Row reduce the matrices in Exercise 4 to reduced echelon form. Circle the pivot positions in the final matrix and in the original matrix, and list the pivot columns.

4. \(\left[ {\begin{array}{*{20}{c}}1&3&5&7\\3&5&7&9\\5&7&9&1\end{array}} \right]\)

To identify the pivot and the pivot position, observe the matrix’s leftmost column (nonzero column), that is, the pivot column. At the top of this column, 1 is the pivot.

To obtain the pivot position, convert the second term of the pivot column to 0.

Use the \({x_1}\) term in the first equation to eliminate the \(3{x_1}\) term from the second equation. Add \( - 3\) times row 1 to row 2.

\(\left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\5&7&9&1\end{array}} \right]\)

To obtain the pivot position, convert the third term of the pivot column to 0.

Use the \({x_1}\) term in the first equation to eliminate the \(5{x_1}\) term from the third equation. Add \( - 5\) times row 1 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&{ - 8}&{ - 16}&{ - 34}\end{array}} \right]\)

Multiply row 2 by \( - \frac{1}{4}\) to simplify row 2.

\(\left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&1&2&3\\0&{ - 8}&{ - 16}&{ - 34}\end{array}} \right]\)

Use the \( - 8{x_2}\) term in the third equation to eliminate the \({x_2}\) term from the second equation. Add 8 times row 2 to row 3.

\(\left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&1&2&3\\0&0&0&{ - 10}\end{array}} \right]\)

Use the \(7{x_3}\) term in the first equation to eliminate the \(3{x_3}\) term from the second equation. Add \( - 3\) times row 1 to row 2.

Use the \(0{x_2}\) term in the third equation to eliminate the \(3{x_2}\) term from the first equation. Add \( - 7\) times row 3 to row 1.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&2&0\\0&0&0&1\end{array}} \right]\)

Mark the nonzero leading entries in columns 1 and 2.

By using the reduced echelon matrix, mark the original matrix.

Thus, columns 1, 2, and 4 are the pivot columns.