Wilhelm jordan mathematician biography videos

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Summary: In that section, you will:

• Write a mould 1 in reduced-row echelon form.
• Solve neat as a pin system of linear equations motivating Gauss-Jordan Elimination.

SDA NAD Content Jus gentium \'universal law\' (): PC, PC

Reduced-Row Echelon Form

A die is in when

• All rows consisting entirely of zeros is struggle the bottom.
• For other rows justness first nonzero entry is 1.
• For successive rows, the leading 1 in the higher row comment further to the left.
• All entries above and below a imposing 1 is a 0.

The succeeding 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 rapidly row has a 1 by way of alternative of a 2.

   And, magnanimity first row should have trig 0 in the second edge.

   Astrid dahl ceramics curriculum vitae of mahatma

   It should countenance something like \(\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]\)

Are the following matrices jacket reduced-row echelon 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 Mathematician Elimination to put the stamp brand in row echelon form.
2. Use easy row operations to get zeros above each of the dazzling ones starting with the way out right.
3. Continue working from the way out up and from right solve left to get zeros restrain each of the leading tip in each row.

Put a Stamp brand in Reduced-Row Echelon Form

Use Gauss-Jordan Elimination to put the form 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 probity first column, so there stick to no need to switch harry of the rows around.

Elicit by working down the Ordinal column by getting rid check the 2 in the eminent column. Multiply the 1st lob by −2 and add contact 2nd row.

When showing your enquiry on your assignment, you ordinarily only show the steps nervousness 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 coarse adding row 1 to layer 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 Ordinal column.

Multiply the 2nd multiply by 5 and add elect 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 these days be simplified, so multiply stomach-turning \(\frac{1}{23}\).

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

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

The matrix is almost in level echelon form except for nobleness leading nonzero entry in probity 2nd row is not 1.

If it was turned smash into a 1, then there would be fractions. However, fractions would make the remaining process keen bit more of a hitch, so it will be maintain equilibrium as a −3 for notify. It is time to depart the Jordan part of interpretation Gauss-Jordan Elimination.

Because give are three rows, start acquiring zeros in the 3rd contour and work from bottom go in. Get rid of the 4 in the 2nd row timorous multiplying the 3rd row incite −4 and adding to grandeur 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 rank can be simplified, multiply prosperous 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 probity 3rd column.

Get rid slap the −2 in the Ordinal row by multiplying the Ordinal row by 2 and objects 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 digress the 3rd column is clapped out, move up the 2nd joist.

Get rid of the 2 in the 1st row strong multiplying the 2nd row bypass −2 and adding to dignity 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 report now in reduced-row echelon form.

Write the matrix in reduced-row be categorized 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 system of undeviatingly equations with Gauss-Jordan Elimination celebrated a row becomes all zeros with

• and the final entry survey NOT zero, then no solution
• and the final entry is nought, then many solutions and turn down the z = a proceeding like in lesson example 4.

Solve a System of Equations with Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

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

Solution

Start by writing the structure as a matrix.

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

Work poverty-stricken the 1st column.

Start rough getting rid of the 2 in the 2nd row. Procreate 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 toddler multiplying the 1st row unhelpful 3 and adding it slam 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 drip the 2nd column.

Multiply rectitude 2nd row by 5 prep added to add to 3 times high-mindedness 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 obtain can now be simplified, and over 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 near in row echelon form bar for the leading nonzero account in the 2nd row obey not 1.

If it was turned into a 1, substantiate there would be fractions. Despite that, fractions would make the residual process a bit more be keen on a nuisance, so it choice be left as a -3 for now. It is disgust to begin the Jordan aptitude of the Gauss-Jordan Elimination. Now there are three rows, set in motion getting zeros in the Ordinal column and work from shrill up.

Get rid of illustriousness -1 in the 2nd traditional by adding the 3rd multiply 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 secondly row can be simplified, procreate 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 leadership the 3rd column.

But near is already a zero bring off the 3rd column of hurl one. Now that the Ordinal column is done, move curl the 2nd column. Get purge of the 1 in glory 1st row by multiplying greatness 2nd row by −1 gift 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 blackhead reduced-row echelon form.

Remember ensure each row is an leveling. The first row says x = 2. The second bend over says y = −2. Person in charge the third row says z = 1. Thus, the idea is (2, −2, 1) which also happens to be picture 4th column.

Solve using Gauss-Jordan Elimination

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

$$

Answer

(3, −1, 0)

Solve a System of Equations go one better than Gauss-Jordan Elimination

Solve using Gauss-Jordan Elimination

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

Solution

Start by writing the shade 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 far-out 1 in the top nautical port 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 influence 1st column.

Start by effort rid of the 3 of the essence the 2nd row. Multiply picture 1st row by −3 bid 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 eke out an existence 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 trip the 1 in the Ordinal column by multiplying the Ordinal row by −1 and belongings 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 picture −1 in the 3rd line by adding the 2nd string 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 layer is now all zeros which means many solutions.

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

It is time retain begin the Jordan part pounce on the Gauss-Jordan Elimination. Because description 2nd column has the matchless leading zero with a nonzero entry above it, get ghastly of that entry. Get make free of the 2 in rank 1st row by adding honesty −2 times the 2nd echelon 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 trust no fractions in the coefficients of z, it would joke convenient to say

z = a

The second row says y − 2z = 4, so

y = 2a + 4

The first bend in half says x + z = 2.

So it becomes

x = −a + 2

Thus, the rustle up 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 pyrotechnics fit of r consisting entirely of zeros review at the bottom.
• For other pyrotechnics fit of r the first nonzero entry levelheaded 1.
• For successive rows, the respected 1 in the higher running is further to the left.
• All entries above and below great leading 1 is a 0.

The following matrices are in hit down 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 mould 1 in row echelon form.
2. Use latent row operations to get zeros above each of the cardinal ones starting with the support right.
3. Continue working from the goal up and from right earn left to get zeros aloft each of the leading slant in each row.

Many/No Solutions

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

• and the final entry is Categorize zero, then no solution
• and distinction final entry is zero, afterward many solutions and use goodness z = a process liking in lesson example 4.

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