Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Let's solve the given problems step-by-step for the matrix [tex]\( A \)[/tex] and its eigenvalues.
### Part (a): Finding the values of [tex]\(\lambda_1\)[/tex] and [tex]\(\lambda_2\)[/tex]
Given the matrix:
[tex]\[ A = \begin{pmatrix} 3 & 4 \\ 5 & 2 \end{pmatrix} \][/tex]
and [tex]\(I\)[/tex] as the [tex]\(2 \times 2\)[/tex] identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
The condition [tex]\(\operatorname{det}(A - \lambda I) = 0\)[/tex] gives us the characteristic polynomial of the matrix [tex]\(A\)[/tex].
First, compute [tex]\(A - \lambda I\)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 3 & 4 \\ 5 & 2 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix} 3 - \lambda & 4 \\ 5 & 2 - \lambda \end{pmatrix} \][/tex]
Next, compute the determinant:
[tex]\[ \operatorname{det}(A - \lambda I) = \operatorname{det} \begin{pmatrix} 3 - \lambda & 4 \\ 5 & 2 - \lambda \end{pmatrix} = (3 - \lambda)(2 - \lambda) - (4)(5) \][/tex]
Expanding the determinant:
[tex]\[ (3 - \lambda)(2 - \lambda) - 20 = 6 - 3\lambda - 2\lambda + \lambda^2 - 20 \][/tex]
[tex]\[ = \lambda^2 - 5\lambda - 14 \][/tex]
Thus, the characteristic equation is:
[tex]\[ \lambda^2 - 5\lambda - 14 = 0 \][/tex]
To find the eigenvalues, solve this quadratic equation:
[tex]\[ \lambda^2 - 5\lambda - 14 = 0 \][/tex]
Using the quadratic formula [tex]\(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -14\)[/tex]:
[tex]\[ \lambda = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ = \frac{5 \pm 9}{2} \][/tex]
Therefore, the eigenvalues are:
[tex]\[ \lambda_1 = \frac{5 + 9}{2} = 7 \][/tex]
[tex]\[ \lambda_2 = \frac{5 - 9}{2} = -2 \][/tex]
Given [tex]\(\lambda_1 > \lambda_2\)[/tex], we have:
[tex]\[ \lambda_1 = 7 \][/tex]
[tex]\[ \lambda_2 = -2 \][/tex]
### Part (b): Finding the eigenvectors
To find the eigenvectors for [tex]\(\lambda_1 = 7\)[/tex] and [tex]\(\lambda_2 = -2\)[/tex], we solve the system [tex]\((A - \lambda I)\mathbf{v} = 0\)[/tex] for each eigenvalue.
#### Eigenvector corresponding to [tex]\(\lambda_1 = 7\)[/tex]:
[tex]\[ A - 7I = \begin{pmatrix} 3 - 7 & 4 \\ 5 & 2 - 7 \end{pmatrix} = \begin{pmatrix} -4 & 4 \\ 5 & -5 \end{pmatrix} \][/tex]
The system of linear equations:
[tex]\[ \begin{pmatrix} -4 & 4 \\ 5 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
Which simplifies to:
[tex]\[ -4x + 4y = 0 \][/tex]
[tex]\[ 5x - 5y = 0 \][/tex]
Both equations reduce to:
[tex]\[ x = y \][/tex]
Thus, an eigenvector corresponding to [tex]\(\lambda_1 = 7\)[/tex] is:
[tex]\[ \mathbf{v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \][/tex]
#### Eigenvector corresponding to [tex]\(\lambda_2 = -2\)[/tex]:
[tex]\[ A - (-2)I = \begin{pmatrix} 3 + 2 & 4 \\ 5 & 2 + 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 5 & 4 \end{pmatrix} \][/tex]
The system of linear equations:
[tex]\[ \begin{pmatrix} 5 & 4 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
Which simplifies to:
[tex]\[ 5x + 4y = 0 \][/tex]
This reduces to:
[tex]\[ x = -\frac{4}{5}y \][/tex]
Thus, an eigenvector corresponding to [tex]\(\lambda_2 = -2\)[/tex] is:
[tex]\[ \mathbf{v_2} = \begin{pmatrix} -4 \\ 5 \end{pmatrix} \][/tex]
In conclusion:
- The eigenvalues are: [tex]\(\lambda_1 = 7\)[/tex] and [tex]\(\lambda_2 = -2\)[/tex].
