Question

Solve the System using Cramer's Rule
** Open to see the picture of the system of equations**

Answer 1

I know how to solve them when they only have 2 variable, but the 3 variable ones are confusing me

Answer 2

do you realize how long Cramer's rule is?

Answer 3

yeah..?

Answer 4

I don't see how three variables is harder. It's just longer

Answer 5

Do you know how to calculate the determinate?

Answer 6

you can't use cramer's rule on 4 x 4 matrices so at least those types of assignments are impossible to even create. But 3 x 3 is longer.

Answer 7

yeah but for the 2x2 i just have to cross multiply a lot.. i don't get how I can cross multiply 3 numbers?

Answer 8

http://www.purplemath.com/modules/cramers.htm

Answer 9

oh hey @Lyrae long time no see

Answer 10

You are finding the determinate of the matrix

Answer 11

In the 2 by 2 case, it's just cross multiplication and adding.

Answer 12

yeah i got that it's the 3x3 that's confusing me

Answer 13

can't i just get the same answers b simple elimination?

Answer 14

no.. I gave a link just follow the example and it apply it to your problem or something .

Answer 15

@UsukiDoll um thanks for the help but i didn't ask for the attitude. Thanks.

Answer 16

oh wow.. I wasn't giving attitude.. I was giving supplement material ... read what the example is doing and try to understand it x.x

Answer 17

in most cultures when you add "or something ." it means you're giving attitude. Sorry if i misunderstood.

Answer 18

what the! Well I will ignore all countries who believe that because plenty people use or something like that.....

Answer 19

\[
\begin{bmatrix}
a&b&c\\
d&e&f\\
g&h&i
\end{bmatrix}
\]First thing you do to make it easier is put the matrix side by side: \[

\begin{bmatrix}
a&b&c&a&b&c\\
d&e&f&d&e&f\\
g&h&i&g&h&i
\end{bmatrix}
\]Then recognize the diagonals: \[
\begin{bmatrix}
\color{red}a&b&c&\color{blue}a&b&c\\
d&\color{red}e&\color{blue}f&d&e&f\\
g&\color{blue} h&\color{red}i&g&h&i
\end{bmatrix}
\]For our \(a\) part, we have \(aei - afh = a(ei-fh)\)\[
\begin{bmatrix}
a&\color{red}b&c&a&\color{blue}b&c\\
d&e&\color{red}f&\color{blue}d&e&f\\
g& h&\color{blue}i&\color{red}g&h&i
\end{bmatrix}
\]For our \(b\) part, we have \(bfg- bdi = b(fg-di)\)\[
\begin{bmatrix}
a&b&\color{red}c&a&b&\color{blue}c\\
d&e&f&\color{red}d&\color{blue}e&f\\
g& h&i&\color{blue}g&\color{red}h&i
\end{bmatrix}
\]For our \(c\) part, we have \(cdh- ceg= c(dh-eg)\)
We make the we add the odd column ones \(a\) and \(c\), thensubtract the even column one \(b\).
\[
\begin{vmatrix}
a&b&c\\
d&e&f\\
g&h&i
\end{vmatrix}=a(ei-fh)-b(fg-di)+c(dh-eg)
\]

Answer 20

OH MY GOD!! THANKYOU SO MUCH!!