9-02 Gaussian Elimination

Identify Reduced-Row Echelon Form

Are the following matrices in reduced-row echelon form?

1. \(\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}\right]\)
2. \(\left[\begin{matrix} 1 & 2 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\)

Solution

1. Yes
2. No, the second row has a 1 instead of a 2. And, the first row should have a 0 in the second column. It should look something like \(\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\)

Put a Matrix in Reduced-Row Echelon Form

Use Gauss-Jordan Elimination to put the matrix in reduced-row echelon form.

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 2 & 1 & 0 & 3 \\ -1 & 3 & 3 & 4 \end{matrix}\right] $$

Solution

There are no zeros in the first column, so there is no need to switch any of the rows around. Start by working down the 1st column by getting rid of the 2 in the first column. Multiply the 1st row by −2 and add to 2nd row.

When showing your work on your assignment, you typically only show the steps with the red numbers in them.

$$ \begin{matrix} \swarrow ×\left(-2\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{blue}{2} & 1 & 0 & 3 \\ -1 & 3 & 3 & 4 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{-3} & \color{red}{4} & \color{red}{15} \\ -1 & 3 & 3 & 4 \end{matrix}\right] $$

Get rid of the −1 in the 1st column by adding row 1 to row 3.

$$ \begin{matrix} \swarrow \\ \downarrow \quad \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{blue}{-1} & 3 & 3 & 4 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{5} & \color{red}{1} & \color{red}{-2} \end{matrix}\right] $$

Now work down the 2nd column. Multiply the 2nd row by 5 and add to 3 times the 3rd row.

$$ \begin{matrix} \quad \\ \swarrow ×5 \\ +\rightarrow ×3 \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ 0 & \color{blue}{5} & 1 & -2 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{0} & \color{red}{23} & \color{red}{69} \end{matrix}\right] $$

The 3rd row can now be simplified, so multiply by \(\frac{1}{23}\).

$$ \begin{matrix} \quad \\ \quad \\ ×\tfrac{1}{23} \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ 0 & 0 & 23 & 69 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 4 & 15 \\ \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{red}{3} \end{matrix}\right] $$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd row is not 1. If it was turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so it will be left as a −3 for now. It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because there are three rows, start getting zeros in the 3rd column and work from bottom up. Get rid of the 4 in the 2nd row by multiplying the 3rd row by −4 and adding to the 2nd row.

$$ \begin{matrix} \quad \\ +\rightarrow \qquad \\ \nwarrow ×\left(-4\right) \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & \color{blue}{4} & 15 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{-3} & \color{red}{0} & \color{red}{3} \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

The second row can be simplified, multiply it by \(-\frac{1}{3}\).

$$ \begin{matrix} \quad \\ ×\left(-\tfrac{1}{3}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -2 & -6 \\ 0 & -3 & 0 & 3 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -2 & -6 \\ \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{-1} \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

Continue working up the the 3rd column. Get rid of the −2 in the 1st row by multiplying the 3rd row by 2 and adding to the 1st row.

$$ \begin{matrix} +\rightarrow\quad \\ \uparrow\qquad \\ \nwarrow ×2 \end{matrix} \left[\begin{matrix} 1 & 2 & \color{blue}{-2} & -6 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{2} & \color{red}{0} & \color{red}{0} \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

Now that the 3rd column is done, move up the 2nd column. Get rid of the 2 in the 1st row by multiplying the 2nd row by −2 and adding to the 1st row.

$$ \begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-2\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{2} & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{2} \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{matrix}\right] $$

The matrix is now in reduced-row echelon form.

Solve a System of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

$$ \left\{\begin{align} x + y \qquad &= 0 \\ 2x - y - z &= 5 \\ -3x + 2y + z &= -9 \end{align}\right. $$

Solution

Start by writing the system as a matrix.

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 2 & -1 & -1 & 5 \\ -3 & 2 & 1 & -9 \end{matrix}\right] $$

Work down the 1st column. Start by getting rid of the 2 in the 2nd row. Multiply the 1st row by −2 and add to 2nd row.

$$ \begin{matrix} \swarrow ×\left(-2\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{blue}{2} & -1 & -1 & 5 \\ -3 & 2 & 1 & -9 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-3} & \color{red}{-1} & \color{red}{5} \\ -3 & 2 & 1 & -9 \end{matrix}\right] $$

Get rid of the −3 in the 1st column by multiplying the 1st row by 3 and adding it to row 3.

$$ \begin{matrix} \swarrow ×3 \\ \downarrow \quad \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{blue}{-3} & 2 & 1 & -9 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{5} & \color{red}{1} & \color{red}{-9} \end{matrix}\right] $$

Now work down the 2nd column. Multiply the 2nd row by 5 and add to 3 times the 3rd row.

$$ \begin{matrix} \quad \\ \swarrow ×5 \\ +\rightarrow ×3 \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ 0 & \color{blue}{5} & 1 & -9 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{0} & \color{red}{-2} & \color{red}{-2} \end{matrix}\right] $$

