give answer in 3 step with explanation at the end of each step and final answer at the end:Solve the following eigenvalue problem t 300 10 —4 0 wilo 2 2|{X}=|-4 6 -=2|{x} 0 0 1 0 -2 2
Question:
give answer in 3 step with explanation at the end of each step and final answer at the end:
Solve the following eigenvalue problem t
300 10 —4 0
wilo 2 2|{X}=|-4 6 -=2|{x}
0 0 1 0 -2 2
This is AI generated Response.
Asked by: saravjeet
Created at: 2025-05-26 03:58:22
Asked by: saravjeet
Created at: 2025-05-26 03:58:22
Final Answer:
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Let's solve the given eigenvalue problem step by step:
### Step 1: Write the equation in standard eigenvalue form
The equation is:
\[
\omega^2
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 2 \\
0 & 0 & 1
\end{bmatrix}
\{X\}
=
\begin{bmatrix}
10 & -4 & 0 \\
-4 & 6 & -2 \\
0 & -2 & 2
\end{bmatrix}
\{X\}
\]
Rewriting:
\[
\begin{bmatrix}
10 & -4 & 0 \\
-4 & 6 & -2 \\
0 & -2 & 2
\end{bmatrix}
\{X\}
=
\omega^2
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 2 \\
0 & 0 & 1
\end{bmatrix}
\{X\}
\]
Move all terms to one side:
\[
\left(
\begin{bmatrix}
10 & -4 & 0 \\
-4 & 6 & -2 \\
0 & -2 & 2
\end{bmatrix}
-
\omega^2
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 2 \\
0 & 0 & 1
\end{bmatrix}
\right)\{X\} = 0
\]
**Explanation:**
We have rewritten the problem in the form of a generalized eigenvalue problem: \( [A] - \omega^2 [B] \).
---
### Step 2: Set up the characteristic equation
The equation for non-trivial solutions is:
\[
\det\left(
\begin{bmatrix}
10 & -4 & 0 \\
-4 & 6 & -2 \\
0 & -2 & 2
\end{bmatrix}
-
\omega^2
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 2 \\
0 & 0 & 1
\end{bmatrix}
\right) = 0
\]
Compute the matrix:
\[
\begin{bmatrix}
10 - 3\omega^2 & -4 & 0 \\
-4 & 6 - 2\omega^2 & -2 - 2\omega^2 \\
0 & -2 & 2 - \omega^2
\end{bmatrix}
\]
**Explanation:**
We substitute and subtract the matrices, forming a determinant equation whose roots will give us the eigenvalues (\(\omega^2\)).
---
### Step 3: Solve the determinant for eigenvalues \(\omega^2\)
Compute the determinant:
\[
\left|
\begin{array}{ccc}
10 - 3\omega^2 & -4 & 0 \\
-4 & 6 - 2\omega^2 & -2 - 2\omega^2 \\
0 & -2 & 2 - \omega^2
\end{array}
\right| = 0
\]
Expand along the first row:
\[
(10 - 3\omega^2)
\left|
\begin{array}{cc}
6 - 2\omega^2 & -2 - 2\omega^2 \\
-2 & 2 - \omega^2
\end{array}
\right|
- (-4)
\left|
\begin{array}{cc}
-4 & -2 - 2\omega^2 \\
0 & 2 - \omega^2
\end{array}
\right| = 0
\]
Let's compute:
- First minor:
\((6 - 2\omega^2)(2 - \omega^2) - (-2 - 2\omega^2)(-2)\)
\(= (6 - 2\omega^2)(2 - \omega^2) - 2(2 + 2\omega^2)\)
\(= [12 - 6\omega^2 - 4\omega^2 + 2\omega^4] - [4 + 4\omega^2]\)
\(= 12 - 10\omega^2 + 2\omega^4 - 4 - 4\omega^2\)
\(= 8 - 14\omega^2 + 2\omega^4\)
- Second minor:
\((-4)[(-4)(2 - \omega^2) - 0(-2 - 2\omega^2)]\)
\(= -4 [-4(2 - \omega^2)]\)
\(= -4(-8 + 4\omega^2)\)
\(= 32 - 16\omega^2\)
Thus, the determinant simplifies to:
\[
(10 - 3\omega^2)[8 - 14\omega^2 + 2\omega^4] + 32 - 16\omega^2 = 0
\]
Expand:
\[
(10)(8 - 14\omega^2 + 2\omega^4) - 3\omega^2(8 - 14\omega^2 + 2\omega^4) + 32 - 16\omega^2 = 0
\]
\[
80 - 140\omega^2 + 20\omega^4 - 24\omega^2 + 42\omega^4 - 6\omega^6 + 32 - 16\omega^2 = 0
\]
Combine like terms:
- Constants: \(80 + 32 = 112\)
- \(\omega^2\): \(-140 - 24 - 16 = -180\)
- \(\omega^4\): \(20 + 42 = 62\)
- \(\omega^6\): \(-6\)
So,
\[
-6\omega^6 + 62\omega^4 - 180\omega^2 + 112 = 0
\]
Divide by -2:
\[
3\omega^6 - 31\omega^4 + 90\omega^2 - 56 = 0
\]
**Explanation:**
We have formed and simplified the characteristic polynomial equation. The solutions to this equation are the eigenvalues \(\omega^2\).
