What is the reduced row echelon form of $A$?

Question

Let $$A = \left( \begin{array}{cccc} 7 & 7 & 9 & -17\\ 6 & 6 & 1 & -2 \\ -12 & -12 & -27 & 1 \\ 7& 7 & 17 & -15\end{array} \right)$$

1. What is the reduced row echelon form of $A$?

2. What is the rank of $A$?

Answer 1

$1.$ Multiply the first row by $\frac{1}{7}:$ $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 6 & 6 & 1 & -2 \\ -12 & -12 & -27 & 1 \\ 7 & 7 & 17 & -15 \end{pmatrix} $$ $2.$ Multiply first row by $6$ and add to row $2$: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & -\frac{47}{7} & \frac{88}{7} \\ -12 & -12 & -27 & 1 \\ 7 & 7 & 17 & -15 \end{pmatrix} $$ $3.$ Multiply first row by $12$ to the third row: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & -\frac{47}{7} & \frac{88}{7} \\ 0 & 0 & -\frac{81}{7} & -\frac{197}{7} \\ 7 & 7 & 17 & -15 \end{pmatrix} $$ $4.$ Multiply first row by $-7$ and add to the fourth row: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & -\frac{47}{7} & \frac{88}{7} \\ 0 & 0 & -\frac{81}{7} & -\frac{197}{7} \\ 0 & 0 & 8 & 2 \end{pmatrix} $$ $5.$ Multiply the second row by $-\frac{7}{47}$: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & -\frac{88}{47} \\ 0 & 0 & -\frac{81}{7} & -\frac{197}{7} \\ 0 & 0 & 8 & 2 \end{pmatrix} $$ $6.$ Multiply second row by $\frac{81}{7}$ and add to the third row: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & -\frac{88}{47} \\ 0 & 0 & 0 & -\frac{2341}{47} \\ 0 & 0 & 8 & 2 \end{pmatrix} $$ $7.$ Multiply the second row by $-8$ to the fourth row: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & -\frac{88}{47} \\ 0 & 0 & 0 & -\frac{-2341}{47} \\ 0 & 0 & 0 & \frac{798}{47} \end{pmatrix} $$ $8.$ Multiply the third row by $-\frac{47}{2341}$: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & -\frac{88}{47} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & \frac{798}{47} \end{pmatrix} $$ $9.$ Multiply the third row by $-\frac{798}{47}$ and add to the fourth row: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & -\frac{88}{47} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ This is the row echelon form. For the reduced row echelon form, $10.$ Multiply the third row by $\frac{88}{47}$ and add to the second row to get: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & -\frac{17}{7} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ $11.$ Multiply the third row by $\frac{17}{7}$ and add it to the first row to get: $$ \begin{pmatrix} 1 & 1 & \frac{9}{7} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ $12.$ Multiply the second row by $-\frac{9}{7}$ and add it to the first row to get: $$ \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ This has rank $3$ (ie, the number of non-zero rows).