Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
2 & -1\\
-4 & 2
\end{bmatrix}$
We must find all solutions to $Ax = 0$. Reducing the augmented matrix of this system yields:
$A^\#=\begin{bmatrix}
2 & -1\\
-4 & 2
\end{bmatrix} \approx A=\begin{bmatrix}
1 & -\frac{1}{2}\\
0 & 0
\end{bmatrix}$
Consider vectors $(x,y) $ in nullspace $(A)$ satisfies $x-\frac{1}{2}y=0$. Let $x=t, y=2t$
So:
$nullspace(A)
=\{(t,2t): t \in R\\
=span\{(\frac{1}{2})\}$
In this problem, according to the Rank-Nullity Theorem we have $n=1$. Further, from the foregoing row-echelon form of the augmented matrix of the system $Ax = 0$, we see that $rank(A) = 1$. Hence:
$$rank(A)+nullity(A)=1+1=2=n$$
Hence the Rank-Nullity Theorem is verified.
