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Particular name for matrices


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#1 AllCombinations

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Posted 11 October 2016 - 03:34 PM

Hello.

 

If you have a matrix, a 2x2 say

 

 M = \left(\begin{array}{ll}a_{1} & a_{2}\\b_{1} & b_{2}\end{array}\right)

 

and supposing it is in some way expanded into a 3x3 matrix so that the third row or column is formed from some kind of multiplication of the first two, such as

 

 N = \left(\begin{array}{lll}a_{1} & a_{2} & a_{1}a_{2}\\b_{1} & b_{2} & b_{1}b_{2}\\c_{1} & c_{2} & c_{1}c_{2}\end{array}\right)

 

I was wondering if this process has a name because I am having trouble finding anything about it online. The reason for including a new row is because I am studying determinants and wanted to keep the matrix square. The determinant won't be zero because we multiplied two rows together rather than adding them.

 

Does this have a name and, if it is common enough to have a name, what does this process mean geometrically? There is a rule for adding rows or columns but not for multiplying.

 

Thanks. If I've not been clear I can try to be more so.

 

 


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#2 studiot

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Posted 12 October 2016 - 07:50 AM

 

Allcombinations

 

The determinant won't be zero because we multiplied two rows together rather than adding them.

 

Have a care

 

What is the value of the determinant if a1 = b1 = c1 = 1?

 

 

 
 
 
 

 

 


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#3 AllCombinations

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Posted 12 October 2016 - 02:45 PM

Then two columns would be the same. I didn't say the determinant would never be zero. The point was that there is no immediate property of determinants (that I know of) concerning multiplication of rows or columns. :-)

 

Just a question about terminologies so I could learn more about this sort of matrix.


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#4 Mordred

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Posted 12 October 2016 - 02:54 PM

Those matrix definitions depends on the associations. For example a cummutative matrix is  a*b= b*a . Only the sign changes in this type. When you multiply a matrix the resultant matrix must have the same number of columns as the previous had rows. In matrix multiplication the commutative property is important. Key rule when you multiply two matrix, they are not cummutave. (Ab) does not equal (ba).

Rule two is association given 3 matrixes (ab)c=a(bc) however as its not commutative (ab)c does not equal (ac)b.
Rule 3) matrix multiplication is distributed across addition. A (b+c)=(ab+ac)

So I'm not really what you have done in the above. From what I can tell the second matrix isn't valid. However I could be wrong, but I don't think so

Edited by Mordred, 12 October 2016 - 03:47 PM.

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#5 studiot

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Posted 12 October 2016 - 05:20 PM

There is nothing inherently wrong with the second matrix,or creating a column of products of other terms present elsewhere in the matrix, but it depends upon what you are doing.

 

Matrix theory is part of linear algebra.

Matrix operations are linear.

 

Remember that each row of a matrix represents a linear combination of the entries (written as an equation if you like) so

 

there has to be a linear combination a+ a2 + a1a2 implied somewhere.  You cannot use matrix operations to create the non linear a1a2 cross product.

 

But you can include it a a specific term in the matrix.

 

This is done in many branches of Engineering to create systems of (linear) equations involving non linear quantities.

 

There is no special name for it since there is no opportunity to create these terms in matrix operations, they have to be 'planted into' the matrix.


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#6 Mordred

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Posted 13 October 2016 - 12:18 AM

Well I did leave the option open that I wasn't sure. Its a type I don't recognize.

Oh wait doh I see what he's doing. Lol not sure how I missed it. I looked at the determinents of |M|.  A_1b_2-A_2b_1 and trying to figure out how he got to N

Edited by Mordred, 13 October 2016 - 03:47 AM.

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http://www.einsteins.../LightCone.html
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http://cosmocalc.wikidot.com/start
If you wish to change the rules, you must first understand the rules.

#7 studiot

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Posted 13 October 2016 - 07:37 AM

Well I did leave the option open that I wasn't sure. Its a type I don't recognize.

Oh wait doh I see what he's doing. Lol not sure how I missed it. I looked at the determinents of |M|.  A_1b_2-A_2b_1 and trying to figure out how he got to N

 

Yes but he that is at the expense of introducing c1 and c2 without saying what they are or where they come from.

 

Only square matrices have determinants and by introducing the cross products into the original 2 X 2 matrix he has lost the squareness property.

 

Note that a determinant is a scalar (usually a pure number). It is not a matrix.


