Question

Solve each system of linear equations using matrices.
x+_y=1
x-2y=4

Answer 1

what os _y?

Answer 2

theres a space there

Answer 3

?
x+y = 1
x-2y = 4
?

Answer 4

yes

Answer 5

Ok, so the matrix would looke like
[1,1,1;1, -2, 4] correct? where [first row;second row]

Answer 6

yes correct

Answer 7

\[\begin{bmatrix} 1& 1\\ 1& -2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 1\\ 4 \end{bmatrix}\]

Answer 8

what do you need to add to r1c2 to get it to be 0?

Answer 9

I don't know that's why I'm asking for help

Answer 10

i asked what you need to add to 1 to get 0

Answer 11

\[A\begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} 1\\ 4 \end{bmatrix}\]
\[A^{-1}A\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} x\\ y \end{bmatrix}= A^{-1}\begin{bmatrix} 1\\ 4 \end{bmatrix}\]

Answer 12

this is about row reduction henpen

Answer 13

Basically, just like in normal algebra, we 'divide by A' (or multiply by the inverse matrix of a) to get what x and y actually ARE.

Answer 14

I doupt they have got to inverse yet

Answer 15

I'm so confused

Answer 16

are you doing row reduction in class?

Answer 17

I think so

Answer 18

have you been to class?

Answer 19

lol

Answer 20

ok would you be able to solve this with elelmination method we learn in algebra class?

Answer 21

I'm sorry that math isn't an easy subject for me. Thanks for making it a joke though. Nice. I'll just find someone else to explain it to me.

Answer 22

I did not make a joke, I asked what you are doing in class. You need not understnad exactly how to do what you are doing in class, but you should know what it is you are doing.

Answer 23

if you want help. there are many ways to solve such a question....so knowing what you are working on could be very helpfull.

Answer 24

but ok good luck m8

Answer 25

what you say we start over?:) would you be able to solve this with the elimitation method we learn in algebra @mluna1218 ?

Answer 26

I figured this out by myself. Thanks though