- The corresponding eigenvectors are:
[tex]\[ \mathbf{v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{v_2} = \begin{pmatrix} -4 \\ 5 \end{pmatrix} \][/tex]
### Part (a): Finding the values of [tex]\(\lambda_1\)[/tex] and [tex]\(\lambda_2\)[/tex]
Given the matrix:
[tex]\[ A = \begin{pmatrix} 3 & 4 \\ 5 & 2 \end{pmatrix} \][/tex]
and [tex]\(I\)[/tex] as the [tex]\(2 \times 2\)[/tex] identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
The condition [tex]\(\operatorname{det}(A - \lambda I) = 0\)[/tex] gives us the characteristic polynomial of the matrix [tex]\(A\)[/tex].
First, compute [tex]\(A - \lambda I\)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 3 & 4 \\ 5 & 2 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix} 3 - \lambda & 4 \\ 5 & 2 - \lambda \end{pmatrix} \][/tex]
Next, compute the determinant:
[tex]\[ \operatorname{det}(A - \lambda I) = \operatorname{det} \begin{pmatrix} 3 - \lambda & 4 \\ 5 & 2 - \lambda \end{pmatrix} = (3 - \lambda)(2 - \lambda) - (4)(5) \][/tex]
Expanding the determinant:
[tex]\[ (3 - \lambda)(2 - \lambda) - 20 = 6 - 3\lambda - 2\lambda + \lambda^2 - 20 \][/tex]
[tex]\[ = \lambda^2 - 5\lambda - 14 \][/tex]
Thus, the characteristic equation is:
[tex]\[ \lambda^2 - 5\lambda - 14 = 0 \][/tex]
To find the eigenvalues, solve this quadratic equation:
[tex]\[ \lambda^2 - 5\lambda - 14 = 0 \][/tex]
Using the quadratic formula [tex]\(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -14\)[/tex]:
[tex]\[ \lambda = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ = \frac{5 \pm 9}{2} \][/tex]
Therefore, the eigenvalues are:
[tex]\[ \lambda_1 = \frac{5 + 9}{2} = 7 \][/tex]
[tex]\[ \lambda_2 = \frac{5 - 9}{2} = -2 \][/tex]
Given [tex]\(\lambda_1 > \lambda_2\)[/tex], we have:
[tex]\[ \lambda_1 = 7 \][/tex]
[tex]\[ \lambda_2 = -2 \][/tex]
### Part (b): Finding the eigenvectors
To find the eigenvectors for [tex]\(\lambda_1 = 7\)[/tex] and [tex]\(\lambda_2 = -2\)[/tex], we solve the system [tex]\((A - \lambda I)\mathbf{v} = 0\)[/tex] for each eigenvalue.
#### Eigenvector corresponding to [tex]\(\lambda_1 = 7\)[/tex]:
[tex]\[ A - 7I = \begin{pmatrix} 3 - 7 & 4 \\ 5 & 2 - 7 \end{pmatrix} = \begin{pmatrix} -4 & 4 \\ 5 & -5 \end{pmatrix} \][/tex]
The system of linear equations:
[tex]\[ \begin{pmatrix} -4 & 4 \\ 5 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
Which simplifies to:
[tex]\[ -4x + 4y = 0 \][/tex]
[tex]\[ 5x - 5y = 0 \][/tex]
Both equations reduce to:
[tex]\[ x = y \][/tex]
Thus, an eigenvector corresponding to [tex]\(\lambda_1 = 7\)[/tex] is:
[tex]\[ \mathbf{v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \][/tex]
#### Eigenvector corresponding to [tex]\(\lambda_2 = -2\)[/tex]:
[tex]\[ A - (-2)I = \begin{pmatrix} 3 + 2 & 4 \\ 5 & 2 + 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 5 & 4 \end{pmatrix} \][/tex]
The system of linear equations:
[tex]\[ \begin{pmatrix} 5 & 4 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
Which simplifies to:
[tex]\[ 5x + 4y = 0 \][/tex]
This reduces to:
[tex]\[ x = -\frac{4}{5}y \][/tex]
Thus, an eigenvector corresponding to [tex]\(\lambda_2 = -2\)[/tex] is:
[tex]\[ \mathbf{v_2} = \begin{pmatrix} -4 \\ 5 \end{pmatrix} \][/tex]
In conclusion:
- The eigenvalues are: [tex]\(\lambda_1 = 7\)[/tex] and [tex]\(\lambda_2 = -2\)[/tex].
- The corresponding eigenvectors are:
[tex]\[ \mathbf{v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{v_2} = \begin{pmatrix} -4 \\ 5 \end{pmatrix} \][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
