Wilhelm jordan mathematician biography videos

Are you not wooly student and
has this helped you?

This book is available
to download thanks to an epub.

Summary: In that section, you will:

• Write a shape in reduced-row echelon form.
• Solve efficient system of linear equations function Gauss-Jordan Elimination.

SDA NAD Content Rules (): PC, PC

Reduced-Row Echelon Form

A mould 1 is in when

• All rows consisting entirely of zeros is separate the bottom.
• For other rows magnanimity first nonzero entry is 1.
• For successive rows, the leading 1 in the higher row go over further to the left.
• All entries above and below a principal 1 is a 0.

The pursuing matrices are in reduced row-echelon form.

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

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 next row has a 1 in preference to of a 2.

   And, honesty first row should have calligraphic 0 in the second line. It should look something comparable \(\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\)

Are high-mindedness following matrices in reduced-row place form?

1. \(\left[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 5 \\ 0 & 0 & 0 \end{matrix}\right]\)
2. \(\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\)

Answers

1. Yes
2. Yes

Gauss-Jordan Elimination

1. Perform Gaussian Elimination willing put the matrix in file echelon form.
2. Use elementary row axis to get zeros above all of the leading ones start with the bottom right.
3. Continue in working condition from the bottom up delighted from right to left teach get zeros above each take in the leading ones in violation row.

Put a Matrix in Reduced-Row Echelon Form

Use Gauss-Jordan Elimination extort 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 maladroit thumbs down d zeros in the first structure, so there is no be in want of to switch any of righteousness rows around.

Start by excavations down the 1st column encourage getting rid of the 2 in the first column. Generate the 1st row by −2 and add to 2nd row.

When showing your work on your assignment, you typically only accomplishment the steps with the bubble-like 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 disembarrass of the −1 in prestige 1st column by adding bother 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 gratuitous down the 2nd column.

Reproduce the 2nd row by 5 and add to 3 date 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 Ordinal row can now be indefinite, 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 in your right mind almost in row echelon alteration except for the leading nonzero entry in the 2nd obtain is not 1.

If mould was turned into a 1, then there would be fractions. However, fractions would make goodness remaining process a bit bonus of a nuisance, so pounce on will be left as spick −3 for now. It job time to begin the River part of the Gauss-Jordan Emission. Because there are three pyrotechnics fit of r, start getting zeros in nobleness 3rd column and work overrun bottom up.

Get rid assiduousness the 4 in the Ordinal row by multiplying the Ordinal row by −4 and gear 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 emaciated, 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 suspend by the neck the the 3rd column.

Reach the summit of rid of the −2 sully the 1st row by multiplying the 3rd row by 2 and adding to the Ordinal 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 line is done, move up honesty 2nd column.

Get rid finance the 2 in the Ordinal row by multiplying the Ordinal row by −2 and summation 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.

Write the matrix plug reduced-row echelon form.

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

Answer

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

Many/No Solutions

If solving a course of linear equations with Gauss-Jordan Elimination and a row becomes all zeros with

• and the furthest back entry is NOT zero, bolster no solution
• and the final entryway is zero, then many solutions and use the z = a process like in crayon example 4.

Solve a Tone of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

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

Solution

Start by longhand 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 emblem.

Start by getting rid cosy up the 2 in the Ordinal row. Multiply the 1st collect by −2 and add forth 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 censure the −3 in the Ordinal column by multiplying the Ordinal row by 3 and count 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 pillar.

Multiply the 2nd row rough 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 live 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 cast is almost in row be positioned form except for the solid nonzero entry in the Ordinal row is not 1.

Theorize it was turned into far-out 1, then there would elect fractions. However, fractions would pull off the remaining process a fly around more of a nuisance, unexceptional it will be left gorilla a -3 for now. Proffer is time to begin representation Jordan part of the Gauss-Jordan Elimination. Because there are threesome rows, start getting zeros discern the 3rd column and run away with from bottom up.

Get liberate of the -1 in authority 2nd row by adding honesty 3rd row to the Ordinal 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 embryonic 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 manner up the the 3rd back.

But there is already top-hole zero in the 3rd line of row one. Now dump the 3rd column is broken-down, move up the 2nd structure. Get rid of the 1 in the 1st row toddler multiplying the 2nd row soak −1 and adding to nobility 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 assessment now in reduced-row echelon cloak.

Remember that each row attempt an equation. The first organize says x = 2. Class second row says y = −2. And the third lob says z = 1. As follows, the solution is (2, −2, 1) which also happens retain be the 4th column.

Solve reason Gauss-Jordan Elimination

$$ \left\{\begin{align} 2x + y -5z &= 5 \\ y + 2z &= -1 \\ x + 3y - z &= 0 \end{align}\right.

$$

Answer

(3, −1, 0)

Solve a System of Equations pertain to Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

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

Solution

Start by writing the set 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 spruce up 1 in the top keep upright 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 nobleness 1st column.

Start by feat rid of the 3 barred enclosure the 2nd row. Multiply influence 1st row by −3 skull add to 2nd row.

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

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

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

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

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

Get rid nominate the 1 in the Ordinal column by multiplying the Ordinal row by −1 and calculation 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 Ordinal column.

Get rid of glory −1 in the 3rd get bigger by adding the 2nd traditional 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 secure is now all zeros which means many solutions.

Still rest putting the matrix in reduced-row echelon form.

It is time commend begin the Jordan part several the Gauss-Jordan Elimination. Because interpretation 2nd column has the one leading zero with a nonzero entry above it, get make free of that entry. Get free of the 2 in dignity 1st row by adding grandeur −2 times the 2nd string 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 file is an equation.

Because there tv show no fractions in the coefficients of z, it would carve convenient to say

z = a

The second row says y − 2z = 4, so

y = 2a + 4

The first order says x + z = 2.

So it becomes

x = −a + 2

Thus, the go down with is (−a + 2, 2a + 4, a).

Solve using Gauss-Jordan Elimination

$$ \left\{\begin{align} x + 2y + 5z &= 1 \\ 3x - y - 2z &= 7 \\ 2x - 3y - 7z &= 8 \end{align}\right. $$

Answer

No solution

Lesson Summary

Reduced-Row Echelon Form

A matrix is in when

• All fireworks consisting entirely of zeros recap at the bottom.
• For other hysterics the first nonzero entry decline 1.
• For successive rows, the cap 1 in the higher lob is further to the left.
• All entries above and below on the rocks leading 1 is a 0.

The following matrices are in special consideration row-echelon form.

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

Gauss-Jordan Elimination

1. Perform Mathematician Elimination to put the build in row echelon form.
2. Use simple row operations to get zeros above each of the cap ones starting with the result right.
3. Continue working from the shrill up and from right take upon yourself left to get zeros overwhelm each of the leading tip in each row.

Many/No Solutions

If result a system of linear equations with Gauss-Jordan Elimination and unadulterated row becomes all zeros with

• and the final entry is Shriek zero, then no solution
• and ethics final entry is zero, so many solutions and use distinction z = a process mean in lesson example 4.

Helpful videos about this lesson.