Pre-Caluclus Question http://prntscr.com/127afe
B is what I picked and got it wrong lol
\[x+y+z=9\\2x-3y+4z=7\\x-4y+3z=-2\] \[\begin{bmatrix}1&1&1\\2&-3&4\\1&-4&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\7\\-2\end{bmatrix}\]
what do i do here?
(I have to teach myself. So thats why i come on here. people like you are fantastic in times like these. and I learn a lot)
R?
\(R\) for row. so if i do the first row operation \[\left[\begin{array}{ccc|c}1&1&1&9\\2&-3&4&7\\1&-4&3&-2\end{array}\right]\\\sim\left[\begin{array}{ccc|c}1&1&1&9\\2-2(1)&-3-2(1)&4-2(1)&7-2(9)\\1&-4&3&-2\end{array}\right]R_2→R_2−2R_1\\\sim\left[\begin{array}{ccc|c}1&1&1&9\\0&-5&2&-11\\1&-4&3&-2\end{array}\right]\]
keep performing row operations until you get \[\sim\left[\begin{array}{ccc|c}1&0&0&x_1\\0&1&0&y_1\\0&0&1&z_1\end{array}\right]\]
the second row operatio i would make is \[R_3→R_3−R_1\]
So the first row is R2,R2,,R1 and the 2nd is R3,R3,R1?
?
I cant make out what the row operation is for the 2nd one. is that a 3 beside the R
\[\Large R_3→R_3−R_1\]
\[\large\sim\left[\begin{array}{ccc|c}1&1&1&9\\0&-5&2&-11\\1-(1)&-4-(1)&3-(1)&-2-(9)\end{array}\right] R_3→R_3−R_1\]
Im trying to figure out what you did in the first one
for the first row operation, i just took away (two times) the first row from the second row
okay starting to get there
On the 3rd row i see what you did in the 2nd column
What did you do at the bottom?
pardon ?
is it A?
after my second row operation, i found that two rows were equal, which makes the system unsolveable
or did i do something wrong
i dont know , i can't see you working,
So if its unsolvable. What do i put as my answere? I only have the A.CD choices.
Oh I got it!!! duh! I had a blockage there for a moment.
you got it?
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