Help me with matrices? :)
Question? :)
Take all the coefficients and form a 3 x 3 matrix.
In all honesty i have no idea whats going on in my class im trying to catch up fast:/
AX=B is the matrix equation. A is coefficient matrix and B is constant term matrix.
First make a matrix A using the coefficients of x,y,z
ahhh i think i'm just gunna guess on this one I literally know nothing :p
Guess? You taking a test?
It's just a small quiz type thing
Well getting help during the test is a bad idea. Come back after it and we can do the review!
It's a quick check so its not a test but thanks anyways! :)
\[0x+2y+7z\\ -3x-7y+0z\\ 7x+0y-8z\] coefficients are \[0,2,7\\ -3,-7,0\\ 7,0,-8\]
something like \[ \left[ {\begin{array}{cccc} 0 & 2 &-7\\ -3 & -3 &0 \\ 7 & 0 &-8 \end{array} } \right] \]
im not sure how you form the 3x3 matrix
oh wait is that it?
you need a column vector with the right hand side of the equations
is that my answer? or does it have to be simplified more?
that is it, but it is not done you need the column of the answers
-3 3 3
not sure how to draw one here
yeah, like that
so is that it? thank you for helping me btw :)
yeah, that's it easy right? yw
ahhh can you help me with another :)
sure if i can
to hell with matrices this is much easier than that multiply the first equation all the way across by \(3\) and the second one all the way across by \(-2\) you go from \[-3x+2y=10\\ -4x+3y=2\] to \[-9x+6y=30\\ 8x-6y=-4\] then add them up and the \(y\) terms drop out leaving you with \[-x=26\] and so \(x=-26\)
you good with that? you still need \(y\)
Yes that was i need to figure out now.. how do i do that do i multiply?
is it -34???
replace \(x\) by \(-26\) in either equation for example in the first one you would turn \[-3x+2y=10\] so \[-3\times (-26)+2y=10\] or \[78+2y=10\] and that is easy to solve
yes
Ahh thank you so much <3 :D
\[78+2y=10\\ 2y=-68\\ y=-34\] you win yw
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