Gaussian Elimination And Gauss Jordan Method English Language Essay

The following are well-known methods of work outing a system of additive equations:

Gaussian Elimination Method

Gauss-Seidel Iterative Method

Gaussian Elimination

The simplest method for work outing a additive system is Gaussian riddance, which uses three types of simple row operations:

Multiplying a row by a non-zero invariable ( kEij )

Adding a multiple of one row to another ( Eij + kEkj )

Exchanging two rows ( Eij & lt ; — & gt ; kEkj )

Each row operation corresponds to a measure in the solution of the system of equationsA where the equations are combined together. It is of import to observe that none of those operations changes the solution. There are two parts to this method: riddance and back-substitution. The intent of the procedure of riddance is to extinguish the matrix entries below the chief diagonal, utilizing row operations, to obtain a upper triangular matrix with the augmented column. Then, you will be able to continue with back-substitution to happen the values of the terra incognitas.

Introduction: Gaussian riddance method is an exact method which solves a given system of equations in n terra incognitas by transforming the coff. Matrix, into an upper triangular matrix and so work out for the terra incognitas by back permutation.

See a system of n equations in n terra incognitas

a11x1+a12x2+a13x3+aˆ¦aˆ¦aˆ¦.+a1nxn = an+1

a21x1+a22x2+a23x3+aˆ¦aˆ¦aˆ¦.+a2nxn=a2, n+1


an1x1+an2x2+an3x3+aˆ¦aˆ¦aˆ¦aˆ¦.annxn=an, n+1

extinguish the unknown x1 from the ( n-1 ) equations viz. 2,3, aˆ¦aˆ¦ ( n-1 ) , n by substracting the multiply ai1/ai1 of the first equation from the ith equation from the ith equation, for i=2,3,4aˆ¦aˆ¦ , n.Now eliminate x2 from the ( n-2 ) equations of the end point system. By this process, we arrive at a derived system as follows:

a11x1+a12x2+aˆ¦aˆ¦..+a1nxn=al, n+1

a22+x2+aˆ¦aˆ¦aˆ¦aˆ¦..a2nxn=a2, n+1

a33x3+aˆ¦aˆ¦aˆ¦aˆ¦..+a3nxn=a3, n+1


ann ( n-1 ) xn =an, n+1 ( n-1 )

Gaussian riddance is a method for work outing matrix equation of the signifier:

( 1 )

To execute Gaussian riddance get downing with the system of equations

( 2 )

compose the “ augmented matrix equation ”

( 3 )

Here, the column vector in the variables is carried along for labeling the matrix rows. Now, execute simple row operations to set the augmented matrix into the upper triangular signifier

( 4 )

Solve the equation of the Thursday row for, so replace back into the equation of the st row to obtain a solution for, etc. , harmonizing to the expression

( 5 )

Performs a version of Gaussian riddance, with the equation being solved by Gaussian riddance

L U decomposition of a matrix is often used as portion of a Gaussian riddance procedure for work outing a matrix equation.

A matrix that has undergone Gaussian riddance is said to be in echelon signifier.

Example 1, see the matrix equation:

( 6 )

In augmented signifier, this becomes

( 7 )

Switch overing the first and 3rd rows ( without exchanging the elements in the right-hand column vector ) gives

( 8 )

Subtracting 9 times the first row from the 3rd row gives

( 9 )

Subtracting 4 times the first row from the 2nd row gives

( 10 )

Finally, adding times the 2nd row to the 3rd row gives

( 11 )

Restoring the transformed matrix equation gives

( 12 )

which can be solved instantly to give, back-substituting to obtain ( which really follows trivially in this illustration ) , and so once more back-substituting to happen

Solving three-variable, three-equation additive systems is more hard, at least ab initio, than work outing the two-variable systems, because the calculations involved are more mussy. You will necessitate to be really orderly in your working, and you should be after to utilize tonss of scratch paper. The method for work outing these systems is an extension of the two-variable solving-by-addition method, so make sure you know this method good and can utilize it systematically right.

