times the identity matrix. To solve for the eigenvector, we perform matrix operations that include row reductions to echelon form, resulting in solutions for 惫鈧 and 惫鈧. This process, fundamental in linear algebra, reveals the eigenvectors that correspond to each eigenvalue. It shows us how a system can collapse, dilate, or alter based on certain scalars (eigenvalues) and their directional components (eigenvectors).

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Most popular questions from this chapter

Show that all solutions of the system $$ x^{\prime}=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right) \mathbf{x} $$ approach zero as \(t \rightarrow \infty\) if and only if \(a+d<0\) and \(a d-b c>0 .\) Compare this result with that of Problem 38 in Section \(3.5 .\)

Show that if \(\mathbf{A}\) is a diagonal matrix with diagonal elements \(a_{1}, a_{2}, \ldots, a_{n},\) then \(\exp (\mathbf{A} t)\) is also a diagonal matrix with diagonal elements \(\exp \left(a_{1} t\right), \exp \left(a_{2} t\right), \ldots, \exp \left(a_{n} t\right)\)

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{3}{4}} & {\frac{1}{2}} \\\ {\frac{1}{8}} & {-\frac{1}{4}}\end{array}\right) \mathbf{x} $$

The electric circuit shown in Figure 7.6 .6 is described by the system of differential equations \(\frac{d}{d t}\left(\begin{array}{l}{I} \\\ {V}\end{array}\right)=\left(\begin{array}{cc}{0} & {\frac{1}{L}} \\\ {-\frac{1}{C}} & {-\frac{1}{R C}}\end{array}\right)\left(\begin{array}{l}{I} \\\ {V}\end{array}\right)\) where \(I\) is the current through the inductor and \(V\) is the voltage drop across the capacitor. These differential equations were derived in Problem 18 of Section \(7.1 .\) (a) Show that the eigenvalues of the coefficient matrix are real and different if \(L>4 R^{2} C\); show they are complex conjugates if \(L<4 R^{2} C .\) (b) Suppose that \(R=1\) ohm, \(C=\frac{1}{2}\) farad, and \(L=1\) henry. Find the general solution of the system (i) in this case. (c) Find \(I(t)\) and \(V(t)\) if \(I(0)=2\) amperes and \(V(0)=1\) volt (d) For the circuit of part (b) determine the limiting values of \(I(t)\) and \(V(t)\) as \(t \rightarrow \infty\) Do these limiting values depend on the initial conditions?

Find the solution of the given initial value problem. Draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) - space and also draw the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{-\frac{5}{2}} & {1} & {1} \\ {1} & {-\frac{5}{2}} & {1} \\ {1} & {1} & {-\frac{5}{2}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{r}{2} \\ {3} \\\ {-1}\end{array}\right) $$

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