Here's the question you clicked on:
Hayatcakes
http://i729.photobucket.com/albums/ww292/akatushi101/algeeeeeeeeeee.jpg I need help on the Four questions .
=) Give me just a minute to write them down and snap a pic.
also these are Matrices
\[\left[\begin{matrix}2 & 3 \\ 5 & 1\end{matrix}\right]-\left[\begin{matrix}-1 & 7 \\ 9 & 8\end{matrix}\right]=\left[\begin{matrix}2-(-1) & 3-7 \\ 5-9 & 1-8\end{matrix}\right]=\left[\begin{matrix}3 & -4 \\ -4 & -7\end{matrix}\right]\]
Are the first/third ones division or addition?
the first one is addition and the third one is addition
oh u do the same, just change the minus sign to plus
It's uploading.. la dee da!
Here you go (the first problem is last, because I thought it was a division sign!) http://postimage.org/image/kmf3y8d8n/
why is the third problem not possible?
Because the two matrices are of different dimensions. If you go ahead and try to attempt it, you'll see quickly that you've run out of numbers and have two left over. It simply doesn't work. =)
I have three questions left from this matrice ..can you help me :S?
http://i729.photobucket.com/albums/ww292/akatushi101/ALGEBROOO.jpg
the 6th question there is no "-" or " + " so i think we can not operate it?
That would assume that you multiply them.
but it is set like this [ 1 2 ][1] [3 7] [5]
It's uploading! You can only multiply matrices if the rows of the first equals the columns of the second. Basically, how that works is you take the terms as such: [ a b ] [1] [ c d ] [5] so that the new matrix is: [ a*1 + b*5 ] [ c*1 + d*5 ]
Here ya go: http://postimage.org/image/p1sw2isxt/
Do you understand?
Yes.. :) I'm sorry these were the last ones. . these are the last remaining problems http://i729.photobucket.com/albums/ww292/akatushi101/ALGEE222.jpg
I'm sorry, but I have to head off to class now!=( You can try re-posting the new set for someone else to answer or wait a few hours for my response.
okay thanks for all the help!