Though the method of solution is based on addition/elimination, seeking to make existent add-on tends to acquire really mussy, so there is a systematized method for work outing the three-or-more-variables systems. This method is called “ Gaussian riddance ” ( with the equations stoping up in what is called “ row-echelon signifier ” ) .


Solve the undermentioned system of equations.

5x + 4y – A A z = 0

A A A A A A 10y – 3z = 11

A A A A A A A A A A A A A A A A A z = 3

It ‘s reasonably easy to see how to continue in this instance. I ‘ll merely back-substitute the z-value from the 3rd equation into the 2nd equation, work out the consequence for Y, and so stop up omega and yA into the first equation and work out the consequence for ten.

10y – 3 ( 3 ) = 11

10y – 9 = 11

10y = 20

Y = 2

5x + 4 ( 2 ) – ( 3 ) = 0

5x + 8 – 3 = 0

5x + 5 = 0

5x = -5

ten = -1

Then the solution is ( ten, y, omega ) = ( -1, 2, 3 ) .

The ground this system was easy to work out is that the system was “ triangular ” ; this refers to the equations holding the signifier of a trigon, because of the lower equations incorporating merely the ulterior variables.

The point is that, in this format, the system is simple to work out. And Gaussian riddance is the method we ‘ll utilize to change over systems to this upper triangular signifier, utilizing the row operations we learned when we did the add-on method.

Solve the undermentioned system of equations utilizing Gaussian riddance.

-3x + 2y – 6z = 6

A A 5x + 7y – 5z = 6

A A A A x + 4y – 2z = 8

No equation is solved for a variable, so I ‘ll hold to make the multiplication-and-addition thing to simplify this system. In order to maintain path of my work, I ‘ll compose down each measure as I go. But I ‘ll make my calculations on scratch paper. Here is how I did it:

The first thing to make is to acquire rid of the taking x-terms in two of the rows. For now, I ‘ll merely look at which rows will be easy to unclutter out ; I can exchange rows subsequently to acquire the system into “ upper triangular ” signifier. There is no regulation that says I have to utilize the x-term from the first row, and, in this instance, I think it will be simpler to utilize the x-term from the 3rd row, since its coefficient is merely “ 1 ” . So I ‘ll multiply the 3rd row by 3, and add it to the first row. I do the calculations on scratch paper:

… and so I write down the consequences:

( When we were work outing two-variable systems, weA could multiply a row, rewriting the system off to the side, and so add down. There is no infinite for this in a three-variable system, which is why we need the scratch paper. )

Warning: Since I did n’t really make anything to the 3rd row, I copied it down, unchanged, into the new matrix of equations. I used the 3rd row, but I did n’t really alter it. Do n’t confound “ utilizing ” with “ altering ” .

To acquire smaller Numberss for coefficients, I ‘ll multiply the first row by one-half:

Now I ‘ll multiply the 3rd row by -5 and add this to the 2nd row. I do my work on scratch paper:

… and so I write down the consequences: Copyright A© Elizabeth Stapel 1999-2009 All Rights Reserved

I did n’t make anything with the first row, so I copied it down unchanged. I worked with the 3rd row, but I merely worked on the 2nd row, so the 2nd row is updated and the 3rd row is copied over unchanged.

Okay, now the x-column is cleared out except for the prima term in the 3rd row. So following I have to work on the y-column.

Warning: Since the 3rd equation has an x-term, I can non utilize it on either of the other two equations any more ( or I ‘ll undo my advancement ) . I can work on the equation, but non with it.

