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

Short Answer

Expert verified

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

Step by step solution

01

Identify the pivot position

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.

02

Apply row operation

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.

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

03

Apply row operation

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.

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

04

Apply row operation

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

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

05

Apply row operation

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.

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

06

Convert the matrix into the row echelon form

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.

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

07

Mark the pivot positions in the matrix

Mark the nonzero leading entries in columns 1 and 2.

08

Mark the positions in the original matrix

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


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

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Most popular questions from this chapter

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

Determine whether the statements that follow are true or false, and justify your answer.

18: [111315171921][-13-1]=[131921]

In Exercises 11 and 12, determine if \({\rm{b}}\) is a linear combination of \({{\mathop{\rm a}\nolimits} _1},{a_2}\) and \({a_3}\).

11.\({a_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\0\end{array}} \right],{a_2} = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right],{a_3} = \left[ {\begin{array}{*{20}{c}}5\\{ - 6}\\8\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\6\end{array}} \right]\)

Solve the linear system of equations. You may use technology.

|3x+5y+3z=257X+9y+19z=654X+5y+11z=5|

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]
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