Question

Need Help with 2 questions!

Answer 1

what are they i can try and help?

Answer 2

@amistre64

Answer 3

Can not help sorry :'(

Answer 4

what are your ideas?

Answer 5

Would we combine like terms for the equation?

Answer 6

Matrix :D

Answer 7

no, just strip the equation of everything but the coefficients ...

Answer 8

for the next one, add in the 'variables'

Answer 9

3x-2y = 4

has the matrix

[3 -2 4]

why?

Answer 10

you removed the coefficients?

Answer 11

Variables*

Answer 12

yep ... i stripped it down to the bare essentials

Answer 13

so for number 1 would it be A?

Answer 14

if we have ore than one equation to play with, we simply line up the like terms

x -2y +5z = 1
4x + 3y = 3
-2x + z = 5

x y z c
1 -2 5 1
4 3 0 3
-2 0 1 5

Answer 15

yes, A is the coefficient matrix for the system

Answer 16

So i got it correct? Thanks.. lets move onto #2

Answer 17

ive given a hint for #2, any ideas how we can apply it to 'undo' a matrix?

Answer 18

no take me through lol no idea

Answer 19

the options suggest we need variables of xyz, and some constant ...

so, top it off with x y z c, and distribute them down the column

Answer 20

would it be C?

Answer 21

x y z c
-----------
1 -2 5 1
4 3 0 3
-2 0 1 5


1x -2y 5z =1
4x 3y 0z =3
-2x 0y 1z =5

Answer 22

yes, C works out fine

Answer 23

okay can you help me with a few more but they're not complicated ones like these, they're simple

Answer 24

sure, why not

Answer 25

simplest approach, trial and error, see what works and what doesnt

Answer 26

im not sure would it be C or D for the first one, what do you think?

Answer 27

neither, they dont fit the system ....

Answer 28

aren't we combing like terms?

Answer 29

how would i find the solution?

Answer 30

not really, you are taking a point (a,b) and testing to see if it makes both statements true.

i can tell right away that 11-8 is not 6 so D fails to be true for both parts

Answer 31

8-4 is not 6, so C fails as well

Answer 32

x+3y = 5
x+4y = 6
1 5


1+15 = 5
1+20 = 6

1,5 is not a solution either.

Answer 33

if your objective is to determine the correct option, then work what is simplest for you.

if your desire is to know how to work with a matrix ... then a longer approach is required.

Answer 34

B?

Answer 35

is 1,5 a solution to both equations?

Answer 36

Im not typing anymore its just stuck on the screen lol.

Answer 37

i think that is a client side affect, but yeah .... 'user' is typing seems to be a glitch

Answer 38

Where are you getting 15 and 20 from?

Answer 39

let x=1, and y=5 of course

Answer 40

Lets start fresh move onto the 2nd one..

Answer 41

simplest test is to let q=0 .. since that is an option

Answer 42

do you know how to do matrix multiplication? inner product?

Answer 43

yes matrix multiplication

Answer 44

\[\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}n\\m\end{pmatrix}\color{red}{\implies}\begin{matrix}ax+by=n\\cx+dy=m\end{matrix}\]

therefore the options given, some (x,y) will satisfy the system. test them out.

or work it in whatever way is simplest for you. subsitution maybe? elimination? trial/error?

Answer 45

i forgot to divide by a ...

Answer 46

or by generalization of subsitution to get a formula if

x=(n-by)/a

c(n-by)/a + dy = m

cn/a -bcy/a + dy = m

y(d-bc/a) = m-cn/a

y = (m-cn/a)/(d-bc/a)

y = (am-cn)/(ad-bc)

Answer 47

Would it be C? -1,0

Answer 48

show me if C works or not

Answer 49

im getting to much lag on the site .... C works for me. good luck