If I add twice the first row to the 2nd row, this will give me a taking 1 in the 2nd row. I wo n’t hold gotten rid of the taking y-term in the 2nd row, but I will hold converted it ( without acquiring involved in fractions ) to a signifier that is simpler to cover with. ( You should maintain an oculus out for this kind of simplification. ) First I do the abrasion work:

… and so I write down the consequences:

Now I can utilize the 2nd row to unclutter out the y-term in the first row. I ‘ll multiply the 2nd row by -7 and add. First I do the abrasion work:

… and so I write down the consequences:

I can state what omega is now, but, merely to be thorough, I ‘ll split the first row by 43. Then I ‘ll rearrange the rows to set them in upper-triangular signifier:

Now I can get down the procedure of back-solving:

y – 7 ( 1 ) = -4

y – 7 = -4

Y = 3

ten + 4 ( 3 ) – 2 ( 1 ) = 8

ten + 12 – 2 = 8

ten + 10 = 8

ten = -2

Then the solution is ( ten, y, omega ) = ( -2, 3, 1 ) .

Note: There is nil sacred about the stairss I used in work outing the above system ; there was nil particular about how I solved this system. You could work in a different order or simplify different rows, and still come up with the right reply. These systems are sufficiently complicated that there is improbable to be one right manner of calculating the reply. So do n’t emphasize over “ how did she cognize to make that next? “ , because there is no regulation. I merely did whatever struck my illusion ; I did whatever seemed simplest or whatever came to mind first. Do n’t worry if you would hold used wholly different stairss. Equally long as each measure along the manner is right, you ‘ll come up with the same reply.

This manner, I can merely read off the values of ten, Y, and omega, and I do n’t hold to trouble oneself with the back-substitution. This more-complete method of work outing is called “ Gauss-Jordan riddance ” ( with the equations stoping upA in what is called “ reduced-row-echelon signifier ” ) . Many texts merely travel every bit far as Gaussian riddance, but I ‘ve ever found it easier to go on on and make Gauss-Jordan.

Note that I did two row operations at one time in that last measure before exchanging the rows. Equally long as I ‘m non working with and working on the same row in the same measure, this is all right. In this instance, I was working with the first row and working on the 2nd and 3rd rows.

Gauss Jordan riddance method

Gauss-Jordan Elimination is a discrepancy of Gaussian Elimination. Again, we are transforming the coefficient matrix into another matrix that is much easier to work out, and the system represented by the new augmented matrix has the same solution set as the original system of additive equations. In Gauss-Jordan Elimination, the end is to transform the coefficient matrix into a diagonal matrix, and the nothing are introduced into the matrix one column at a clip. We work to extinguish the elements both above and below the diagonal component of a given column in one base on balls through the matrix.

The general process for Gauss-Jordan Elimination can be summarized in the undermentioned stairss:

Gauss-Jordan Elimination Stairss

Write the augmented matrix for the system of additive equations.

Use simple row operations on the augmented matrix [ A|b ] to transform A into diagonal signifier. If a nothing is located on the diagonal, exchange the rows until a nonzero is in that topographic point. If you are unable to make so, halt ; the system has either infinite or no solutions.

By spliting the diagonal component and the right-hand-side component in each row by the diagonal component in that row, do each diagonal component equal to one.

An Example.

We will use Gauss-Jordan Elimination to the same illustration that was used to show Gaussian Elimination. Remember, in Gauss-Jordan Elimination we want to present nothings both below and above the diagonal.

1. Write the augmented matrix for the system of additive equations.

As earlier, we use the symbol to bespeak that the matrix predating the pointer is being changed due to the specified operation ; the matrix following the pointer displays the consequence of that alteration.

2. Use simple row operations on the augmented matrix [ A|b ] to transform A into diagonal signifier.

At this point we have a diagonal coefficient matrix. The concluding measure in Gauss-Jordan Elimination is to do each diagonal component equal to one. To make this, we divide each row of the augmented matrix by the diagonal component in that row.

3. By spliting the diagonal component and the right-hand-side component in each row by the diagonal component in that row, do each diagonal component equal to one.


Our solution is merely the right-hand side of the augmented matrix. Notice that the coefficient matrix is now a diagonal matrix with 1s on the diagonal. This is a particular matrix called the individuality matrix.

When executing computations by manus, many persons choose Gauss-Jordan Elimination over Gaussian Elimination because it avoids the demand for back permutation. However, we will demo subsequently that Gauss-Jordan riddance involves somewhat more work than does Gaussian riddance, and therefore it is non the method of pick for work outing systems of additive equations on a computing machine.

