[1 2 3] 4 5 6 7. 7 8 9 14. Find a symmetric matrix Q such that q(7) = T" Qu = "

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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A problem from linear algebra:

**Problem Statement:**

14. Find a symmetric matrix \( Q \) such that \( q(\vec{x}) = \vec{x}^T Q \vec{x} = \vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x} \).

**Explanation:**

The problem is to find a symmetric matrix \( Q \) such that the quadratic form \( q(\vec{x}) \) given by \( \vec{x}^T Q \vec{x} \) matches the quadratic form given by \( \vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x} \).

A matrix \( Q \) is symmetric if and only if \( Q = Q^T \). This means that the matrix should be equal to its transpose. 

\[
\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}
\]

The quadratic form \( q(\vec{x}) \) can be expanded as:

\[
\vec{x}^T Q \vec{x} = x_1 x_1 Q_{11} + x_1 x_2 (Q_{12} + Q_{21}) + x_1 x_3 (Q_{13} + Q_{31}) + x_2 x_2 Q_{22} + x_2 x_3 (Q_{23} + Q_{32}) + x_3 x_3 Q_{33}
\]

We need to equate this with:

\[
\vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x}
\]

Which expands to:

\[
= 1 x_1^2 + (2 + 4)x_1 x_2 + (3 + 7)x_1 x_3 + 5 x_2^2 + (6 + 8)x_2 x_3 + 9 x_3^2
\]

So, \(
Transcribed Image Text:**Problem Statement:** 14. Find a symmetric matrix \( Q \) such that \( q(\vec{x}) = \vec{x}^T Q \vec{x} = \vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x} \). **Explanation:** The problem is to find a symmetric matrix \( Q \) such that the quadratic form \( q(\vec{x}) \) given by \( \vec{x}^T Q \vec{x} \) matches the quadratic form given by \( \vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x} \). A matrix \( Q \) is symmetric if and only if \( Q = Q^T \). This means that the matrix should be equal to its transpose. \[ \vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \] The quadratic form \( q(\vec{x}) \) can be expanded as: \[ \vec{x}^T Q \vec{x} = x_1 x_1 Q_{11} + x_1 x_2 (Q_{12} + Q_{21}) + x_1 x_3 (Q_{13} + Q_{31}) + x_2 x_2 Q_{22} + x_2 x_3 (Q_{23} + Q_{32}) + x_3 x_3 Q_{33} \] We need to equate this with: \[ \vec{x}^T \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \vec{x} \] Which expands to: \[ = 1 x_1^2 + (2 + 4)x_1 x_2 + (3 + 7)x_1 x_3 + 5 x_2^2 + (6 + 8)x_2 x_3 + 9 x_3^2 \] So, \(
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