Solving a system of equations by Gauss's method

[Physicaa]

Homework Statement

So I have the following :

AX=B, where

$$A=\begin{bmatrix}
a_{ij}
\end{bmatrix}_{mXn}$$

with $$a_{ij}=i$$

$$
X=\begin{bmatrix}
X_{1} &X_{2} &X_{3} &... &X_{n}
\end{bmatrix}^T$$

and

$$B=\begin{bmatrix}
1 &2 &3 &... &m
\end{bmatrix}^T$$

Homework Equations

Gauss' method of solving system of linear equations

The Attempt at a Solution

Now the thing is, when I usually solve a system of equation with Gauss I usually get 3 normal equations. It's the first time I get this and I'm not even sure of what I have in my hands...

Could somebody help?
[/B]

 

[Svein]

If $a_{i,j}=i$ then the two first rows are $\begin{pmatrix} 1 & 1 & ... \\ 2 & 2 & ... \\ ... & ... & ... \\ \end{pmatrix}$ What does that tell you about the determinant of A?

 

[SteamKing]

Homework Statement

So I have the following :

AX=B, where

$$A=\begin{bmatrix}
a_{ij}
\end{bmatrix}_{mXn}$$

with $$a_{ij}=i$$

$$
X=\begin{bmatrix}
X_{1} &X_{2} &X_{3} &... &X_{n}
\end{bmatrix}^T$$

and

$$B=\begin{bmatrix}
1 &2 &3 &... &m
\end{bmatrix}^T$$

Homework Equations

Gauss' method of solving system of linear equations

The Attempt at a Solution

Now the thing is, when I usually solve a system of equation with Gauss I usually get 3 normal equations. It's the first time I get this and I'm not even sure of what I have in my hands...

Could somebody help?[/B]

It's not clear what you mean when you say you get 3 normal equations when you use Gauss' method. Normal equations are formed when you multiply A by its transpose, and this method of forming normal equations is frequently used to solve for the unknown coefficients in a regression equation. Even then, the system of normal equations must be solved using some technique, but because the resulting system is usually only a 2 x 2 or a 3 x 3, Cramer's Rule can be employed rather than Gaussian elimination.

Your present system of equations is a little more general. The matrix of coefficients A is composed of elements where each element in a particular row is equal to its row number. That's what $a_{ij} = i$ means. Ditto for the RHS: the value of each element in the B vector is also equal to the row number.

Thus, your system looks like this (for n = 3):

| 1   1   1 |   | X1 |    | 1 |
| 2   2   2 | * | X2 |  = | 2 |
| 3   3   3 |   | X3 |    | 3 |

Applying Gaussian elimination to this system of equations will produce an upper triangular matrix of coefficients with just a single non-zero coefficient in the n-th equation. The n-th unknown can then be solved and the values of the other n-1 unknowns can be determined in turn by proceeding from the n-th equation back to the first equation, using back-substitution.

The technique is illustrated here:

http://mathworld.wolfram.com/GaussianElimination.html

 

[Physicaa]

It's not clear what you mean when you say you get 3 normal equations when you use Gauss' method. Normal equations are formed when you multiply A by its transpose, and this method of forming normal equations is frequently used to solve for the unknown coefficients in a regression equation. Even then, the system of normal equations must be solved using some technique, but because the resulting system is usually only a 2 x 2 or a 3 x 3, Cramer's Rule can be employed rather than Gaussian elimination.

Your present system of equations is a little more general. The matrix of coefficients A is composed of elements where each element in a particular row is equal to its row number. That's what $a_{ij} = i$ means. Ditto for the RHS: the value of each element in the B vector is also equal to the row number.

Thus, your system looks like this (for n = 3):

| 1   1   1 |   | X1 |    | 1 |
| 2   2   2 | * | X2 |  = | 2 |
| 3   3   3 |   | X3 |    | 3 |

Applying Gaussian elimination to this system of equations will produce an upper triangular matrix of coefficients with just a single non-zero coefficient in the n-th equation. The n-th unknown can then be solved and the values of the other n-1 unknowns can be determined in turn by proceeding from the n-th equation back to the first equation, using back-substitution.

The technique is illustrated here:

http://mathworld.wolfram.com/GaussianElimination.html

So I'm supposed to solve this generally ? I just meant that it's the first time I get something like this. Never saw this.

How do I even begin this in the first place ? I know how to use Gauss, it's just the setting up that's causig problems.

 

[Physicaa]

It's not clear what you mean when you say you get 3 normal equations when you use Gauss' method. Normal equations are formed when you multiply A by its transpose, and this method of forming normal equations is frequently used to solve for the unknown coefficients in a regression equation. Even then, the system of normal equations must be solved using some technique, but because the resulting system is usually only a 2 x 2 or a 3 x 3, Cramer's Rule can be employed rather than Gaussian elimination.

Your present system of equations is a little more general. The matrix of coefficients A is composed of elements where each element in a particular row is equal to its row number. That's what $a_{ij} = i$ means. Ditto for the RHS: the value of each element in the B vector is also equal to the row number.

Thus, your system looks like this (for n = 3):

| 1   1   1 |   | X1 |    | 1 |
| 2   2   2 | * | X2 |  = | 2 |
| 3   3   3 |   | X3 |    | 3 |

Applying Gaussian elimination to this system of equations will produce an upper triangular matrix of coefficients with just a single non-zero coefficient in the n-th equation. The n-th unknown can then be solved and the values of the other n-1 unknowns can be determined in turn by proceeding from the n-th equation back to the first equation, using back-substitution.