Example 2: Using Gauss-Jordan Elimination

( 1 ) We start with a system of additive equations:

x1 – 2×2 + 4×3 = 12

2×1 – x2 + 5×3 = 18

-x1 + 3×2 – 3×3 = -8

( 2 ) We represent them in augmented matrix signifier:

( 3 ) We use simple row operations to set this matrix into reduced echelon signifier ( I will utilize R1, R2, R3 for Row 1, Row 2, and Row 3 ) :

( a ) Let R2 = R2 + ( -2 ) R1 [ Operation # 3 ]

( B ) Let R3 = R3 + R1 [ Operation # 3 ]

( degree Celsius ) Let R2 = ( 1/3 ) R2 [ Operation # 2 ]

( vitamin D ) Let R1 = R1 + 2*R2 [ Operation # 3 ]

( vitamin E ) Let R3 = R3 + ( -1 ) *R2 [ Operation # 3 ]

( degree Fahrenheit ) Let R3 = ( 1/2 ) R3 [ Operation # 2 ]

( g ) Let R1 = R1 + ( -2 ) *R3 [ Operation # 3 ]

( H ) Let R2 = R2 + R3 [ Operation # 3 ]

( 4 ) We now write down the system of additive equations matching to the reduced echelon signifier:

x1 = 2

x2 = 1

x3 = 3

Of class, we need to be certain that we can ever acquire to cut down echelon signifier. Here is the theorem that guarantees this:

Theorem 1: Every matrix is row equivalent to a reduced echelon signifier matrix


( 1 ) The Gauss-Jordan Elimination Algorithm works for any 1 tens n matrix.

( a ) Let A = 1 x N matrix

( B ) Assume that A is non all nothings ( if it were, so A is already in reduced echelon signifier )

( degree Celsius ) Let a1, i be the first nonzero component in A.

( vitamin D ) Let A = ( 1/a1, I ) *A

( degree Fahrenheit ) A now has a taking 1 and since this is a one-row matrix, A is now in reduced echelon signifier.

( 2 ) Assume that the Gauss-Jordan Elimination Algorithm works for any matrix up to k rows.

( 3 ) We can now demo that it will work for a matrix with k+1 rows.

( a ) Let A = a nonzero matrix with k+1 rows [ We can presume it is nonzero since if it were zero, it would already be in decreased echelon signifier ]

( B ) Assume that we ‘ve run the Gauss-Jordan Elimination Algorithm up to the kth row so that if we define B such that:

( degree Celsius ) We can see that B is in decreased echelon signifier since

( I ) B has k rows

( two ) We can presume by our premise in measure # 2 that the algorithm works for any matrix of K rows or less

( three ) The algorithm leaves B unchanged if we run it once more.

( vitamin D ) We can besides see that Ak+1 consists of a individual row which is zero for every column where there is a taking 1 in B. [ Since these are zero ‘d out by the algorithm above ]

( vitamin E ) If there is a nonzero column in Ak+1, it must hence be to the right of all the taking 1s in B.

( degree Fahrenheit ) Assume that there is a nonzero column in Ak+1 [ If there were non, we would be done, since so Ak+1 would dwell of all nothing ‘s and A would so be in decreased echelon signifier ]

( g ) In this instance, none of the rows in B are all nothings. [ In order for Ak+1 to be nonzero, each of the old rows must hold had at least one nonzero column. If one was all nothing, it would hold been exchanged with Ak+1 and so Ak+1 would be all zeros which it is non ]

( H ) Let ak+1, x = the first nonzero component in Ak+1

( I ) Let Ak+1 = ( 1/ak+1, ten ) Ak+1

( J ) For all rows i, 1 thru K, allow Ri = Ri + ( -ai, x ) *Rk+1.

( K ) Clearly, this will zero out each of the staying rows.

( cubic decimeter ) A is now in reduced echelon signifier

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