Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
6\\
-1\\
9
\end{bmatrix}$
We must find all solutions to $Ax = 0$. Reducing the augmented matrix of this system yields:
$A^\#=\begin{bmatrix}
6 | 0\\
-1 | 0\\
9 | 0
\end{bmatrix} \approx \begin{bmatrix}
1 | 0\\
-1 | 0\\
9 | 0
\end{bmatrix} \approx \begin{bmatrix}
1 | 0\\
0 | 0\\
0 | 0
\end{bmatrix}$
We can notice that $nullspace (A)=span \{(0)\}$. In this problem, $A$ is a $3 \times 1$ matrix, so, in the Rank-Nullity Theorem, $n=1$. Further, from the foregoing row-echelon form of the augmented matrix of the system $Ax = 0$, we see that $rank(A) = 1$. Hence:
$$rank(A)+nullity(A)=1+0=1=n$$
Hence the Rank-Nullity Theorem is verified.
