(1 point) Let

𝐴=⎡⎣⎢⎢−3−20640−6−40⎤⎦⎥⎥.A=[−36−6−24−4000].

A basis for the null space of 𝐴A is


We are tasked with finding a basis for the null space of the matrix \( A \). The matrix is given as:

\[
A = \begin{bmatrix}
-3 & 6 & -6 \\
-2 & 4 & -4 \\
0 & 0 & 0
\end{bmatrix}
\]

### Step 1: Set up the null space equation
The null space consists of all vectors \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \) such that:

\[
A \mathbf{x} = \mathbf{0}
\]

This gives us the following system of equations:

\[
\begin{aligned}
-3x_1 + 6x_2 - 6x_3 &= 0 \\
-2x_1 + 4x_2 - 4x_3 &= 0 \\
0x_1 + 0x_2 + 0x_3 &= 0
\end{aligned}
\]

### Step 2: Simplify the system
The first two equations are:

\[
-3x_1 + 6x_2 - 6x_3 = 0 \quad \text{(Equation 1)}
\]
\[
-2x_1 + 4x_2 - 4x_3 = 0 \quad \text{(Equation 2)}
\]

Equation 1 can be simplified by dividing through by \(-3\):

\[
x_1 - 2x_2 + 2x_3 = 0 \quad \text{(Simplified Equation 1)}
\]

Similarly, Equation 2 can be simplified by dividing through by \(-2\):

\[
x_1 - 2x_2 + 2x_3 = 0 \quad \text{(Simplified Equation 2)}
\]

Since both simplified equations are identical, we are left with the single equation:

\[
x_1 = 2x_2 - 2x_3
\]

### Step 3: Express the solution in parametric form
Let \( x_2 = t \) and \( x_3 = s \), where \( t \) and \( s \) are free parameters. From the equation \( x_1 = 2x_2 - 2x_3 \), we have:

\[
x_1 = 2t - 2s
\]

Thus, the solution vector \( \mathbf{x} \) is:

\[
\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2t - 2s \\ t \\ s \end{bmatrix}
\]

This can be rewritten as:

\[
\mathbf{x} = t \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}
\]

### Step 4: Determine the basis
The null space is spanned by the vectors:

\[
\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}
\]

Therefore, a basis for the null space of \( A \) is:

\[
\left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right\}
\]

Would you like further details or clarification on this solution?

Here are some related questions:
1. What is the rank of matrix \( A \)?
2. How do you find the nullity of a matrix?
3. What is the geometric interpretation of the null space of a matrix?
4. Can a matrix have a trivial null space?
5. How do you find the null space of larger matrices?

**Tip**: The null space of a matrix represents all the solutions to the homogeneous system \( A\mathbf{x} = 0 \).

A basis for the null space of the matrix A=⎡⎣⎢−3−20640−6−40⎤⎦. A basis for the null space of A is?

This problem involves finding a basis for the null space of a 3x3 matrix using linear algebra techniques. Step-by-step solutions cover matrix simplifications, parametric solutions, and basis determination, suitable for undergraduate students studying linear algebra.

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