|
|
|
|
|
|
|
|
|
|
|
|
Null space of $$$\left[\begin{array}{ccccc}2 & 4 & 3 & 7 & -1\\1 & 2 & 2 & 5 & 3\\-3 & -6 & -1 & 0 & 1\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{ccccc}2 & 4 & 3 & 7 & -1\\1 & 2 & 2 & 5 & 3\\-3 & -6 & -1 & 0 & 1\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{ccccc}1 & 2 & 0 & -1 & 0\\0 & 0 & 1 & 3 & 0\\0 & 0 & 0 & 0 & 1\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{ccccc}1 & 2 & 0 & -1 & 0\\0 & 0 & 1 & 3 & 0\\0 & 0 & 0 & 0 & 1\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right].$$$
If we take $$$x_{2} = t$$$, $$$x_{4} = s$$$, then $$$x_{1} = s - 2 t$$$, $$$x_{3} = - 3 s$$$, $$$x_{5} = 0$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}s - 2 t\\t\\- 3 s\\s\\0\end{array}\right] = \left[\begin{array}{c}-2\\1\\0\\0\\0\end{array}\right] t + \left[\begin{array}{c}1\\0\\-3\\1\\0\end{array}\right] s.$$$
This is the null space.
The nullity of a matrix is the dimension of the basis for the null space.
Thus, the nullity of the matrix is $$$2$$$.
Answer
The basis for the null space is $$$\left\{\left[\begin{array}{c}-2\\1\\0\\0\\0\end{array}\right], \left[\begin{array}{c}1\\0\\-3\\1\\0\end{array}\right]\right\}$$$A.
The nullity of the matrix is $$$2$$$A.