Null space of $\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right]$

Find the null space of $\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right]$.

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 $1$.

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.