Null space of [123417]\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right]

The calculator will find the null space of the 22x33 matrix [123417]\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right], with steps shown.
×\times
A

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

Your Input

Find the null space of [123417]\left[\begin{array}{ccc}1 & 2 & 3\\4 & 1 & 7\end{array}\right].

Solution

The reduced row echelon form of the matrix is [101170157]\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 [101170157][x1x2x3]=[00].\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 x3=tx_{3} = t, then x1=11t7x_{1} = - \frac{11 t}{7}, x2=5t7x_{2} = - \frac{5 t}{7}.

Thus, x=[11t75t7t]=[117571]t.\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 11.

Answer

The basis for the null space is {[117571]}{[1.5714285714285710.7142857142857141]}.\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 11A.