Chapter 3.5, Problem 18PS

To determine

The combination of rows of A that produced zero. Also, the vectors in the nullspace of $A^{\text{T}}$ and nullspace of A.

The combination that gives zero rows is: $\left(Row3\right)-2\times \left(Row2\right)+\left(Row1\right)$

The nullspace of $A^{\text{T}}$ and the nullspace of A is spanned by the vector $\left(1,-2,1\right)$.

Given:

$\left[Ab\right]=\left[\begin{array}{l}123b_1\\ 456b_2\\ 789b_3\end{array}\right]\to \left[\begin{array}{l}1-23b_1\\ 0-3-6b_2-4b_1\\ 000b_3-2b_2+b_1\end{array}\right]$

Calculation:

Consider the matrix $\left[Ab\right]=\left[\begin{array}{l}123b_1\\ 456b_2\\ 789b_3\end{array}\right]\to \left[\begin{array}{l}1-23b_1\\ 0-3-6b_2-4b_1\\ 000b_3-2b_2+b_1\end{array}\right]$

The row that gives zero row is $b_3$that means the combination is $\left(Row3\right)-2\times \left(Row2\right)+\left(Row1\right)$.

As $\left(Row3\right)-2\times \left(Row2\right)+\left(Row1\right)=0$, this implies that the vectors $c\left(1,-2,1\right)$are present in the left nullspace.

Thus, the combination is $\left(Row3\right)-2\times \left(Row2\right)+\left(Row1\right)$and the nullspace of $A^{\text{T}}$ and the nullspace of A is spanned by the vector $\left(1,-2,1\right)$.