The technique is illustrated here:

http://mathworld.wolfram.com/GaussianElimination.html

I drew this : http://imgur.com/5lQZNvk

How do I get my system now ? I m going to go sleep, I hope I get an answer :(

 

[SteamKing]

I drew this : http://imgur.com/5lQZNvk

How do I get my system now ? I m going to go sleep, I hope I get an answer :(

Look at the clue given to you by Svein in Post #2.

What can you say about the determinant of A in general, given its composition?

 

[Physicaa]

Look at the clue given to you by Svein in Post #2.

What can you say about the determinant of A in general, given its composition?

Well it's going to be the "i" that we place in it. like 1,1,1,1...,1

2,2,2,...,2

etc.

Oh I didn't notice his post. Sorry

 

[Physicaa]

If $a_{i,j}=i$ then the two first rows are $\begin{pmatrix} 1 & 1 & ... \\ 2 & 2 & ... \\ ... & ... & ... \\ \end{pmatrix}$ What does that tell you about the determinant of A?

I didn't lean about determinants yet

 

[SteamKing]

I didn't lean about determinants yet

That makes it difficult to determine if a system of linear equations has a solution, if you don't know anything about determinants.

 

[Physicaa]

That makes it difficult to determine if a system of linear equations has a solution, if you don't know anything about determinants.

So there's no other way for me to determine the answer ?

 

[SteamKing]

So there's no other way for me to determine the answer ?

I don't know what to tell you.

Studying matrix algebra without studying determinants is like studying arithmetic but omitting discussions of multiplication.

If you don't know what a determinant is, Cramer's Rule is off the table, for example, since that method is based entirely on calculating determinants.

While you don't need to explicitly calculate the determinant of the matrix of coefficients to use Gaussian elimination, knowledge that the determinant is not zero suggests that there is a unique solution to the associated system of linear equations.

 

[Physicaa]

I don't know what to tell you.

Studying matrix algebra without studying determinants is like studying arithmetic but omitting discussions of multiplication.

If you don't know what a determinant is, Cramer's Rule is off the table, for example, since that method is based entirely on calculating determinants.

While you don't need to explicitly calculate the determinant of the matrix of coefficients to use Gaussian elimination, knowledge that the determinant is not zero suggests that there is a unique solution to the associated system of linear equations.

I didn't say we won'T learn it, we just didn't at the moment. Now what I wonder is why did I get this problem if it's not possible for me to solve it without any tools...

 

[SteamKing]

I didn't say we won'T learn it, we just didn't at the moment. Now what I wonder is why did I get this problem if it's not possible for me to solve it without any tools...

Well, skip determinants for the time being.

Hve you learned how to tell if a system of linear equations is independent?

 

[Ray Vickson]

I didn't say we won'T learn it, we just didn't at the moment. Now what I wonder is why did I get this problem if it's not possible for me to solve it without any tools...

You do not need determinants; you just need to actually know how Gaussian elimination works. Then you need to carry out the actual Gaussian elimination steps. I mean: do it instead of agonizing about it.

To help you get going, I suggest very strongly that you do it from start to finish first for the case of n = 2 and then for the case of n = 3. Both of these are small enough that you should have no difficulties. Come back with more questions (if you need to) after you have done what I suggested.

 

[Physicaa]

You do not need determinants; you just need to actually know how Gaussian elimination works. Then you need to carry out the actual Gaussian elimination steps. I mean: do it instead of agonizing about it.

To help you get going, I suggest very strongly that you do it from start to finish first for the case of n = 2 and then for the case of n = 3. Both of these are small enough that you should have no difficulties. Come back with more questions (if you need to) after you have done what I suggested.

Ok I'll go try and see if i can figure it out.

 

[Physicaa]

Ok I'll go try and see if i can figure it out.

You do not need determinants; you just need to actually know how Gaussian elimination works. Then you need to carry out the actual Gaussian elimination steps. I mean: do it instead of agonizing about it.

To help you get going, I suggest very strongly that you do it from start to finish first for the case of n = 2 and then for the case of n = 3. Both of these are small enough that you should have no difficulties. Come back with more questions (if you need to) after you have done what I suggested.

ok, so for 2x2 I get the following equations :

x1+x2=1

2x1+2x2=2

These are good right ? I multiplied the matrices

 

[SteamKing]

ok, so for 2x2 I get the following equations :

x1+x2=1

2x1+2x2=2

These are good right ? I multiplied the matrices

To use elimination, you don't need to multiply the matrices together. It's better if you leave them as original.

Maybe you missed reading this link:

http://mathworld.wolfram.com/GaussianElimination.html

But it contains an example of how to use elimination to solve a system of equations. You should study this example carefully.

 

[Physicaa]

To use elimination, you don't need to multiply the matrices together. It's better if you leave them as original.

Maybe you missed reading this link:

http://mathworld.wolfram.com/GaussianElimination.html

But it contains an example of how to use elimination to solve a system of equations. You should study this example carefully.

Ok thanks, going to read this. It looks like what I have.

 

[Svein]

ok, so for 2x2 I get the following equations :

x1+x2=1

2x1+2x2=2

Then divide all elements in row 2 by 2. What do you get?