Row Reduction

A system of equations can be solved more easily using linear algebra.

The Essentials

Three row operations exist to reduce a matrix: scaling rows, swapping rows, and adding rows to other rows. Systems of equations can be solved by converting the equations into a matrix and using row operations to reduce the matrix to triangular form.

Example

Consider this system of equations:

\[ \begin{cases} 3x+4y-2z-w=11\\ x-y+z-w=-4\\ 2x+5y+w=14\\ x+y-2z=5 \end{cases} \]

This system of equations can be represented by an augmented matrix:

\[ \left[ \begin{array}{cccc|c} 3 & 4 & -2 & -1 & 11\\ 1 & -1 & 1 & -1 & -4\\ 2 & 5 & 0 & 1 & 14\\ 1 & 1 & -2 & 0 & 5 \end{array} \right] \]

Now we can do row operations:
\( R_1=R_1-3R_2 \)
\( R_3=R_3-2R_2 \)
\( R_4=R_4-R_2 \)

\[ \left[ \begin{array}{cccc|c} 0 & 7 & -5 & 2 & 23\\ 1 & -1 & 1 & -1 & -4\\ 0 & 7 & -2 & 3 & 22\\ 0 & 2 & -3 & 1 & 9 \end{array} \right] \]

\( R_3=R_3-R_1 \)
\( R_4=7R_4 \)

\[ \left[ \begin{array}{cccc|c} 0 & 7 & -5 & 2 & 23\\ 1 & -1 & 1 & -1 & -4\\ 0 & 0 & 3 & 1 & -1\\ 0 & 14 & 21 & 7 & 63 \end{array} \right] \]

\( R_4=R_4-2R_1 \)

\[ \left[ \begin{array}{cccc|c} 0 & 7 & -5 & 2 & 23\\ 1 & -1 & 1 & -1 & -4\\ 0 & 0 & 3 & 1 & -1\\ 0 & 0 & -11 & 3 & 17 \end{array} \right] \]

\( R_1\leftrightarrow R_2 \)
\( R_4=3R_4 \)

\[ \left[\begin{array}{cccc|c} 1 & -1 & 1 & -1 & -4\\ 0 & 7 & -5 & 2 & 23\\ 0 & 0 & 3 & 1 & -1\\ 0 & 0 & -33 & 9 & 51 \end{array}\right] \]

\( R_4 = R_4 + 11R_3 \)

\[ \left[\begin{array}{cccc|c} 1 & -1 & 1 & -1 & -4\\ 0 & 7 & -5 & 2 & 23\\ 0 & 0 & 3 & 1 & -1\\ 0 & 0 & 0 & 20 & 40 \end{array} \right] \]

Now that the matrix is in triangular form, we can convert the matrix back into a system of equations:

\[ \begin{cases} x-y+z-w=-4\\ 7y-5z+2w=23\\ 3z+w=-1\\ 20z=40 \end{cases} \]

Starting from the bottom and working our way up, we can find each variable in the equation:

\[ \begin{cases} x=1\\ y=2\\ z=-1\\ w=2 \end{cases} \]

Practice

Solve this system of equations using row operations:

\[ \begin{cases} x+y+z=1\\ 5x-2y-6z=3\\ 2x+3y-5z=-23 \end{cases} \]

Solution:

\[ \begin{cases} x=1\\ y=-5\\ z=2 \end{cases} \]