The 3rd row can now be simplified, so multiply by \(-\frac{1}{2}\).

$$ \begin{matrix} \quad \\ \quad \\ ×\left(-\tfrac{1}{2}\right) \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ 0 & 0 & -2 & -2 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & -1 & 5 \\ \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{red}{1} \end{matrix}\right] $$

The matrix is almost in row echelon form except for the leading nonzero entry in the 2nd row is not 1. If it was turned into a 1, then there would be fractions. However, fractions would make the remaining process a bit more of a nuisance, so it will be left as a -3 for now. It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because there are three rows, start getting zeros in the 3rd column and work from bottom up. Get rid of the -1 in the 2nd row by adding the 3rd row to the 2nd row.

$$ \begin{matrix} \quad \\ +\rightarrow \quad \\ \nwarrow \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & \color{blue}{-1} & 5 \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{-3} & \color{red}{0} & \color{red}{6} \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

The second row can be simplified, multiply it by \(-\frac{1}{3}\).

$$ \begin{matrix} \quad \\ ×\left(-\tfrac{1}{3}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & -3 & 0 & 6 \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 1 & 0 & 0 \\ \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{-2} \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

Continue working up the the 3rd column. But there is already a zero in the 3rd column of row one. Now that the 3rd column is done, move up the 2nd column. Get rid of the 1 in the 1st row by multiplying the 2nd row by −1 and adding to the 1st row.

$$ \begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-1\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{1} & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{2} \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{matrix}\right] $$

The matrix is now in reduced-row echelon form. Remember that each row is an equation. The first row says x = 2. The second row says y = −2. And the third row says z = 1. Thus, the solution is (2, −2, 1) which also happens to be the 4th column.

Solve a System of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

$$ \left\{\begin{align} 3x + y + z &= 10 \\ x + 2y - 3z &= 10 \\ x + y - z &= 6 \end{align}\right. $$

Solution

Start by writing the system as a matrix.

$$ \left[\begin{matrix} 3 & 1 & 1 & 10 \\ 1 & 2 & -3 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

It would be nice to have a 1 in the top left entry, so switch rows 1 and 2.

$$ \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 3 & 1 & 1 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

Work down the 1st column. Start by getting rid of the 3 in the 2nd row. Multiply the 1st row by −3 and add to 2nd row.

$$ \begin{matrix} \swarrow ×\left(-3\right) \\ +\rightarrow \qquad \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{blue}{3} & 1 & 1 & 10 \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{red}{0} & \color{red}{-5} & \color{red}{10} & \color{red}{-20} \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

Notice the 2nd row can be simplified by multiplying by \(-\frac{1}{5}\).

$$ \begin{matrix} \quad \\ ×\left(-\tfrac{1}{5}\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & -5 & 10 & -20 \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -3 & 10 \\ \color{red}{0} & \color{red}{1} & \color{red}{-2} & \color{red}{4} \\ 1 & 1 & -1 & 6 \end{matrix}\right] $$

Get rid of the 1 in the 1st column by multiplying the 1st row by −1 and adding it to row 3.

$$ \begin{matrix} \swarrow ×\left(-1\right) \\ \downarrow \qquad \\ +\rightarrow \quad \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{blue}{1} & 1 & -1 & 6 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{red}{0} & \color{red}{-1} & \color{red}{2} & \color{red}{-4} \end{matrix}\right] $$

Now work down the 2nd column. Get rid of the −1 in the 3rd row by adding the 2nd row to the 3rd row.

$$ \begin{matrix} \quad \\ \swarrow \\ +\rightarrow \end{matrix} \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ 0 & \color{blue}{-1} & 2 & -4 \end{matrix}\right] $$

$$ \left[\begin{matrix} 1 & 2 & -3 & 10 \\ 0 & 1 & -2 & 4 \\ \color{red}{0} & \color{red}{0} & \color{red}{0} & \color{red}{0} \end{matrix}\right] $$

The 3rd row is now all zeros which means many solutions. Still finish putting the matrix in reduced-row echelon form.

It is time to begin the Jordan part of the Gauss-Jordan Elimination. Because the 2nd column has the only leading zero with a nonzero entry above it, get rid of that entry. Get rid of the 2 in the 1st row by adding the −2 times the 2nd row to the 1st row.

$$ \begin{matrix} +\rightarrow \qquad \\ \nwarrow ×\left(-2\right) \\ \quad \end{matrix} \left[\begin{matrix} 1 & \color{blue}{2} & -3 & 10 \\ 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 \end{matrix}\right] $$

$$ \left[\begin{matrix} \color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{2} \\ 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 \end{matrix}\right] $$

The matrix is now in reduced-row echelon form.

Remember that each row is an equation.

Because there are no fractions in the coefficients of z, it would be convenient to say

z = a

The second row says y − 2z = 4, so

y = 2a + 4

The first row says x + z = 2. So it becomes

x = −a + 2

Thus, the solution is (−a + 2, 2a + 4, a).