Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
1 & 4 & -1 & 3\\
2 & 9 & -1 & 7 \\
2 & 8 & -2 & 6
\end{bmatrix}$
After reducing the augmented matrix of this system yields we have:
$A^\#=\begin{bmatrix}
1 & 4 & -1 & 3\\
2 & 9 & -1 & 7 \\
0 & 0 & 0 & 0
\end{bmatrix}$
Consequently, there are two free variables, $x_3= t$ and $x_4 = s$, so that:
$nullspace (A)\\
=\{5t+s,-t-s,t,s: s,t \in R\}\\
=span \{(5,-1,1,0),(1,-1,0,1)\}$
In this problem, according to the Rank-Nullity Theorem we have $n=2$. Further we see that $rank(A) = 2$. Hence:
$$rank(A)+nullity(A)=2+2=4=n$$
Hence the Rank-Nullity Theorem is verified.
