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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 $A$?

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closed as off-topic by Hakim, Davide Giraudo, user26857, Yagna Patel, graydad Sep 4 '15 at 15:10

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  • $\begingroup$ click edit to see how to write matrix in a clear way $\endgroup$ – Petite Etincelle Aug 16 '14 at 8:38
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    $\begingroup$ Surely $rank A \leq 3$ because the first two columns are equal $\endgroup$ – Ella Smith Aug 16 '14 at 8:43
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    $\begingroup$ To help us help you: Do you know what reduced row echelon form and rank mean? Do you know what an elementary row operation is? $\endgroup$ – Rebecca J. Stones Aug 16 '14 at 8:59
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$1.$ 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).

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