Hi,

pleane any one give me proof for cramers method

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- Aug 27th 2007, 03:59 AMksssudhanvamatrices
Hi,

pleane any one give me proof for cramers method - Aug 27th 2007, 04:40 AMTKHunny
Have you tried an arbitrary solution of a 2 variable system? Just set up two equations with ALL arbitrary parameters and solve them using substitution or something else. Then do it again using Cramer's Method.

ax + by = c

dx + ey = f

Solve that every way you can and see what it looks like. - Aug 27th 2007, 09:32 AMThePerfectHacker
$\displaystyle \left\{ \begin{array}{cc}a_{11}x+a_{12}y=k_1\\a_{21}x+a_{2 2}y=k_2 \end{array} \right.$

Multiply the top equation by $\displaystyle a_{21}$ and bottom by $\displaystyle a_{12}$. Now subtract these equations and solve for $\displaystyle y$. (Really messy). Then do the same for $\displaystyle x$. - Aug 28th 2007, 12:56 AMksssudhanvamatrices
Thanks

but how can you prove that cramers method is correct?You may prove it by taking a system of arbitary equation.But how did Cramer invented this method?(how did he know that he can solve a system of equations by doing the steps we follow to solve the system of equations using cramers maethd?) - Aug 28th 2007, 04:57 AMTKHunny
Mr. Cramer probably solved a system of equations with arbitrary coefficients. Give it a try.

There are two ways to know for sure:

1) Ask Mr. Cramer. Too bad he no longer is with us.

2) Find the original text in which it first appears and for which the method was assigned his name. - Aug 28th 2007, 05:21 AMSoroban
Hello, ksssudhanva!

Cramer probably got tired of solving every system separately

. . and sought a generalized solution.

You could have done it, too.

We have: .$\displaystyle \begin{array}{cccc}ax + by & = & h & [1]\\ cx + dy & = & k & [2]\end{array}$

$\displaystyle \begin{array}{cccc}\text{Multiply [1] by }d: & adx + bdy & = & dh \\

\text{Multiply [2] by -}b: & \text{-}bcx - bdy & = & \text{-}bk \end{array}$

Add: .$\displaystyle adx - bcx \:=\:dh - bk\quad\Rightarrow\quad(ad-bc)x \:=\:dh-bk\quad\Rightarrow\quad\boxed{x \:=\:\frac{dh-bk}{ad-bc}}$

$\displaystyle \begin{array}{cccc}\text{Multiply [1] by -}c: & \text{-}acx - bcy & = & \text{-}ch \\

\text{Multiply [2] by }a: & acx + ady & = & ak\end{array}$

Add: .$\displaystyle ady - bcy \:=\:ak-ch\quad\Rightarrow\quad(ad-bc)y \:=\:ak-ch\quad\Rightarrow\quad\boxed{ y \:=\:\frac{ak-ch}{ad-bc}}$

The following is my speculation on what happened.

Then he wondered, "How am I going to memorize those formulas?"

He noticed that the denominators are the determinant of the coefficients:

. . . . . . $\displaystyle ad - bc \:=\:\begin{vmatrix}a & b \\ c & d\end{vmatrix}$

Then he did some mental juggling to see that:

. . . . . . $\displaystyle dh - bk \:=\:\begin{vmatrix}{\color{blue}h} & b \\{\color{blue}k} & d\end{vmatrix}$ . and . $\displaystyle ak-ch \:=\:\begin{vmatrix}a & {\color{blue}h} \\ c & {\color{blue}k}\end{vmatrix}$

. . where the constants replace the respective coefficients.

Then he said, "Hey, I may be onto something here . . . "

. . and tested this pattern for larger systems

. . and eventually proving the procedure in general.

Then he and his buddies traded high-fives and said, "It's Miller time!"

But, of course, I'm guessing . . .

- Aug 28th 2007, 06:48 AMThePerfectHacker
Here is the way to prove the

**generalized**version of Cramer's rule. :eek:

Given,

$\displaystyle A\bold{x}=\bold{b}$,

Where, $\displaystyle A=\left[ \begin{array}{cccc}a_{11}&a_{12}&...&a_{1n}\\a_{21 }&a_{22}&...&a_{2n}\\...&...&...&...\\a_{n1}&a_{n2 }&...&a_{nn}\end{array} \right]$ and $\displaystyle \bold{b} = \left[ \begin{array}{c}b_1\\b_2\\...\\b_n\end{array} \right]$.

Since $\displaystyle \det(A)\not = 0\implies A \mbox{ invertible }$.

So, there exists a unique solution given by,

$\displaystyle \bold{x} = A^{-1}\bold{b}=\frac{1}{\det(A)}\cdot \mbox{adj}(A)\bold{b} = \frac{1}{\det(A)}\left[ \begin{array}{cccc}C_{11}&C_{21}&...&C_{n1}\\C_{12 }&C_{22}&...&C_{n2}\\...&...&...&...\\C_{1n}&C_{2n }&...&C_{nn}\end{array} \right]\left[ \begin{array}{c}b_1\\b_2\\...\\b_n\end{array} \right]$.

Let me explained what just happened. There is a rule that $\displaystyle A^{-1}$ is equal to its "adjoint" matrix divided by its determinant. Now the "adjoint" matrix is the transpose of the "cofactor matrix". That is what those $\displaystyle C$'s area. They are cofactors. But they are written backwarks because of the transpose operator.

Multiply them out,

$\displaystyle \bold{x} = \frac{1}{\det(A)} \left[ \begin{array}{c}b_1C_{11}+b_2C_{21}+...+b_nC_{n1}\ \b_1C_{12}+b_2C_{22}+...+b_nC_{n2}\\...+...+...+.. .\\b_1C_{1n}+b_2C_{2n}+...+b_nC_{nn} \end{array}\right]$.

Now $\displaystyle \bold{x}$ is the solution matrix to the system of equations. The $\displaystyle k$-th entry in this matrix is:

$\displaystyle x_k = \frac{b_1C_{1k}+b_2C_{2k}+...+b_nC_{nk}}{\det(A)}$.

Now consider a matrix $\displaystyle M_k$ obtained from $\displaystyle A$ but its $\displaystyle k$-th coloum is replaced by $\displaystyle \bold{b}$. What is the determinant of this matrix? It is the same as $\displaystyle x_k$! Because if we compute the determinant of this matrix by cofactor expansion along the $\displaystyle k$-th coloum we get just that!

Thus,

$\displaystyle \boxed{ x_k = \frac{\det(M_k)}{\det(A)} }$

Q.E.D.