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8.5 Row Operations and Row Echelon Forms

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    1024
  • Applying row operations to reduce a matrix is a procedural skill that takes lots of writing, rewriting and careful arithmetic.  The payoff for being able to transform a matrix into a simplified form will become clear later.  For now, what does the simplified form mean for a matrix? 

    Heading Row Operations and Row Echelon Forms

    There are only three operations that are permitted to act on matrices.  They are the exact same operations that are permitted when solving a system of equations. 

    Add a multiple of one row to another row.

    Scale a row by multiplying through by a non-zero constant.

    Swap two rows.

    Using these three operations, your job is to simplify matrices into row echelon form.  Row echelon form must meet three requirements. 

    1. The leading coefficient of each row must be a one.

    2. All entries in a column below a leading one must be zero.

    3. All rows that just contain zeros are at the bottom of the matrix.

    Here are some examples of matrices in row echelon form:

    \(\left[\begin{array}{cc}1 & 14 \\ 0 & 1\end{array}\right],\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 4\end{array}\right],\left[\begin{array}{ccccc}1 & 2 & 3 & 5 & 6 \\ 0 & 0 & 1 & 4 & 7 \\ 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\)

    Reduced row echelon form also has one extra stipulation compared with row echelon form.

    4. Every leading coefficient of 1 must be the only non-zero element in that column.

    Here are some examples of matrices in reduced row echelon form:

    \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{lll}1 & 0 & 3 \\ 0 & 1 & 4\end{array}\right],\left[\begin{array}{ccccc}1 & 2 & 0 & 0 & 6 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\)

    Putting a matrix into reduced row echelon form is a result of performing Gauss-Jordan elimination.  The process illustrated in this concept is named after those two mathematicians.

    To put the a matrix into reduced row echelon form, use the row operations to change the matrix. Take the following matrix:

    \(\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\)

    In each step of reducing the matrix, only one of the three row operations will be used.  Specific shorthand will be introduced.

    \(\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\)

    \(3 R_{2} \rightarrow\left[\begin{array}{cc}3 & 7 \\ 6 & 15\end{array}\right]\)

    \(-2 R_{1}+R_{2} \rightarrow\left[\begin{array}{ll}3 & 7 \\ 0 & 1\end{array}\right]\)

    \(-7 R_{2}+R_{1} \rightarrow\left[\begin{array}{ll}3 & 0 \\ 0 & 1\end{array}\right]\)

    \(\frac{1}{3} R \rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

    Note that the \(3 R_{2}\) indicates that the second row of the matrix is scaled by a factor of \(3 .\) The \(-2 R_{1}+R_{2}\) before the third matrix indicates that the second row has two times the first row subtracted from it.

    Row reducing a \(2 \times 2\) matrix to become the identity matrix illustrates the fact that the rows of the original matrix are linearly independent.

    Examples

    Example 1

    Earlier, you were asked what it means for a matrix to be simplified. There are two forms of a matrix that are most simplified.  The most important is reduced row echelon form that follows the four stipulations from the guidance section.  An example of a matrix in reduced row echelon form is:

    \(\left[\begin{array}{ccccc}1 & 0 & 0 & 2 & 43 \\ 0 & 1 & 0 & 2 & 3 \\ 0 & 0 & 1 & 98 & 5\end{array}\right]\)

    Example 2

    Put the following matrix into reduced row echelon form.

    \(\left[\begin{array}{lll}2 & 4 & 0 \\ 0 & 3 & 1 \\ 1 & 2 & 4\end{array}\right]\)

     

    \(\left[\begin{array}{lll}2 & 4 & 0 \\ 0 & 3 & 1 \\ 1 & 2 & 4\end{array}\right]\)

    \(R_{1} \cdot-\frac{1}{2}+R_{3} \rightarrow\left[\begin{array}{lll}2 & 4 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 4\end{array}\right]\)

    \(R_{3} \div 4 \rightarrow\left[\begin{array}{lll}2 & 4 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 1\end{array}\right]\)

    \(R_{1} \div 2, R_{2} \div 3 \rightarrow\left[\begin{array}{lll}1 & 2 & 0 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1\end{array}\right]\)

    \(R_{3} \cdot-\frac{1}{3}+R_{2} \rightarrow\left[\begin{array}{lll}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

    \(R_{2} \cdot-2+R_{1} \rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

    Note that two operations were used in the fourth row to produce the fourth matrix.  This is acceptable when the operations do not interfere or interact with each other.

    Again, row reducing a \(3 \times 3\) matrix to become the identity matrix is just an exercise that illustrates the fact that the rows were linearly independent.

    Example 3

    Reduce the following matrix to reduced row echelon form.

    \(\left[\begin{array}{lll}0 & 4 & 5 \\ 2 & 6 & 8\end{array}\right]\)

     

    \(\left[\begin{array}{lll}0 & 4 & 5 \\ 2 & 6 & 8\end{array}\right]\)

    Switch Rows \(\rightarrow\left[\begin{array}{lll}2 & 6 & 8 \\ 0 & 4 & 5\end{array}\right]\)

    \(R_{1} \div 2 \rightarrow\left[\begin{array}{lll}1 & 3 & 4 \\ 0 & 4 & 5\end{array}\right]\)

    \(R_{2} \cdot \frac{1}{4} \rightarrow\left[\begin{array}{lll}1 & 3 & 4 \\ 0 & 1 & \frac{5}{4}\end{array}\right]\)

    \(R_{2} \cdot-3+R_{1} \rightarrow\left[\begin{array}{lll}1 & 0 & \frac{1}{4} \\ 0 & 1 & \frac{5}{4}\end{array}\right]\)

     

    Example 4

    Reduce the following matrix to row echelon form.

