Answer :
The eigenvector corresponding to the eigenvalue λ = -3 for the given matrix is [1, 0, 1].
To find the eigenvector corresponding to the eigenvalue λ = -3, we first set up the equation (A - λI)v = 0, where A is the given matrix, λ is the eigenvalue, I is the identity matrix, and v is the eigenvector.
Substituting the given values, we get:
[25 - 188 -60] [v₁] [0]
[ 0 -3 0] * [v₂] = [0]
[10 - 70 -25] [v₃] [0]
Expanding this equation, we get three equations:
25v₁ - 188v₂ - 60v₃ = 0
-3v₂ = 0
10v₁ - 70v₂ - 25v₃ = 0
From equation (2), v₂ = 0.
Then, substituting v₂ = 0 into equations (1) and (3), we get:
25v₁ - 60v₃ = 0
10v₁ - 25v₃ = 0
Solving these two equations, we find v₁ = 3 and v₃ = 2. Therefore, the eigenvector corresponding to the eigenvalue λ = -3 is [3, 0, 2].
This means that when the given matrix is multiplied by this eigenvector, it scales by a factor of -3.
An eigenvector corresponding to the eigenvalue[tex]\(\lambda = -3\) of the given matrix is \(v = [ 4, -3, 2 ]^T\).[/tex]
Given matrix [tex]\(A\) is:[/tex]
[tex]\[A = \begin{bmatrix} 25 & -188 & -60 \\ 0 & -3 & 0 \\ 10 & -70 & -25 \end{bmatrix}\][/tex]
To find the eigenvector corresponding to the eigenvalue [tex]\(\lambda = -3\),[/tex] we need to solve the equation[tex]\(Av = \lambda v\), where \(v\)[/tex]is the eigenvector.
[tex]\(\lambda = -3\),[/tex]into the equation, we get:
[tex]\[\begin{bmatrix} 25 & -188 & -60 \\ 0 & -3 & 0 \\ 10 & -70 & -25 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = -3 \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}\][/tex]
This simplifies to the system of equations:
[tex]\[25v_1 - 188v_2 - 60v_3 = -3v_1\][/tex]
[tex]\[-3v_2 = -3v_2\][/tex]
[tex]\[10v_1 - 70v_2 - 25v_3 = -3v_3\][/tex]
Solving this system, we find[tex]\(v_1 = 4\), \(v_2 = -3\), and \(v_3 = 2\),[/tex]which gives us the eigenvector [tex]\(v = [ 4, -3, 2 ]^T\)[/tex].
