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You are not logged in. #1 20070211 11:03:07
sup everybodyname is jonathan, i joined this cuz i love math as much as i like playing my alto sax(which i do like to play a lot). im a sophomore in high school but i learn fast and am fast on solutions within the extenet of my knowledge. i joined mathcounts in middle school and was the ranked 3 within our school. now in high school im really lazy about workin cuz no challenges. i recently found out about how i works like normal numbers, and that kept me entertained for a bit. but what i wanna know and this is the reason i joined, is how matrixs work. like #2 20070211 16:39:24
Re: sup everybodyHi T33N_T1T4N, Character is who you are when no one is looking. #3 20070211 16:54:02
Re: sup everybody*sretches* #4 20070212 03:58:11
Re: sup everybodyHey T33N_T14N, Wassup? Welcome to the Math Is Fun forum, You will have lots of fun on this website. "You don't have to like me but you do need to respect me'' #5 20070216 16:27:57
Re: sup everybodywelcome to the forum T33N_T14N.hope u get the help u need from here "Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"  David O. McKay #7 20070217 07:43:53
Re: sup everybodyHope you have lots of fun and this website is also good if you need some Homework help "You don't have to like me but you do need to respect me'' #8 20070217 08:48:32
Re: sup everybodyTo be honest, I can see why your teacher wouldn't want to explain matrices to you. It would be quite a lengthy explanation, and it's quite possible that you wouldn't have understood most of it. (No offence intended)
This system of equations is represented in matrix form like so: (Sometimes matrices are represtented with normal brackets rather than square ones, but they both represent the same thing) If we denote the left 3x3 matrix as A, the middle matrix as X and the right matrix as B, then our equation is AX = B. We want to find X, and so we need to get rid of the A in that equation. To do this, we need to premultiply both sides by A^{1}. By definition, A^{1}A = I (The identity matrix) and so we will be left with just X on that side. That would work with postmultiplying as well, but if we did that then the operation on the righthand side wouldn't be defined. To find the inverse of A, we need to find its determinant and its adjugate. First, the determinant: To find the adjugate of A, we first find its matrix of minors: Then its matrix of cofactors: And then the transpose of this is the adjugate: Divide the adjugate by the determinant, and we have the inverse: Premultiplying both sides of our original matrix equation by this gives the following: Now all we need to do is perform the matrix multiplication on the righthand side: And there, finally, we have it. Your solution is x = 18/7, y = 12/7, z = 4/7. Checking against the three original equations will verify that this works. Of course, just solving them the 'normal' way may well have been quicker. Why did the vector cross the road? It wanted to be normal. #9 20070302 07:50:59
Re: sup everybodyi love the teen titians i like raven the best then beastboy then starfire then cyborg then robin Zappzter  New IM app! Unsure of which room to join? "ZNU" is made to help new users. c: 