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Chapter 6: Problem 25

Find all the eigenvalues and the corresponding eigenvectors for the following matrices. $$ \left[\begin{array}{cc} 9.5 & -4.5 \\ 15 & -7 \end{array}\right] $$

Short Answer

Expert verified
Eigenvalues are \(2\) and \(0.5\) . Corresponding eigenvectors are \( \begin{bmatrix} 0.6 \ 1 \ \right] \) and \( \begin{bmatrix} 1 \ 2 \ \right] \).

Step by step solution

01

Title - Write down the characteristic equation

The characteristic equation of a matrix is given by \( \text{det}(A - \text{I}\bm{\text{位}}) = 0 \). For the matrix \( A = \begin{bmatrix} 9.5 & -4.5 \ 15 & -7 \ \right] \), we must find \( \text{det}\begin{bmatrix} 9.5 - \bm{\text{位}} & -4.5 \ 15 & -7 - \bm{\text{位}} \ \right] = 0 \).
02

Title - Compute the determinant

To solve the determinant, use: \( \text{det}\begin{bmatrix} a & b \ c & d \ \right] = ad - bc \). Applying this: \( (9.5 - \bm{\text{位}})(-7 - \bm{\text{位}}) - (-4.5)(15)=0 \). \( (9.5\bm{\text{位}} - \bm{\text{位}}^2 - 66.5 + 7\bm{\text{位}}) + 67.5 = 0 \) Simplify to obtain the characteristic polynomial.
03

Title - Solve the characteristic polynomial

Combine like terms to form the equation: \( \bm{\text{位}}^2 - 2.5\bm{\text{位}} + 1 = 0 \). To find the eigenvalues, use the quadratic formula: \( \bm{\text{位}}_{1,2} = \frac{-b \text{卤} \text{鈭殅(b^2 - 4ac)}{2a} \).Here, \( a = 1 \), \( b = -2.5 \), and \( c = 1 \). Substitute and solve for \( \bm{\text{位}} \).
04

Title - Calculate the eigenvalues

Substitute the values into the quadratic formula: \( \bm{\text{位}}_{1,2} = \frac{2.5 \text{卤} \text{鈭殅(2.5^2 - 4(1)(1))}{2(1)} \).Solve this to get: \( \bm{\text{位}}_{1,2} = \frac{2.5 \text{卤} 1.5}{2} \). Thus, the eigenvalues are \( \bm{\text{位}}_1 = 2 \) and \( \bm{\text{位}}_2 = 0.5 \).
05

Title - Find the eigenvectors

For each eigenvalue \( \bm{\text{位}} \), solve the equation \( (A - \bm{\text{位}}I)\bm{x} = \bm{0} \). For \( \bm{\text{位}} = 2 \), solve \( \begin{bmatrix} 7.5 & -4.5 \ 15 & -9 \ \right]\begin{bmatrix}x_1 \ x_2 \ \right]=\begin{bmatrix}0 \ 0\right] \). For \( \bm{\text{位}} = 0.5 \), solve \( \begin{bmatrix} 9 & -4.5 \ 15 & -7.5 \ \right]\begin{bmatrix}x_1 \ x_2 \ \right]=\begin{bmatrix}0 \ 0\right] \).
06

Title - Solve the linear systems

Solve each as a system of linear equations. For \( \bm{\text{位}} = 2 \), we get: \( 7.5x_1 - 4.5x_2 = 0 \), solve to get eigenvector \( \bm{x}_1 = \begin{bmatrix} 0.6 \ 1 \ \right] \) (up to scalar multiplication).For \( \bm{\text{位}} = 0.5 \), solve the system to get \( 9x_1 - 4.5x_2 = 0 \), leading to eigenvector \( \bm{x}_2 = \begin{bmatrix} 1 \ 2 \ \right] \) (up to scalar multiplication).

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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 crucial in finding eigenvalues. It involves the determinant of the matrix subtracted by a scalar multiple of the identity matrix, set equal to zero. For a given matrix \( A \), the formula looks like this: \( \text{det}(A - \bm{\text{位}}I) = 0 \). Here, \( \bm{\text{位}} \) represents the eigenvalues. In simpler terms, the characteristic equation transforms the problem of finding eigenvalues into finding the roots of a polynomial.
determinant
The determinant plays a pivotal role in calculating the characteristic equation. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \ \right] \), the determinant is computed as \( ad - bc \). When finding eigenvalues, we use the determinant of the matrix \( A - \bm{\text{位}}I \). For our example matrix: \( \begin{bmatrix} 9.5 - \bm{\text{位}} & -4.5 \ 15 & -7 - \bm{\text{位}} \ \right] \). After solving the determinant and setting it to zero, we get the quadratic characteristic polynomial.
quadratic formula
The quadratic formula is extremely useful when solving the characteristic polynomial. The standard quadratic equation is \( ax^2 + bx + c = 0 \). The quadratic formula, \( \bm{\text{x}} = \frac{-b \text{卤} \text{鈭殅(b^2 - 4ac)}{2a} \), gives the solutions for \( x \). In our case, we convert the determinant into the polynomial \( \bm{\text{位}}^2 - 2.5\bm{\text{位}} + 1 = 0 \) and apply the formula with \( a = 1 \), \( b = -2.5 \), and \( c = 1 \) to find the eigenvalues.
linear systems
Once eigenvalues are determined, the next step is finding their corresponding eigenvectors. This involves solving a system of linear equations formed by \( (A - \bm{\text{位}}I)\bm{x} = 0 \). For each eigenvalue \( \bm{\text{位}} \), substitute it back into the equation to get a new matrix. Solving this linear system provides the eigenvectors, which are the non-zero solutions. For example, substituting \( \bm{\text{位}} = 2 \) and \( \bm{\text{位}} = 0.5 \) results in two different sets of equations that give the eigenvectors up to scalar multiples.

