3. Determine if b is a linear combination of a₁, a2, and a3 2 1 2 3 a1 = 1 a2 = 2 , az 1 4 1 0 9 - = 2 b =

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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## Problem Statement

### 3. Determine if \( b \) is a linear combination of \( a_1, a_2, \) and \( a_3 \).

Given vectors:
\[ a_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \quad a_2 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \quad a_3 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \]

To determine if vector \( b \) is a linear combination of vectors \( a_1, a_2, \) and \( a_3 \), we need to find scalars \( x, y, \) and \( z \) such that:
\[ x \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + y \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + z \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \]

This can be represented as the matrix equation:
\[ x \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + y \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + z \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \]

In matrix form, this becomes:
\[ \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \]

To solve this system of linear equations, we can use methods such as Gaussian elimination or matrix row reduction. The task is to find if there exists a solution
Transcribed Image Text:## Problem Statement ### 3. Determine if \( b \) is a linear combination of \( a_1, a_2, \) and \( a_3 \). Given vectors: \[ a_1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \quad a_2 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \quad a_3 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \] To determine if vector \( b \) is a linear combination of vectors \( a_1, a_2, \) and \( a_3 \), we need to find scalars \( x, y, \) and \( z \) such that: \[ x \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + y \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + z \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \] This can be represented as the matrix equation: \[ x \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + y \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + z \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \] In matrix form, this becomes: \[ \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix} \] To solve this system of linear equations, we can use methods such as Gaussian elimination or matrix row reduction. The task is to find if there exists a solution
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