Finding the rank of a 4 by 5 matrix

Question

Problem:

Find the rank of the following matrix: $$ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} $$ Answer:
\begin{align*} \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 1 & 5 & -1 & -1 \\ 0 & -1 & 1 & -4 & 5 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 0 & 2 & 1 & -2 \\ 0 & 0 & 4 & -6 & 6 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 0 & 2 & 1 & -2 \\ 0 & 0 & 0 & -8 & 10 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 0 & 2 & 1 & -2 \\ 0 & 0 & 0 & 1 & -\frac{5}{4} \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 0 & 2 & 0 & -\frac{3}{4} \\ 0 & 0 & 0 & 1 & -\frac{5}{4} \\ \end{bmatrix} \\ \end{align*} \begin{align*} \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & -2 & 1 \\ 0 & 0 & 1 & 0 & -\frac{3}{8} \\ 0 & 0 & 0 & 1 & -\frac{5}{4} \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 0 & 1 & 3 & 0 & -\frac{3}{2} \\ 0 & 0 & 1 & 0 & -\frac{3}{8} \\ 0 & 0 & 0 & 1 & -\frac{5}{4} \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 0 & -2 & 5 & \dfrac{41}{8} \\ 0 & 1 & 0 & 0 & -\frac{3}{8} \\ 0 & 0 & 1 & 0 & -\frac{3}{8} \\ 0 & 0 & 0 & 1 & -\frac{5}{4} \\ \end{bmatrix} \end{align*} If I were to continue the process it seems clear to me that would put the matrix in reduced row echelon form and have four independent vectors. Hence the rank of this matrix is $4$.

However, the book gets $3$. Where did I go wrong?

Answer 1

The answer is wrong because there was a mistake in putting the matrix in reduced row echelon form. Here is a correct solution.

Using an online calculator, I find that: $$ \begin{bmatrix} 1 & 3 & -2 & 5 & 4 \\ 1 & 4 & 1 & 3 & 5 \\ 1 & 4 & 2 & 4 & 3 \\ 2 & 7 & -3 & 6 & 13 \\ \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 22 & -21 \\ 0 & 1 & 0 & -5 & 7 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Hence the rank of the matrix is $3$.