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January 13th, 2014, 06:09 AM   #11
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Re: Matrices

Overkill:
Let

Then

Which gives
a^2 + bc = 4 [1]
b(a + d) = 0 [2]
c(a + d) = 0 [3]
d^2 + bc = 4 [4]
[1]-[4]: a^2 - d^2 = 0 so

a = d gives with [2] and [3]:


So
Which might also be found by using

Now we have a = -d
which gives 4-a^2 = bc (and a + d = 0)
So
(for non-zero b)
For a = -d, b = 0, we find
,, and
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January 13th, 2014, 07:30 AM   #12
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Re: Matrices

Hey! That's pretty good!

You found an infinite number of distinct matrices. I think you found them all.

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January 13th, 2014, 07:45 AM   #13
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Re: Matrices

Thanks
By the way, for the first problem, ,

I tried to let
and
and to express e, f, g and h in terms of a, b, c and d without any luck so far.
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January 13th, 2014, 07:55 AM   #14
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Re: Matrices

I think you read a post that i deleted part of because of me not paying attention enough to your formulas. Disregard me saying there are two more please.

Yes , i also tried to set up 8 variables for the OP but it turned into a mess so i switched focus and re-wrote the problem using the SUM to my advantage , like this ,..





From here it was easy to get an answer to the OP , so , now you know my secret.

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January 13th, 2014, 08:04 AM   #15
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Re: Matrices

I came back to see what solutions I had and saw that you had edited your post so I edited the part that said I'd look further. No worries!

Cool secret. I'll try my plan again but you have inspired me to look at





and see what happens. However, I have some homework..
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January 13th, 2014, 08:23 AM   #16
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Re: Matrices

Quote:
Originally Posted by Hoempa
...However, I have some homework..


After you're done , take a swig of this , captain.


[attachment=0:1aa31jm4]131_metaxa.jpg[/attachment:1aa31jm4]


There is a 3 star , a 5 star and a 7 star. I like the 5 star the best , smoothest brandy i've ever had.

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File Type: jpg 131_metaxa.jpg (27.9 KB, 298 views)
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January 13th, 2014, 08:36 AM   #17
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Re: Matrices

Ah, good idea while playing a round of cards with my roommates :P
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January 13th, 2014, 09:15 AM   #18
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Re: Matrices

Have you ever seen anyone multiply matrices from right to left instead of from left to right?

1)

2)

So that way the answers on the LHS have square roots? I've introduced the because we have to go strictly from the RIGHT. So , using 1) , to get the entry 1 in the position we would do

(first row of matrix on the right of )(second column of matrix on the left of ) =

0*0 + 1*1 = 1

We go right to left and top to bottom , contrary to the usual way where we go left to right and top to bottom.

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January 16th, 2014, 11:54 PM   #19
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Re: Matrices

Quote:
Originally Posted by agentredlum
Have you ever seen anyone multiply matrices from right to left instead of from left to right?

1)

2)

So that way the answers on the LHS have square roots? I've introduced the because we have to go strictly from the RIGHT. So , using 1) , to get the entry 1 in the position we would do

(first row of matrix on the right of )(second column of matrix on the left of ) =

0*0 + 1*1 = 1

We go right to left and top to bottom , contrary to the usual way where we go left to right and top to bottom.

I have started a discussion about this at the following link.

http://math.stackexchange.com/questions ... redirect=1

My thanx go to anyone who shows an interest in this matter.

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January 31st, 2014, 08:23 AM   #20
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Re: Matrices

Hey Agent!

Maybe we should set up a formula for your definition.

The usual matrixmultiplication A*B where A has the same amount of columns as B has rows, gives the product, say C.
A has rows and columns, B has rows and columns.
Let be the element in the i-th row and j-th column, where and so and likewise for B. C will have rows and columns, so C consists of elements c_{i,j} where and
Now we must have .

then

Now for your product, C = A<<B we seem to have

C with rows and columns requiring such that
.

Do you get the same result?
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