Answer :
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]
