<s>[INST] Find the rank of matrix. "Find rank of matrix $\begin{bmatrix}
                    0&0&-3\\ 
                    9&3&5\\
                    3&1&1\\
                    \end{bmatrix}$
using echelon form?"
"I am getting an answer equal to 2."
"When I reduce this I get $\begin{bmatrix}
                          1&1/3&0\\
                          0&0&1\\
                           0&0&0\\
\end{bmatrix}$ so two pivots so rank must me 2 right ? However the book I am using says rank =1."
 [/INST] It is indeed a rank 2 matrix.
$\begin{bmatrix}
0&0&-3\\
9&3&5\\
3&1&1\\
\end{bmatrix}\overset{r_1\leftrightarrow r_3}{\rightarrow}
\begin{bmatrix}
3&1&1\\
9&3&5\\
0&0&-3\\
\end{bmatrix}
\overset{r_2=r_2-3\times r_1}{\rightarrow}
\begin{bmatrix}
3&1&1\\
0&0&2\\
0&0&-3\\
\end{bmatrix}
\overset{r_3=r_3+\frac{3}{2}\times r_2}{\rightarrow}
\begin{bmatrix}
3&1&1\\
0&0&2\\
0&0&0\\
\end{bmatrix}
\overset{r_1=r_1/3}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&\frac{1}{3}\\
0&0&2\\
0&0&0\\
\end{bmatrix}
\overset{r_2=r_2/2}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&\frac{1}{3}\\
0&0&1\\
0&0&0\\
\end{bmatrix}
\overset{r_1=r_1-r_2/3}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&0\\
0&0&1\\
0&0&0\\
\end{bmatrix}
$
Your book has made a mistake.
P.S. Welcome to Math.SE! Next time please learn to use some basic MathJaX to prevent people from downvoting you and misunderstanding you. You may start from this tutorial on Math.SE.Meta.
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