Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
