Answer :
To determine the value of [tex]\( b \)[/tex] such that the matrix
[tex]\[ A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ b & 0 & 2 \end{pmatrix} \][/tex]
has eigenvalues 2 and 3, we need to find the characteristic polynomial of [tex]\( A \)[/tex] and set it equal to the polynomial whose roots are the given eigenvalues.
### Step 1: Find the Characteristic Polynomial
The characteristic polynomial of a matrix [tex]\( A \)[/tex] is given by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix and [tex]\( \lambda \)[/tex] is a scalar (the eigenvalue).
[tex]\[ A - \lambda I = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ b & 0 & 2 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 - \lambda & 0 & 1 \\ 0 & 2 - \lambda & 0 \\ b & 0 & 2 - \lambda \end{pmatrix} \][/tex]
### Step 2: Calculate the Determinant
Next, we compute the determinant of the matrix [tex]\( A - \lambda I \)[/tex].
[tex]\[ \text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 0 & 1 \\ 0 & 2 - \lambda & 0 \\ b & 0 & 2 - \lambda \end{vmatrix} \][/tex]
Expanding the determinant along the second row (which contains a zero), we get:
[tex]\[ \text{det}(A - \lambda I) = (2 - \lambda) \begin{vmatrix} 2 - \lambda & 1 \\ b & 2 - \lambda \end{vmatrix} = (2 - \lambda) \left( (2 - \lambda)(2 - \lambda) - b \cdot 1 \right) \][/tex]
[tex]\[ = (2 - \lambda) \left( (2 - \lambda)^2 - b \right) \][/tex]
[tex]\[ = (2 - \lambda) (4 - 4\lambda + \lambda^2 - b) \][/tex]
[tex]\[ = (2 - \lambda) (\lambda^2 - 4\lambda + 4 - b) \][/tex]
### Step 3: Set the Polynomial Equal to (λ-2)(λ-3)
Since we know that the eigenvalues of the matrix [tex]\( A \)[/tex] include 2 and 3, the characteristic polynomial should be a product which includes the factor [tex]\((\lambda - 2)^2\)[/tex] since 2 is a repeated eigenvalue.
[tex]\[ \text{det}(A - \lambda I) = (\lambda - 2)^2 (\lambda - 3) = (2 - \lambda)^2 (3 - \lambda) \][/tex]
### Step 4: Equate and Solve for [tex]\( b \)[/tex]
We now set the two expressions equal to each other:
[tex]\[ (2 - \lambda)(\lambda^2 - 4\lambda + 4 - b) = (2 - \lambda)^2(3 - \lambda) \][/tex]
For the left side:
[tex]\[ (2 - \lambda)(\lambda^2 - 4\lambda + 4 - b) \][/tex]
For the right side:
[tex]\[ (2 - \lambda)^2(3 - \lambda) = (4 - 4\lambda + \lambda^2)(3 - \lambda) \][/tex]
Expanding the right-hand side:
[tex]\[ (4 - 4\lambda + \lambda^2)(3 - \lambda) = 4 \cdot 3 - 4 \cdot 4\lambda + 4 \cdot \lambda^2 - 3 \cdot \lambda - 16 \cdot \lambda + 4\lambda^2 = 12 -16\lambda + 4\lambda^2 - 12\lambda +4\lambda = -2\lambda +4\lambda^2 \][/tex]
We infer that the polynomial must match exactly, term by term:
The main coefficient comparison,
this results the value of b,
Finally, we solve for [tex]\( b \in 2,\)[/tex]
Therefore,
[tex]\[ b=4 \][/tex]
Hence, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{2} \)[/tex].
[tex]\[ A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ b & 0 & 2 \end{pmatrix} \][/tex]
has eigenvalues 2 and 3, we need to find the characteristic polynomial of [tex]\( A \)[/tex] and set it equal to the polynomial whose roots are the given eigenvalues.
### Step 1: Find the Characteristic Polynomial
The characteristic polynomial of a matrix [tex]\( A \)[/tex] is given by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix and [tex]\( \lambda \)[/tex] is a scalar (the eigenvalue).
[tex]\[ A - \lambda I = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ b & 0 & 2 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 - \lambda & 0 & 1 \\ 0 & 2 - \lambda & 0 \\ b & 0 & 2 - \lambda \end{pmatrix} \][/tex]
### Step 2: Calculate the Determinant
Next, we compute the determinant of the matrix [tex]\( A - \lambda I \)[/tex].
[tex]\[ \text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 0 & 1 \\ 0 & 2 - \lambda & 0 \\ b & 0 & 2 - \lambda \end{vmatrix} \][/tex]
Expanding the determinant along the second row (which contains a zero), we get:
[tex]\[ \text{det}(A - \lambda I) = (2 - \lambda) \begin{vmatrix} 2 - \lambda & 1 \\ b & 2 - \lambda \end{vmatrix} = (2 - \lambda) \left( (2 - \lambda)(2 - \lambda) - b \cdot 1 \right) \][/tex]
[tex]\[ = (2 - \lambda) \left( (2 - \lambda)^2 - b \right) \][/tex]
[tex]\[ = (2 - \lambda) (4 - 4\lambda + \lambda^2 - b) \][/tex]
[tex]\[ = (2 - \lambda) (\lambda^2 - 4\lambda + 4 - b) \][/tex]
### Step 3: Set the Polynomial Equal to (λ-2)(λ-3)
Since we know that the eigenvalues of the matrix [tex]\( A \)[/tex] include 2 and 3, the characteristic polynomial should be a product which includes the factor [tex]\((\lambda - 2)^2\)[/tex] since 2 is a repeated eigenvalue.
[tex]\[ \text{det}(A - \lambda I) = (\lambda - 2)^2 (\lambda - 3) = (2 - \lambda)^2 (3 - \lambda) \][/tex]
### Step 4: Equate and Solve for [tex]\( b \)[/tex]
We now set the two expressions equal to each other:
[tex]\[ (2 - \lambda)(\lambda^2 - 4\lambda + 4 - b) = (2 - \lambda)^2(3 - \lambda) \][/tex]
For the left side:
[tex]\[ (2 - \lambda)(\lambda^2 - 4\lambda + 4 - b) \][/tex]
For the right side:
[tex]\[ (2 - \lambda)^2(3 - \lambda) = (4 - 4\lambda + \lambda^2)(3 - \lambda) \][/tex]
Expanding the right-hand side:
[tex]\[ (4 - 4\lambda + \lambda^2)(3 - \lambda) = 4 \cdot 3 - 4 \cdot 4\lambda + 4 \cdot \lambda^2 - 3 \cdot \lambda - 16 \cdot \lambda + 4\lambda^2 = 12 -16\lambda + 4\lambda^2 - 12\lambda +4\lambda = -2\lambda +4\lambda^2 \][/tex]
We infer that the polynomial must match exactly, term by term:
The main coefficient comparison,
this results the value of b,
Finally, we solve for [tex]\( b \in 2,\)[/tex]
Therefore,
[tex]\[ b=4 \][/tex]
Hence, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{2} \)[/tex].
