Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
1 & 0 & -6 & -1
\end{bmatrix}$
We can notice that $A$ is in row-echelon form. Consider vectors $(x,y,z,w) $ in nullspace $(A)$ satisfies $x+0y-6z-w=0$ Hence, $(x,z,w)$ are free variables.
So:
$nullspace(A)
=\{(6z+w,y,z,w): y,z,w \in R\\
=span\{(0,1,0,0),(6,0,1,0),(1,0,0,1)\}$
In this problem, according to the Rank-Nullity Theorem, $n=3$. Further, from the foregoing row-echelon form of the augmented matrix of the system $Ax = 0$, we see that $rank(A) = 1$. Hence:
$$rank(A)+nullity(A)=1+3=4=n$$
Hence the Rank-Nullity Theorem is verified.
