Find a basis of null A for A=\left[\begin{matrix}1&-1&2&4&6\\0&1&2&1&-1\\0&0&0&1&0 \end{matrix}\right]

To find a basis for the null space of the matrix \( A \), we need to solve the equation \( A \mathbf{x} = \mathbf{0} \), where \( \mathbf{x} \) is a vector in the null space of \( A \). The matrix \( A \) is given by:

\[
A = \left[\begin{matrix}
1 & -1 & 2 & 4 & 6 \\
0 & 1 & 2 & 1 & -1 \\
0 & 0 & 0 & 1 & 0 
\end{matrix}\right]
\]

Let \(\mathbf{x} = [x_1, x_2, x_3, x_4, x_5]^T\). We need to solve \( A \mathbf{x} = \mathbf{0} \):

\[
\left[\begin{matrix}
1 & -1 & 2 & 4 & 6 \\
0 & 1 & 2 & 1 & -1 \\
0 & 0 & 0 & 1 & 0 
\end{matrix}\right]
\left[\begin{matrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 
\end{matrix}\right] = 
\left[\begin{matrix}
0 \\
0 \\
0 
\end{matrix}\right]
\]

This gives us the following system of linear equations:

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

From equation 3, we have \( x_4 = 0 \).

Substitute \( x_4 = 0 \) into equations 1 and 2:

1. \( x_1 - x_2 + 2x_3 + 6x_5 = 0 \)
2. \( x_2 + 2x_3 - x_5 = 0 \)

Now solve for \( x_2 \) from the second equation:

\[
x_2 = -2x_3 + x_5
\]

Substitute \( x_2 = -2x_3 + x_5 \) into the first equation:

\[
x_1 - (-2x_3 + x_5) + 2x_3 + 6x_5 = 0 \\
x_1 + 2x_3 - x_5 + 2x_3 + 6x_5 = 0 \\
x_1 + 4x_3 + 5x_5 = 0 \\
x_1 = -4x_3 - 5x_5
\]

Now we can express \( x_1 \), \( x_2 \), \( x_4 \) in terms of the free variables \( x_3 \) and \( x_5 \):

\[
x_1 = -4x_3 - 5x_5 \\
x_2 = -2x_3 + x_5 \\
x_4 = 0
\]

So the vector \(\mathbf{x}\) can be written as:

\[
\mathbf{x} = \left[\begin{matrix}
-4x_3 - 5x_5 \\
-2x_3 + x_5 \\
x_3 \\
0 \\
x_5 
\end{matrix}\right] = x_3 \left[\begin{matrix}
-4 \\
-2 \\
1 \\
0 \\
0 
\end{matrix}\right] + x_5 \left[\begin{matrix}
-5 \\
1 \\
0 \\
0 \\
1 
\end{matrix}\right]
\]

Thus, a basis for the null space of \( A \) is given by the set of vectors:

\[
\left\{ \left[\begin{matrix}
-4 \\
-2 \\
1 \\
0 \\
0 
\end{matrix}\right], \left[\begin{matrix}
-5 \\
1 \\
0 \\
0 \\
1 
\end{matrix}\right] \right\}
\]

Would you like more details or have any questions? Here are some additional questions you might find interesting:

1. How do you determine the column space of a matrix?
2. What is the difference between the row space and the column space of a matrix?
3. How do you find the eigenvalues and eigenvectors of a matrix?
4. What is the significance of the rank of a matrix?
5. Can you explain the process of Gaussian elimination?

**Tip:** When solving for the null space of a matrix, identify the free variables and express the dependent variables in terms of them to find the basis vectors.

Learn how to find a basis for the null space of matrix A = [1 -1 2 4 6; 0 1 2 1 -1; 0 0 0 1 0] using matrix operations and solving linear equations. This problem covers linear algebra concepts such as the null space of a matrix and solving systems of linear equations. Suitable for undergraduate students.

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