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Most popular questions from this chapter

Let \(A=\left[\begin{array}{ll}4 & -1 \\ 7 & -9\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 2 & -2\end{array}\right], C=\left[\begin{array}{rr}8 & 3 \\ 0 & -3\end{array}\right]\) \(D=\left[\begin{array}{rrr}5 & -6 & 1 \\ 10 & 3 & -1\end{array}\right], E=\left[\begin{array}{rr}-7 & 4 \\ -3 & 2 \\ 2 & -1\end{array}\right]\) \(F=\left[\begin{array}{rrr}-4 & 2 & 3 \\ 0 & -1 & 2 \\ -7 & -2 & 5\end{array}\right]\), and \(v=\left[\begin{array}{r}2 \\\ -3\end{array}\right]\). Compute \(\mathbf{A}^{3} \mathbf{v}\).

San Diego fairy shrimp live in ponds that fill and dry many times during a year. While the ponds are dry, the [airy shrimp survive as cysts. The population is divided into three groups. Cysts that survive one dry period are in group 1 cysts that survive two dry periods are in group 2 and cysts that survive three or more dry periods are in group \(3 .\) The Leslie matrix representing the survivability and fecundity is given below. \({ }^{14}\) $$G=\left[\begin{array}{ccc} 3.6 & 0.98 & 0.65 \\ 0.5 & 0 & 0 \\ 0 & 0.5 & 0.49 \end{array}\right]$$ In the third dry period there are 8365,1095, and 310 individuals in groups 1,2, and 3, respectively. a) Find the inverse of the Leslie matrix. b) Estimate the population of each group during the second dry spell. c) Estimate the population of each group during the first dry spell.

Find all the eigenvalues and the corresponding eigenvectors for the following matrices. $$ \left[\begin{array}{rr} \frac{51}{13} & -\frac{20}{13} \\ -\frac{24}{13} & \frac{79}{13} \end{array}\right] $$

We investigate inverses and determinants of matrix products. In this exercise, we verify that \(\operatorname{det}(\mathrm{AB})=\operatorname{det}(\mathrm{A}) \operatorname{det}(\mathrm{B})\) by looking \(\mathrm{at}\) examples. a) Let \(A=\left[\begin{array}{ll}2 & 5 \\ 4 & 7\end{array}\right]\) and \(B=\left[\begin{array}{rr}1 & 5 \\ -2 & 3\end{array}\right]\). Verify that \(\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B).\) b) Let \(A=\left[\begin{array}{rrr}2 & -4 & 6 \\ 1 & 0 & -1 \\ 4 & -2 & 3\end{array}\right]\) and \(B=\left[\begin{array}{rrr}2 & -7 & 2 \\ 5 & 3 & -1 \\\ 4 & 0 & 2\end{array}\right] .\) Verify that \(\operatorname{det}(\mathrm{AB})=\operatorname{det}(\mathrm{A}) \operatorname{det}(\mathrm{B})\). c) Pick another pair of square matrices \(\mathrm{A}\) and \(\mathrm{B}\). Verify that \(\operatorname{det}(\mathbf{A B})=\operatorname{det}(\mathbf{A}) \operatorname{det}(\mathbf{B})\).

Let \(\begin{aligned} A &=\left[\begin{array}{rrr}1 & 6 & -2 \\ 4 & -2 & -1 \\\ 0 & 3 & -5\end{array}\right], & B=\left[\begin{array}{llr}0 & 4 & -2 \\ 5 & 0 & -3 \\ 6 & 2 & 1\end{array}\right], \text { and } \\ C &=\left[\begin{array}{rrr}4 & 4 & 0 \\ -2 & 3 & 8 \\ 1 & -3 & 6\end{array}\right] \end{aligned}\). Compute \(A(B-C)\) and \(A B-A C\) to verify that these matrices satisfy the distributive property.
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