Null space of $$$\left[\begin{array}{cccc}\frac{5}{2} & 120 & \frac{6}{5} & \frac{37}{10}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$

Find the null space of $$$\left[\begin{array}{cccc}\frac{5}{2} & 120 & \frac{6}{5} & \frac{37}{10}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$.

The reduced row echelon form of the matrix is $$$\left[\begin{array}{cccc}1 & 0 & 0 & \frac{1}{15}\\0 & 1 & \frac{1}{100} & \frac{53}{1800}\\0 & 0 & 0 & 0\end{array}\right]$$$ (for steps, see rref calculator).

To find the null space, solve the matrix equation $$$\left[\begin{array}{cccc}1 & 0 & 0 & \frac{1}{15}\\0 & 1 & \frac{1}{100} & \frac{53}{1800}\\0 & 0 & 0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right].$$$

If we take $$$x_{3} = t$$$, $$$x_{4} = s$$$, then $$$x_{1} = - \frac{s}{15}$$$, $$$x_{2} = - \frac{53 s}{1800} - \frac{t}{100}$$$.

Thus, $$$\mathbf{\vec{x}} = \left[\begin{array}{c}- \frac{s}{15}\\- \frac{53 s}{1800} - \frac{t}{100}\\t\\s\end{array}\right] = \left[\begin{array}{c}0\\- \frac{1}{100}\\1\\0\end{array}\right] t + \left[\begin{array}{c}- \frac{1}{15}\\- \frac{53}{1800}\\0\\1\end{array}\right] s.$$$

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 $$$2$$$.

The basis for the null space is $$$\left\{\left[\begin{array}{c}0\\- \frac{1}{100}\\1\\0\end{array}\right], \left[\begin{array}{c}- \frac{1}{15}\\- \frac{53}{1800}\\0\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}0\\-0.01\\1\\0\end{array}\right], \left[\begin{array}{c}-0.066666666666667\\-0.029444444444444\\0\\1\end{array}\right]\right\}.$$$A

The nullity of the matrix is $$$2$$$A.