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Null space of $$$\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right]$$$
Your Input
Find the null space of $$$\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right]$$$.
Solution
The reduced row echelon form of the matrix is $$$\left[\begin{array}{ccc}1 & 0 & \frac{11}{7}\\0 & 1 & \frac{5}{7}\end{array}\right]$$$ (for steps, see rref calculator).
To find the null space, solve the matrix equation $$$\left[\begin{array}{ccc}1 & 0 & \frac{11}{7}\\0 & 1 & \frac{5}{7}\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$$$
If we take $$$x_{3} = t$$$, then $$$x_{1} = - \frac{11 t}{7}$$$, $$$x_{2} = - \frac{5 t}{7}$$$.
Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}- \frac{11 t}{7}\\- \frac{5 t}{7}\\t\end{array}\right] = \left[\begin{array}{c}- \frac{11}{7}\\- \frac{5}{7}\\1\end{array}\right] t.$$$
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 $$$1$$$.
Answer
The basis for the null space is $$$\left\{\left[\begin{array}{c}- \frac{11}{7}\\- \frac{5}{7}\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}-1.571428571428571\\-0.714285714285714\\1\end{array}\right]\right\}.$$$A
The nullity of the matrix is $$$1$$$A.