To find an explicit description of the null space [tex]\( \text{Nul } A \)[/tex] for the matrix [tex]\( A \)[/tex], we can follow these steps:
Given the matrix:
[tex]\[ A = \begin{pmatrix}
1 & 3 & -4 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix} \][/tex]
We need to find all vectors [tex]\( \vec{x} = \begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{pmatrix} \)[/tex] such that [tex]\( A \vec{x} = \vec{0} \)[/tex]. This translates to solving the equation [tex]\( A \vec{x} = \vec{0} \)[/tex].
Expressing the system in terms of equations, we have:
[tex]\[
\begin{cases}
1x_1 + 3x_2 - 4x_3 + 0x_4 = 0 \\
0x_1 + 0x_2 + 1x_3 + 0x_4 = 0
\end{cases}
\][/tex]
From the second equation, we immediately get:
[tex]\[ x_3 = 0 \][/tex]
Substituting [tex]\( x_3 = 0 \)[/tex] into the first equation, we obtain:
[tex]\[ x_1 + 3x_2 = 0 \][/tex]
Solving for [tex]\( x_1 \)[/tex] in terms of [tex]\( x_2 \)[/tex], we get:
[tex]\[ x_1 = -3x_2 \][/tex]
Thus, the vector [tex]\( \vec{x} \)[/tex] can be written as:
[tex]\[
\vec{x} = \begin{pmatrix}
-3x_2 \\
x_2 \\
0 \\
0
\end{pmatrix}
= x_2 \begin{pmatrix}
-3 \\
1 \\
0 \\
0
\end{pmatrix}
\][/tex]
Next, we consider [tex]\( x_4 \)[/tex] which is free, meaning it can take any value. So we can also write:
[tex]\[ \vec{x} = x_4 \begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix}
\][/tex]
Combining these observations, the null space [tex]\( \text{Nul } A \)[/tex] is spanned by the vectors:
[tex]\[
\begin{pmatrix}
-3 \\
1 \\
0 \\
0
\end{pmatrix}, \begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix}
\][/tex]
Therefore, the vectors that span the null space [tex]\( \text{Nul } A \)[/tex] are:
[tex]\[ \begin{pmatrix} -3 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \][/tex]
So, a spanning set for [tex]\( \text{Nul } A \)[/tex] is:
[tex]\[ \boxed{\begin{pmatrix} -3 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}} \][/tex]
