For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A. 2 3 8 - 11 A= -8 - 9 - 20 29 8 6 8 - 14 Find a nonzero vector in Nul A.

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### Finding Nonzero Vectors in Null Space and Column Space of a Matrix

#### Problem Statement:
For the matrix \( A \) below, find a nonzero vector in \( \text{Nul}\,A \) (Null Space of A) and a nonzero vector in \( \text{Col}\,A \) (Column Space of A).

Given matrix:
\[ 
A = \begin{bmatrix}
2 & 3 & 8 & -11 \\
-8 & -9 & -20 & 29 \\
8 & 6 & 8 & -14 \\
\end{bmatrix} 
\]

#### Tasks:
- Find a nonzero vector in \( \text{Nul}\,A \).
- Find a nonzero vector in \( \text{Col}\,A \).

### Solution Steps:

**1. Find a Nonzero Vector in \( \text{Nul}\,A \):**

To find a vector in \( \text{Nul}\,A \), we need to solve the equation \( A \mathbf{x} = \mathbf{0} \), where \(\mathbf{x}\) is a vector and \(\mathbf{0}\) is the zero vector.

Let's denote \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}\).

So, we have:
\[ 
A \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 2x_1 + 3x_2 + 8x_3 - 11x_4 \\ -8x_1 - 9x_2 - 20x_3 + 29x_4 \\ 8x_1 + 6x_2 + 8x_3 - 14x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\]

By solving this system of linear equations, we can determine the necessary values of \(x_1, x_2, x_3,\) and \(x_4\) that satisfy the equation.

**2. Find a Nonzero Vector in \( \text{Col}\,A \):**

The column space of \(A
Transcribed Image Text:### Finding Nonzero Vectors in Null Space and Column Space of a Matrix #### Problem Statement: For the matrix \( A \) below, find a nonzero vector in \( \text{Nul}\,A \) (Null Space of A) and a nonzero vector in \( \text{Col}\,A \) (Column Space of A). Given matrix: \[ A = \begin{bmatrix} 2 & 3 & 8 & -11 \\ -8 & -9 & -20 & 29 \\ 8 & 6 & 8 & -14 \\ \end{bmatrix} \] #### Tasks: - Find a nonzero vector in \( \text{Nul}\,A \). - Find a nonzero vector in \( \text{Col}\,A \). ### Solution Steps: **1. Find a Nonzero Vector in \( \text{Nul}\,A \):** To find a vector in \( \text{Nul}\,A \), we need to solve the equation \( A \mathbf{x} = \mathbf{0} \), where \(\mathbf{x}\) is a vector and \(\mathbf{0}\) is the zero vector. Let's denote \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}\). So, we have: \[ A \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 2x_1 + 3x_2 + 8x_3 - 11x_4 \\ -8x_1 - 9x_2 - 20x_3 + 29x_4 \\ 8x_1 + 6x_2 + 8x_3 - 14x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] By solving this system of linear equations, we can determine the necessary values of \(x_1, x_2, x_3,\) and \(x_4\) that satisfy the equation. **2. Find a Nonzero Vector in \( \text{Col}\,A \):** The column space of \(A
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