**Find a Non-Trivial Solution of a Matrix Equation** **Problem Statement:** Find a non-trivial solution of the equation \(A\mathbf{x} = \mathbf{0}\), where \[A = \begin{bmatrix} -15 & -3 \\ 30 & 6 \\ -10 & -2 \end{bmatrix}\] **Options:** - \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\) - \(\begin{bmatrix} -5 \\ 1 \end{bmatrix}\) - \(\begin{bmatrix} 5 \\ -1 \end{bmatrix}\) - \(\begin{bmatrix} 1 \\ -5 \end{bmatrix}\) - \(\begin{bmatrix} 1 \\ 5 \end{bmatrix}\) Choose the correct non-trivial solution from the given options. **Explanation:** In linear algebra, a non-trivial solution to the homogeneous system \(A\mathbf{x} = \mathbf{0}\) is a solution where \(\mathbf{x} \neq \mathbf{0}\). This involves finding a vector \(\mathbf{x}\) which, when multiplied by the matrix \(A\), results in the zero vector. Non-trivial solutions occur when the determinant of \(A\) is zero, indicating that the matrix is singular and has linearly dependent rows or columns. ### Solve the Problem #### Describe all solutions of \( Ax = b \), where \[ A = \begin{bmatrix} 2 & -5 & 3 \\ -2 & 6 & -5 \\ -4 & 7 & 0 \end{bmatrix} \] and \[ b = \begin{bmatrix} -3 \\ 8 \\ -9 \end{bmatrix}. \] #### Describe the general solution in parametric vector form. ### Options: 1. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7/2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} \] 2. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 1 \end{bmatrix} \] 3. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -17 \\ 2 \\ 1 \end{bmatrix} \] 4. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 0 \end{bmatrix} \]
**Find a Non-Trivial Solution of a Matrix Equation** **Problem Statement:** Find a non-trivial solution of the equation \(A\mathbf{x} = \mathbf{0}\), where \[A = \begin{bmatrix} -15 & -3 \\ 30 & 6 \\ -10 & -2 \end{bmatrix}\] **Options:** - \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\) - \(\begin{bmatrix} -5 \\ 1 \end{bmatrix}\) - \(\begin{bmatrix} 5 \\ -1 \end{bmatrix}\) - \(\begin{bmatrix} 1 \\ -5 \end{bmatrix}\) - \(\begin{bmatrix} 1 \\ 5 \end{bmatrix}\) Choose the correct non-trivial solution from the given options. **Explanation:** In linear algebra, a non-trivial solution to the homogeneous system \(A\mathbf{x} = \mathbf{0}\) is a solution where \(\mathbf{x} \neq \mathbf{0}\). This involves finding a vector \(\mathbf{x}\) which, when multiplied by the matrix \(A\), results in the zero vector. Non-trivial solutions occur when the determinant of \(A\) is zero, indicating that the matrix is singular and has linearly dependent rows or columns. ### Solve the Problem #### Describe all solutions of \( Ax = b \), where \[ A = \begin{bmatrix} 2 & -5 & 3 \\ -2 & 6 & -5 \\ -4 & 7 & 0 \end{bmatrix} \] and \[ b = \begin{bmatrix} -3 \\ 8 \\ -9 \end{bmatrix}. \] #### Describe the general solution in parametric vector form. ### Options: 1. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7/2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} \] 2. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 1 \end{bmatrix} \] 3. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -17 \\ 2 \\ 1 \end{bmatrix} \] 4. \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 0 \end{bmatrix} \]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Find a Non-Trivial Solution of a Matrix Equation**
**Problem Statement:**
Find a non-trivial solution of the equation \(A\mathbf{x} = \mathbf{0}\), where
\[A = \begin{bmatrix}
-15 & -3 \\
30 & 6 \\
-10 & -2
\end{bmatrix}\]
**Options:**
- \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
- \(\begin{bmatrix} -5 \\ 1 \end{bmatrix}\)
- \(\begin{bmatrix} 5 \\ -1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ -5 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 5 \end{bmatrix}\)
Choose the correct non-trivial solution from the given options.
**Explanation:**
In linear algebra, a non-trivial solution to the homogeneous system \(A\mathbf{x} = \mathbf{0}\) is a solution where \(\mathbf{x} \neq \mathbf{0}\). This involves finding a vector \(\mathbf{x}\) which, when multiplied by the matrix \(A\), results in the zero vector. Non-trivial solutions occur when the determinant of \(A\) is zero, indicating that the matrix is singular and has linearly dependent rows or columns.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F157e66f5-e794-4af2-988e-1885b818a80a%2Fa3ce6374-4aa0-43f5-b5e8-5075acbd0e37%2Fo0q66n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find a Non-Trivial Solution of a Matrix Equation**
**Problem Statement:**
Find a non-trivial solution of the equation \(A\mathbf{x} = \mathbf{0}\), where
\[A = \begin{bmatrix}
-15 & -3 \\
30 & 6 \\
-10 & -2
\end{bmatrix}\]
**Options:**
- \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
- \(\begin{bmatrix} -5 \\ 1 \end{bmatrix}\)
- \(\begin{bmatrix} 5 \\ -1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ -5 \end{bmatrix}\)
- \(\begin{bmatrix} 1 \\ 5 \end{bmatrix}\)
Choose the correct non-trivial solution from the given options.
**Explanation:**
In linear algebra, a non-trivial solution to the homogeneous system \(A\mathbf{x} = \mathbf{0}\) is a solution where \(\mathbf{x} \neq \mathbf{0}\). This involves finding a vector \(\mathbf{x}\) which, when multiplied by the matrix \(A\), results in the zero vector. Non-trivial solutions occur when the determinant of \(A\) is zero, indicating that the matrix is singular and has linearly dependent rows or columns.
![### Solve the Problem
#### Describe all solutions of \( Ax = b \), where
\[ A = \begin{bmatrix} 2 & -5 & 3 \\ -2 & 6 & -5 \\ -4 & 7 & 0 \end{bmatrix} \]
and
\[ b = \begin{bmatrix} -3 \\ 8 \\ -9 \end{bmatrix}. \]
#### Describe the general solution in parametric vector form.
### Options:
1.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7/2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix}
\]
2.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 1 \end{bmatrix}
\]
3.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -17 \\ 2 \\ 1 \end{bmatrix}
\]
4.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 0 \end{bmatrix}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F157e66f5-e794-4af2-988e-1885b818a80a%2Fa3ce6374-4aa0-43f5-b5e8-5075acbd0e37%2Fb108mbo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solve the Problem
#### Describe all solutions of \( Ax = b \), where
\[ A = \begin{bmatrix} 2 & -5 & 3 \\ -2 & 6 & -5 \\ -4 & 7 & 0 \end{bmatrix} \]
and
\[ b = \begin{bmatrix} -3 \\ 8 \\ -9 \end{bmatrix}. \]
#### Describe the general solution in parametric vector form.
### Options:
1.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 7/2 \\ -2 \\ 1 \end{bmatrix} + x_3 \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix}
\]
2.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 1 \end{bmatrix}
\]
3.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -17 \\ 2 \\ 1 \end{bmatrix}
\]
4.
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 11 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 7/2 \\ 2 \\ 0 \end{bmatrix}
\]
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