    \(\left[\begin{array}{cc}3 & 6 \\ 2 & 4 \\ 5 & 17\end{array}\right]\)

     

    \(\left[\begin{array}{cc}3 & 6 \\ 2 & 4 \\ 5 & 17\end{array}\right]\)

    \(R_{1} \div 3, R_{2} \div 2 \rightarrow\left[\begin{array}{cc}1 & 2 \\ 1 & 2 \\ 5 & 17\end{array}\right]\)

    \(R_{1} \cdot-1+R_{2} \rightarrow\left[\begin{array}{cc}1 & 2 \\ 0 & 0 \\ 5 & 17\end{array}\right]\)

    \(R_{1} \cdot-5+R_{3} \rightarrow\left[\begin{array}{ll}1 & 2 \\ 0 & 0 \\ 0 & 7\end{array}\right]\)

    Switch \(R_{2}\) and \(R_{3} \rightarrow\left[\begin{array}{ll}1 & 2 \\ 0 & 7 \\ 0 & 0\end{array}\right]\)

    \(R_{2} \div 7 \rightarrow\left[\begin{array}{ll}1 & 2 \\ 0 & 1 \\ 0 & 0\end{array}\right]\)

    \(R_{2} \cdot-2+R_{1} \rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1 \\ 0 & 0\end{array}\right]\)

    Example 5

    Reduce the following matrix to reduced row echelon form.

    \(\left[\begin{array}{cccc}3 & 4 & 1 & 0 \\ 5 & -1 & 0 & 1\end{array}\right]\)

     

    \(\left[\begin{array}{cccc}3 & 4 & 1 & 0 \\ 5 & -1 & 0 & 1\end{array}\right]\)

    \(R_{1} \cdot 5 \rightarrow\left[\begin{array}{cccc}15 & 20 & 5 & 0 \\ 5 & -1 & 0 & 1\end{array}\right]\)

    \(R_{2} \cdot 3 \rightarrow\left[\begin{array}{cccc}15 & 20 & 5 & 0 \\ 15 & -3 & 0 & 3\end{array}\right]\)

    \(R_{2}-R_{1} \rightarrow\left[\begin{array}{cccc}15 & 20 & 5 & 0 \\ 0 & -23 & -5 & -3\end{array}\right]\)

    \(R_{1} \div 15 \rightarrow\left[\begin{array}{cccc}1 & \frac{4}{3} & \frac{1}{3} & 0 \\ 0 & -23 & -5 & 3\end{array}\right]\)

    \(R_{2} \div-23 \rightarrow\left[\begin{array}{cccc}1 & \frac{4}{3} & \frac{1}{3} & 0 \\ 0 & 1 & \frac{5}{23} & -\frac{3}{23}\end{array}\right]\)

    \(R_{2} \cdot-\frac{4}{3}+R_{1} \rightarrow\left[\begin{array}{cccc}1 & 0 & \frac{1}{23} & \frac{4}{23} \\ 0 & 1 & \frac{5}{23} & -\frac{3}{23}\end{array}\right]\)

    Notice how fractions were avoided until the final step.  Adding and subtracting large numbers in a matrix is easier to handle than adding and subtracting small numbers because then you don’t need to find a common denominator.

    Review

    1. Give an example of a matrix in row echelon form.

    2. Give an example of a matrix in reduced row echelon form.

    3. What are the three row operations you are allowed to perform when reducing a matrix?

    4. If a square matrix reduces to the identity matrix, what does that mean about the rows of the original matrix?

    Use the following matrix for \(5-6\)

    \(A=\left[\begin{array}{ccc}-3 & -4 & -12 \\ 4 & 4 & 12 \\ -11 & -12 & -35\end{array}\right]\)

    5. Reduce matrix \(A\) to row echelon form.

    6. Reduce matrix \(A\) to reduced row echelon form. Are the rows of matrix \(A\) linearly independent?

    Use the following matrix for \(7-8\).

    \(B=\left[\begin{array}{ccc}3 & -4 & 8 \\ 9 & 0 & 1 \\ 0 & 1 & -2\end{array}\right]\)

    7. Reduce matrix \(B\) to row echelon form.

    8. Reduce matrix \(B\) to reduced row echelon form. Are the rows of matrix \(B\) linearly independent?

    Use the following matrix for \(9-10 .\)

    \(C=\left[\begin{array}{cccc}0 & 0 & -1 & -1 \\ 3 & 6 & -3 & 1 \\ 6 & 12 & -7 & 0\end{array}\right]\)

    9. Reduce matrix \(C\) to row echelon form.

    10. Reduce matrix \(C\) to reduced row echelon form. Are the rows of matrix \(C\) linearly independent?

    Use the following matrix for \(11-12\).

    \(D=\left[\begin{array}{ll}1 & 1 \\ 3 & 4 \\ 2 & 3\end{array}\right]\)

    11. Reduce matrix \(D\) to row echelon form.

    12. Reduce matrix \(D\) to reduced row echelon form. Are the rows of matrix \(D\) linearly independent?

    Use the following matrix for \(13-14\)

    \(E=\left[\begin{array}{ccc}-5 & -6 & -12 \\ -1 & -1 & -2 \\ 2 & 2 & 4\end{array}\right]\)

    13. Reduce matrix \(E\) to row echelon form.

    14. Reduce matrix \(E\) to reduced row echelon form. Are the rows of matrix \(E\) linearly independent?

    Use the following matrix for \(15-16 .\)

    \(F=\left[\begin{array}{ccc}-23 & 6 & 3 \\ 2 & -\frac{1}{2} & 0 \\ -8 & 2 & 1\end{array}\right]\)

    15. Reduce matrix \(F\) to row echelon form.

    16. Reduce matrix \(F\) to reduced row echelon form. Are the rows of matrix \(F\) linearly independent?

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