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January 13th, 2014, 06:09 AM  #11 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  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 4a^2 = bc (and a + d = 0) So (for nonzero b) For a = d, b = 0, we find ,, and 
January 13th, 2014, 07:30 AM  #12 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Matrices
Hey! That's pretty good! You found an infinite number of distinct matrices. I think you found them all. 
January 13th, 2014, 07:45 AM  #13 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  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. 
January 13th, 2014, 07:55 AM  #14 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  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 rewrote 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. 
January 13th, 2014, 08:04 AM  #15 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  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.. 
January 13th, 2014, 08:23 AM  #16  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Matrices Quote:
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.  
January 13th, 2014, 08:36 AM  #17 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: Matrices
Ah, good idea while playing a round of cards with my roommates :P 
January 13th, 2014, 09:15 AM  #18 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  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. 
January 16th, 2014, 11:54 PM  #19  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Matrices Quote:
http://math.stackexchange.com/questions ... redirect=1 My thanx go to anyone who shows an interest in this matter.  
January 31st, 2014, 08:23 AM  #20 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  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 ith row and jth 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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