Question

Algebra 2b help please?
@campbell_st
picture posted below

Answer 1

in your solution I'd expect you have to say row 1 x 3 + row 3

then write the solution matrix... as I did in the previous part

Answer 2

dont the second and last row already have 2 terms? or does the 0 count as a term? and what about the first row?

Answer 3

you may need to check my working... as it would have been nice if the solutions were integers

Answer 4

ooh ok i get it! thanks so much!!

Answer 5

well I've made a mistake somewhere...as I did it longhand ... and it should be

x = 2, y =-1 and z = 3

Answer 6

alrighty, thanks again! i have only one more that i will post in a minute:) thanks!

Answer 7

ok.... the mistake is in the 2nd matrix... after adding row 1 x 3 to row 3 you get

\[\left[\begin{matrix}1 & 1 & 1 \\ 4 & 5&0\\ 3 & 4 & 0\end{matrix}\right] = \left[\begin{matrix}4 \\ 3 \\2\end{matrix}\right]\]

Answer 8

i thought the last row was 0 11 0 -5?

Answer 9

*-11

Answer 10

could you show me how to get y?:)

Answer 11

then multiply row 3 by 3 and row 3 by 4 and subtract them

you get

12 15 = 9
- 12 16 = 8
------------------
-1 = 1

so the matrix is

\[\left[\begin{matrix}1 & 1 & 1 \\ 4 & 5 & 0\\ 0 & -1 & 0\end{matrix}\right] = \left[\begin{matrix}4 \\ 3 \\-1\end{matrix}\right]\]

so y = -1

and you can now find x and z by using the original

Answer 12

dang... multiply row 2 by 3 and then subtract row 3 times 4