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#8 AllCombinations

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Posted 13 October 2016 - 02:43 PM

 

Yes but he that is at the expense of introducing c1 and c2 without saying what they are or where they come from.

 

They are numbers. Nothing more. As you said, I introduced them to keep the matrix square. The focus was less on the third row and more on the third column....


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#9 studiot

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Posted 13 October 2016 - 06:07 PM

What you may be interested in the the following property of determinants.

 

The sum of the elements in one row, i,  by the corresponding cofactors, in another row. m, is zero.

 

 
\sum\limits_{j = 1}^n {{S_{ij}}C{o_{mj}}}  = 0
 
 
i \ne m
 

 

This leads to Laplaces expansion of determinants

 

https://www.google.c...811.h5aMLZkxu6w

 

As regards combinations of matrix element here is a matrix equaltion from structural enfgineering illustrating the point I made in post#5

 

matrixeq1.jpg


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#10 AllCombinations

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Posted 13 October 2016 - 07:36 PM

Studiot, that property of determinants is really interesting. I've never seen that before. As for your comment on nonlinear quantities, and it did catch my attention as I can most certainly see the potential connections with non-linear concepts. As for your picture, I don't know what it is but it looks intriguing.
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#11 studiot

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Posted 14 October 2016 - 09:57 AM

Studiot, that property of determinants is really interesting. I've never seen that before. As for your comment on nonlinear quantities, and it did catch my attention as I can most certainly see the potential connections with non-linear concepts. As for your picture, I don't know what it is but it looks intriguing.

 

The expression in the picture (This site has difficulty with my latex generator for matrices) is one stage in the calculation of the fixed end moments for a beam spanning between two walls and built into each at its ends, by matrix methods.

Restraining the beam by building the ends in considerably stiffens the beam compared to simply resting it on top of the wall because the end restraints apply additional moments to the beam called fixed end moments.

Stiffening means the same beam deflects less under load when built in as opposed to resting on.

 

The point is that if you look carefully it illustrates my comment about linear combinations of non linear expressions nicely.

 

The 3 x 3 matrix is called the stiffness matrix and it premultiplies the deflection vector, which has 3 elements.

So by the usual operation of matrix multiplication the output is another column vector of applied loads (moments in this case), which also has 3 elements.

 

So the first row is equivalent to

 

 
\frac{{4EI}}{a}*{\theta _1} + 0 + \frac{{6EI}}{{{a^2}}}*\Delta  =  - M
 
 
 
 

Where theta is a rotation and delta is the linear deflection and M the applied moment.

Other letters are constants (for that beam)


Edited by studiot, 14 October 2016 - 09:58 AM.

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#12 Mordred

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Posted 14 October 2016 - 06:57 PM

I found using pmatrix a little easier than matrix latex. That might work better for your generator which I don't use.

Edited by Mordred, 14 October 2016 - 06:58 PM.

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http://www.einsteins.../LightCone.html
http://cosmology101.wikidot.com/main
http://cosmocalc.wikidot.com/start
If you wish to change the rules, you must first understand the rules.

#13 AllCombinations

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Posted 18 October 2016 - 05:59 PM

While we are on the subject of determinants, here is an interesting problem I found in an old linear algebra text. It reads that if you have two distinct points (x_{1},y_{1}) and (x_{2},y_{2}), then solving the equation

 

det\left(\begin{array}{lll}1 & x & y\\1 & x_{1} & y_{1}\\1 & x_{2} & y_{2}\end{array}\right)=0

 

for y yields a line that passes through those two points. Why is that and what is this peculiar form of interpolation/curve-fitting?

 

It seems to me that because determinants measure the area of a parallelogram in \mathbb{R}^{2}, then defining a matrix with two variables - in this case, x and y - and setting it equal to zero causes the parallelogram to have a zero area and the result is an empty parallelogram that acts as a line through those two points.

 

Isn't that odd? And then the columnspace (or just space) of the matrix is the plane in \mathbb{R}^{2} and the rowspace (or dualspace) is the set of points along the line, of which two are defined but any combination of x and y for which said matrix's determinant is zero will also be in the rowspace.

 

Where can I learn more about this? I would research it on my own but I don't know what it is called and when I try googling it all I get is stuff that doesn't seem to relate.

 

Thanks a lot. Oh, and the text the problem came from is a book from 2000 by David C. Lay called "Linear Algebra and Its Applications" second edition page 206.


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