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 January 13th, 2014, 05:09 AM #11 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Matrices Overkill: Let $A= $a \ b \\ c \ d$$ Then $A^2= $a^2 + bc \ b(a + d) \\ c(a + d) \ bc + d^2$ = $4 \ 0 \\ 0 \ 4$$ 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 $a= d \vee a = -d$ a = d gives with [2] and [3]: $b= c = 0$ $a^2= 4 \rightarrow a = -2 \vee 2$ So $A= $-2 \ 0 \\ 0 \ -2$ \vee A = $2 \ 0 \\ 0 \ 2$$ Which might also be found by using $A^2= 4I$ Now we have a = -d which gives 4-a^2 = bc (and a + d = 0) So $\large A = \begin{bmatrix} a & b\\ \frac{4 - a^2}{b} & -a \end{bmatrix}$ (for non-zero b) For a = -d, b = 0, we find $\large A = \begin{bmatrix} 2 & 0\\ c & -2 \end{bmatrix}$ ,$\large A = \begin{bmatrix} 2 & c\\ 0 & -2 \end{bmatrix}$, $\large A = \begin{bmatrix} -2 & 0\\ c & 2 \end{bmatrix}$ and $\large A = \begin{bmatrix} -2 & c\\ 0 & 2 \end{bmatrix}$
 January 13th, 2014, 06:30 AM #12 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Matrices Hey! That's pretty good! You found an infinite number of distinct matrices. I think you found them all.
 January 13th, 2014, 06:45 AM #13 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Matrices Thanks By the way, for the first problem, $A^2+B^2=\left( \begin{array}{cc}2&3\\3=&2\end{array}\right)=$, I tried to let $A= \left( \begin{array}{cc}a&b\\c=&d\end{array}\right)=$ and $B= \left( \begin{array}{cc}e&f\\g=&h\end{array}\right)=$ and to express e, f, g and h in terms of a, b, c and d without any luck so far.
 January 13th, 2014, 06:55 AM #14 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Matrices I think you read a post that i deleted part of because of me not paying attention enough to your formulas. Disregard me saying there are two more please. Yes , i also tried to set up 8 variables for the OP but it turned into a mess so i switched focus and re-wrote the problem using the SUM to my advantage , like this ,.. $A^2 \= \ $2 \ 3 \\ 0 \ 0$$ $B^2 \= \ $0 \ 0 \\ 3 \ 2$$ From here it was easy to get an answer to the OP , so , now you know my secret.
 January 13th, 2014, 07:04 AM #15 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Matrices I came back to see what solutions I had and saw that you had edited your post so I edited the part that said I'd look further. No worries! Cool secret. I'll try my plan again but you have inspired me to look at $A^2 \= \ $a \ b \\ c \ d$$ $B^2 \= \ $2-a \ 3-b \\ 3-c \ 2-d$$ and see what happens. However, I have some homework..
January 13th, 2014, 07:23 AM   #16
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 233

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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 January 13th, 2014, 07:36 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, 08:15 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) $\ \ $0 \ 1 \\ 1 \ 0$ \= \ $1 \ 0 \\ 0 \ 1$$ « $$1 \ 0 \\ 0 \ 1$$ 2) $\ \ $0 \ 1 \\ 1 \ 0$ \= \ $0 \ 1 \\ 1 \ 0$$ « $$0 \ 1 \\ 1 \ 0$$ 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 $\ a_{1,2} $/extract_itex] 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, 10:54 PM #19 Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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) $\ \ \[0 \ 1 \\ 1 \ 0$ \= \ $1 \ 0 \\ 0 \ 1$$ « $$1 \ 0 \\ 0 \ 1$$ 2) $\ \ $0 \ 1 \\ 1 \ 0$ \= \ $0 \ 1 \\ 1 \ 0$$ « $$0 \ 1 \\ 1 \ 0$$ 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 $\ a_{1,2} \$ 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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My thanx go to anyone who shows an interest in this matter.

 January 31st, 2014, 07:23 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 $m_A$ rows and $n_A$ columns, B has $m_B$ rows and $n_B$ columns. Let $a_{i,j}$ be the element in the i-th row and j-th column, where $1 \le i \le m_a$ and $1 \le j \le n_a$ so $A= [a_{i,j}]$ and likewise for B. C will have $m_A$ rows and $n_B$ columns, so C consists of elements c_{i,j} where $1 \le i \le m_A$ and $1 \le j \le n_B$ Now we must have $n_A= m_B$. then $c_{i,j}= \sum_{k = 1}^{n_A} a_{i,k} \cdot b_{k, j}$ Now for your product, C = A<

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