
1. Find a basis for the null space of A, where
1 4 3 3 −1 A=−1 −4 −3 −3 1.
0 0 0 0 0 2 8 6 6 −2

We are tasked with finding a basis for the null space of the matrix \( A \), where:

\[
A = \begin{pmatrix}
1 & 4 & 3 & 3 & -1 \\
-1 & -4 & -3 & -3 & 1 \\
0 & 0 & 0 & 0 & 0 \\
2 & 8 & 6 & 6 & -2
\end{pmatrix}
\]

### Step 1: Setting up the system
The null space of \( A \) consists of all vectors \( x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} \) such that \( A x = 0 \). This leads to the system of linear equations:

\[
A x = \begin{pmatrix}
1 & 4 & 3 & 3 & -1 \\
-1 & -4 & -3 & -3 & 1 \\
0 & 0 & 0 & 0 & 0 \\
2 & 8 & 6 & 6 & -2
\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = 0
\]

This gives us the following system of equations:
\[
1x_1 + 4x_2 + 3x_3 + 3x_4 - x_5 = 0
\]
\[
-1x_1 - 4x_2 - 3x_3 - 3x_4 + x_5 = 0
\]
\[
0 = 0 \quad \text{(no information from the third row)}
\]
\[
2x_1 + 8x_2 + 6x_3 + 6x_4 - 2x_5 = 0
\]

### Step 2: Simplifying the system
Note that the second equation is just the negative of the first, so it does not provide any new information. We are left with two independent equations:

1. \( x_1 + 4x_2 + 3x_3 + 3x_4 - x_5 = 0 \)
2. \( 2x_1 + 8x_2 + 6x_3 + 6x_4 - 2x_5 = 0 \)

We can see that the second equation is just twice the first. So again, it is redundant and does not provide any new information. Therefore, we only need to work with the first equation:

\[
x_1 + 4x_2 + 3x_3 + 3x_4 - x_5 = 0
\]

### Step 3: Expressing variables
We can express \( x_1 \) in terms of the other variables:

\[
x_1 = -4x_2 - 3x_3 - 3x_4 + x_5
\]

Thus, the solution to the system can be written as:

\[
x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = \begin{pmatrix} -4x_2 - 3x_3 - 3x_4 + x_5 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix}
\]

### Step 4: Parametrizing the solution
This expression shows that the null space has free variables \( x_2, x_3, x_4, x_5 \). We can express the general solution as a linear combination of four vectors, one for each free variable. Let’s factor out each free variable:

\[
x = x_2 \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -3 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_5 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]

### Step 5: Conclusion
The null space is spanned by the following vectors:

\[
\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} -3 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
\]

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

\[
\left\{ \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}
\]

Would you like more details on the steps, or do you have any further questions?

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Here are some related questions:
1. What is the dimension of the null space for this matrix?
2. How can we interpret the null space geometrically?
3. How does the rank-nullity theorem apply to this problem?
4. How would the basis change if the matrix had different entries?
5. How do we check if a vector belongs to the null space?

**Tip**: The dimension of the null space is also known as the **nullity** of the matrix.

Find a basis for the null space of A, where
1 4 3 3 −1 A=−1 −4 −3 −3 1.
0 0 0 0 0 2 8 6 6 −2

Learn how to find a basis for the null space of the matrix A = [1 4 3 3 −1; -1 -4 -3 -3 1; 0 0 0 0 0; 2 8 6 6 −2]. This detailed step-by-step solution covers essential linear algebra concepts, including the null space and the rank-nullity theorem, suitable for undergraduate students.

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