Prove cramers rule for a system of 2 equations
- Thread starter DogMan
- Start date
Hello, DogMan!
Why is this suddenly showing up in a Calculus course?
No idea? .That's so lame . . .
How about showing that Cramer's Rule produces the solution?
We are given: .\(\displaystyle \begin{array}{cccccc}(1) & ax + by &=& p \\ (2) & cx + dy &=& q \end{array}\)
Solve by Elimination.
\(\displaystyle \begin{array}{cccccccccc}\text{Multiply (1) by }d: & adx + bdy &=& dp \\ \text{Multiply (2) by -}b: & \text{-}bcx - bdy &=& \text{-}bq \end{array}\)
\(\displaystyle \text{Add: } \: (ad-bc)x \:=\:dp - bq \quad\Rightarrow\quad \boxed{x \;=\;\dfrac{dp-bq}{ad-bc}}\)
\(\displaystyle \begin{array}{cccccccccc}\text{Multiply (1) by -}c: & \text{-}acx - bcy &=& \text{-}cp \\ \text{Multiply (2) by }a: & acx + ady &=& aq \end{array}\)
\(\displaystyle \text{Add: } \: (ad-bc)y \:=\:aq-cp \quad\Rightarrow\quad \boxed{y \;=\;\dfrac{aq-cp}{ad-bc}}\)
Solve by Cramer's Rule.
\(\displaystyle \displaystyle x \;=\;\frac{\begin{vmatrix}p & b\\ q&d\end{vmatrix}} {\begin{vmatrix}a&b\\c&d\end{vmatrix}} \quad\Rightarrow\quad \boxed{x \;=\;\dfrac{dp-bq}{ad-bc}}\)
\(\displaystyle \displaystyle y \;=\;\frac{\begin{vmatrix}a & p\\ c&q\end{vmatrix}} {\begin{vmatrix}a&b\\c&d\end{vmatrix}} \quad\Rightarrow\quad \boxed{y \;=\;\dfrac{aq-cp}{ad-bc}}\)
Why is this suddenly showing up in a Calculus course?
No idea? .That's so lame . . .
How about showing that Cramer's Rule produces the solution?
We are given: .\(\displaystyle \begin{array}{cccccc}(1) & ax + by &=& p \\ (2) & cx + dy &=& q \end{array}\)
Solve by Elimination.
\(\displaystyle \begin{array}{cccccccccc}\text{Multiply (1) by }d: & adx + bdy &=& dp \\ \text{Multiply (2) by -}b: & \text{-}bcx - bdy &=& \text{-}bq \end{array}\)
\(\displaystyle \text{Add: } \: (ad-bc)x \:=\:dp - bq \quad\Rightarrow\quad \boxed{x \;=\;\dfrac{dp-bq}{ad-bc}}\)
\(\displaystyle \begin{array}{cccccccccc}\text{Multiply (1) by -}c: & \text{-}acx - bcy &=& \text{-}cp \\ \text{Multiply (2) by }a: & acx + ady &=& aq \end{array}\)
\(\displaystyle \text{Add: } \: (ad-bc)y \:=\:aq-cp \quad\Rightarrow\quad \boxed{y \;=\;\dfrac{aq-cp}{ad-bc}}\)
Solve by Cramer's Rule.
\(\displaystyle \displaystyle x \;=\;\frac{\begin{vmatrix}p & b\\ q&d\end{vmatrix}} {\begin{vmatrix}a&b\\c&d\end{vmatrix}} \quad\Rightarrow\quad \boxed{x \;=\;\dfrac{dp-bq}{ad-bc}}\)
\(\displaystyle \displaystyle y \;=\;\frac{\begin{vmatrix}a & p\\ c&q\end{vmatrix}} {\begin{vmatrix}a&b\\c&d\end{vmatrix}} \quad\Rightarrow\quad \boxed{y \;=\;\dfrac{aq-cp}{ad-bc}}\)
soroban,
if that's "so lame," then certainly don't provide spelled out
solutions for the user!
And, Dogman, please read:
http://www.freemathhelp.com/forum/threads/41538-Read-Before-Posting!!
Silence. Kindly follow the rules. Simple. Don't get on here and act defensive by
deflecting onto me what is to be your personal responsibility.
Act humble and appreciate people who are to help you, not
do your work for you.
Get an attitude adjustment sooner than later.
Edit:
A private message was sent to admin regarding the insolence of cobber82,
as well as the undermining by soroban to the OP of the intent of this math
message board to *assist* users in their problem solving.
Last edited:
I see the fantastic help that is being given on this forum.
I also understand the volunteers who are giving their time to "do work" for others. But in half the threads I see is you being rude man.
Indeed, some many have slightly broken the rules, in order to try and get an answer to their question. But does repetitively posting a link to the forum rules help? In anyway shape or form?
What kind of power hungry random idiot of a math board private message the admin because of one small comment from a random new user?
Do you have a high level of autism, or just lacking in social skills?
I also understand the volunteers who are giving their time to "do work" for others. But in half the threads I see is you being rude man.
Indeed, some many have slightly broken the rules, in order to try and get an answer to their question. But does repetitively posting a link to the forum rules help? In anyway shape or form?
What kind of power hungry random idiot of a math board private message the admin because of one small comment from a random new user?
Do you have a high level of autism, or just lacking in social skills?
D
Deleted member 4993
Guest
Please refrain from personal attacks - any more of these and you will be banned!
There was nothing rude about "lookagain"s response - he called a spade a spade. Most of us dislike spoon-feeding - and dislike lazy people.
So watch your "typing" finger.....
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,990
Thank you, thank you.
Methinks that cobber82 is trouble maker.