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Section4.2Reduced Row–Echelon Matrices

Definition4.2.1Reduced Row–Echelon Form

A matrix \(A,\) is in reduced row–echelon form if it is in row–echelon form and it satisfies:

  1. Each leading entry of \(A\) is 1.

  2. Each leading entry of \(A\) is the only nonzero element in its column.

SubsectionGauss–Jordan Elimination

Recall that an augmented matrix in row–echelon form represents a system of linear equations that can be solved via backsubstitution. Gauss–Jordan elimination extends the Gaussian elimination algorithm so that back–substitution is not necessary.

SubsectionParametrized Solution Sets

Recall that an augmented matrix in row–echelon form represents a system of linear equations that can be solved via backsubstitution. However, when we transform an augmented matrix into row–echelon form, sometimes we will encounter a row of all zeros at the bottom of the matrix. Algebraically, this means that one of the rows corresponded with a linear equation that was equivalent to a linear combination of the previous equations.

SubsectionExercises

1

Compute the reduced row-echelon form of the matrix. \begin{equation*} \left[ \begin{array}{rrr} 4 \amp 8 \amp 4 \\ 1 \amp 0 \amp 4 \\ 5 \amp 16 \amp -4 \end{array} \right] \end{equation*}

Answer
2

The following augmented matrix is in RREF. Parametrize the solution set. Assume the variables are \(x_1, \ldots, x_4\text{,}\) and use parameters, \(s\) and \(t\text{.}\) \begin{equation*} \left[ \begin{array}{rrrr|r} 1 \amp -2 \amp 0 \amp 3 \amp -1 \\ 0 \amp 0 \amp 1 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array} \right] \end{equation*}

3

The following matrix corresponds with a linear system in the variables \(x,y\) and \(z\text{.}\) Use Gauss-Jordan elimination to determine if the system is consistent. If it is consistent find the solution set. \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 1 \amp 0 \amp 2 \\ 0 \amp 2 \amp 1 \amp 1 \\ -1 \amp 3 \amp 5 \amp -3 \end{array} \right] \end{equation*}

4

The following matrix corresponds with a linear system in the variables, \(x_1, x_2, \ldots, x_5\text{.}\) Use Gauss-Jordan elimination to determine if the system is consistent. If it is consistent find the solution set. \begin{equation*} \left[ \begin{array}{rrrrr|r} 4 \amp -3 \amp 2 \amp 5 \amp 7 \amp 3 \\ 0 \amp 1 \amp 2 \amp -1 \amp 1 \amp 0 \\ 2 \amp 1 \amp 6 \amp 3 \amp 9 \amp 0 \end{array} \right] \end{equation*}

5

The following matrix corresponds with a linear system in the variables, \(x, y\text{.}\) Use Gauss-Jordan elimination to determine if the system is consistent. If it is consistent find the solution set. \begin{equation*} \left[ \begin{array}{rr|r} 1 \amp 2 \amp 3 \\ 0 \amp 3 \amp 4 \\ -2 \amp 0 \amp 5 \\ 1 \amp 2 \amp 0 \end{array} \right] \end{equation*}

6

Use Gauss-Jordan elimination to determine if the following linear system is consistent. If it is consistent find the solution set. \begin{equation*} \left\{ \begin{alignedat}{6} x_1 \amp{}+{}\amp 2x_2 \amp{}+{}\amp 3x_3 \amp{}+{}\amp 6x_4 \amp{}={}\amp {-4} \\ \amp{}+{}\amp 3x_2 \amp{}+{}\amp 3x_3 \amp{}+{}\amp 6x_4 \amp{}={}\amp {-6} \\ -2x_1 \amp{}+{}\amp 2x_2 \amp{}~{}\amp \amp{}~{}\amp \amp{}={}\amp {-4} \\ x_1 \amp{}+{}\amp \amp{}~{}\amp x_3 \amp{}+{}\amp 2x_4 \amp{}={}\amp 0 \\ \end{alignedat} \right. \end{equation*}

7

Answer the following and explain your answer. Geometric reasoning is fine.

  1. Consider an under-determined system, i.e. a system with fewer equations than unknowns. Can such a system ever have a unique solution?

  2. Consider an over-determined system, i.e. a system with more equations than unknowns. Does such a system necessarily have no solution?