---
## **Final Answer**
**The eigenvalue problem reduces to solving:**
\[
3\omega^6 - 31\omega^4 + 90\omega^2 - 56 = 0
\]
**The values of \(\omega^2\) that satisfy this equation are the required eigenvalues.** (You can now solve this cubic equation in \(\omega^2\) using analytical or numerical methods.)
Question:
Given me correct answer
This is AI generated Response.
Asked by: saravjeet
Created at: 2025-05-26 04:12:07
Asked by: saravjeet
Created at: 2025-05-26 04:12:07
Final Answer:
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Did you like this answer?
Let's break down the solution step-by-step:
### **Step 1: Formulate the eigenvalue problem**
The given matrices are:
\[
\mathbf{K} = \begin{bmatrix}
10 & -4 & 0 \\
-4 & 6 & -2 \\
0 & -2 & 2
\end{bmatrix}, \quad
\mathbf{M} = \begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 2 \\
0 & 0 & 1
\end{bmatrix}
\]
The eigenvalue problem is:
\[
\mathbf{K} \mathbf{X} = \omega^2 \mathbf{M} \mathbf{X}
\]
or equivalently,
\[
(\mathbf{K} - \omega^2 \mathbf{M}) \mathbf{X} = 0
\]
**Explanation:**
This is a generalized eigenvalue problem where \(\omega^2\) are the eigenvalues, and \(\mathbf{X}\) are the eigenvectors.
---
### **Step 2: Set up the characteristic equation**
The eigenvalues \(\omega^2\) satisfy:
\[
\det(\mathbf{K} - \omega^2 \mathbf{M}) = 0
\]
Compute the matrix:
\[
\mathbf{K} - \omega^2 \mathbf{M} =
\begin{bmatrix}
10 - 3 \omega^2 & -4 & 0 \\
-4 & 6 - 2 \omega^2 & -2 - 2 \omega^2 \\
0 & -2 & 2 - \omega^2
\end{bmatrix}
\]
**Explanation:**
Subtracting \(\omega^2\) times \(\mathbf{M}\) from \(\mathbf{K}\) gives the matrix whose determinant yields the characteristic equation.
---
### **Step 3: Find the eigenvalues by solving the determinant**
The determinant:
\[
\det(\mathbf{K} - \omega^2 \mathbf{M}) = 0
\]
Expanding along the first row:
\[
(10 - 3 \omega^2) \times
\begin{vmatrix}
6 - 2 \omega^2 & -2 - 2 \omega^2 \\
-2 & 2 - \omega^2
\end{vmatrix}
- (-4) \times
\begin{vmatrix}
-4 & -2 - 2 \omega^2 \\
0 & 2 - \omega^2
\end{vmatrix} = 0
\]
Calculating minors:
- First minor:
\[
(6 - 2 \omega^2)(2 - \omega^2) - (-2 - 2 \omega^2)(-2)
\]
\[
= (12 - 6 \omega^2 - 4 \omega^2 + 2 \omega^4) - [2 + 4 \omega^2]
\]
\[
= (12 - 10 \omega^2 + 2 \omega^4) - (2 + 4 \omega^2) = 10 - 14 \omega^2 + 2 \omega^4
\]
- Second minor:
\[
-4 \times [( -4)(2 - \omega^2) - 0 \times (-2 - 2 \omega^2)] = -4 \times (-4)(2 - \omega^2) = 16 (2 - \omega^2) = 32 - 16 \omega^2
\]
Putting it all together:
\[
(10 - 3 \omega^2)(10 - 14 \omega^2 + 2 \omega^4) + 32 - 16 \omega^2 = 0
\]
Expanding:
\[
(10)(10 - 14 \omega^2 + 2 \omega^4) - 3 \omega^2 (10 - 14 \omega^2 + 2 \omega^4) + 32 - 16 \omega^2 = 0
\]
which simplifies to:
\[
100 - 140 \omega^2 + 20 \omega^4 - 30 \omega^2 + 42 \omega^4 - 6 \omega^6 + 32 - 16 \omega^2 = 0
\]
Combine like terms:
\[
(100 + 32) + (-140 - 30 - 16) \omega^2 + (20 + 42) \omega^4 - 6 \omega^6 = 0
\]
\[
132 - 186 \omega^2 + 62 \omega^4 - 6 \omega^6 = 0
\]
Divide through by -2:
\[
3 \omega^6 - 31 \omega^4 + 93 \omega^2 - 66 = 0
\]
**Final polynomial in \(\omega^2\):**
\[
3 \omega^6 - 31 \omega^4 + 93 \omega^2 - 66 = 0
\]
---
### **Final answer:**
**The eigenvalues \(\omega^2\) are solutions to the polynomial:**
\[
\boxed{
3 \omega^6 - 31 \omega^4 + 93 \omega^2 - 66 = 0
}
\]
You can solve this degree-6 polynomial (or reduce it to a cubic in \(\lambda = \omega^2\)) using numerical methods or factorization techniques to find the specific eigenvalues \(\omega\).
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