# Cramer's Rule

• May 30th 2006, 11:45 AM
kbryant05
Cramer's Rule
2x +2y+3z=10
4x-y+z=-5
5x-2y+6z=1

This is a problem using cramer's rule and I am lost trying to get this one to work out.
• May 30th 2006, 12:08 PM
TD!
Could you specify what the problem is? It's just applying Cramer's rule and thus filling in the formula. It comes down to computing 4 determinants, perhaps a lot of work but not difficult.

Do you have trouble in understanding the rule or with the arithmetic of the computations of the determinants?
• May 30th 2006, 12:16 PM
kbryant05
Cramer's Rule
I have problems with all the above on the determinants, arithmatic and getting the equation to work out properly. I have managed to do several other ones but maybe my dilexia is kicking in.
• May 30th 2006, 12:25 PM
TD!
Understanding the rule is not a problem? The determinant of the coefficient matrix is used 3 times, but you only have to compute it once of course. In the nominator, to solve for the i-th unknown, the i-th column is replaced by the column of constants. Its determinant divided by the determinant of the coefficient matrix gives the solution. For x, this is:

$
x = \frac{{\left| {\begin{array}{*{20}c}
{10} & 2 & 3 \\
{ - 5} & { - 1} & 1 \\
1 & { - 2} & 6 \\
\end{array}} \right|}}{{\left| {\begin{array}{*{20}c}
2 & 2 & 3 \\
4 & { - 1} & 1 \\
5 & { - 2} & 6 \\
\end{array}} \right|}} = \frac{{55}}{{ - 55}} = - 1
$

I'll show how to compute a determinant, I'll do the bottom one as an example.

$
\left| {\begin{array}{*{20}c}
2 & 2 & 3 \\
4 & { - 1} & 1 \\
5 & { - 2} & 6 \\
\end{array}} \right| = \left| {\begin{array}{*{20}c}
{10} & 0 & 5 \\
4 & { - 1} & 1 \\
7 & 0 & 9 \\
\end{array}} \right| = - \left| {\begin{array}{*{20}c}
{10} & 5 \\
7 & 9 \\
\end{array}} \right|
$

I added the second row to the last one and I added it twice to the first one, creating two zeroes in column 2 and leaving only a -1. You then expand into the (only one left) minor. We can now easily find:

$
- \left| {\begin{array}{*{20}c}
{10} & 5 \\
7 & 9 \\
\end{array}} \right| = 5 \cdot 7 - 10 \cdot 9 = 35 - 90 = - 55
$
• May 30th 2006, 12:34 PM
kbryant05
Still confused
my book comes up with the answers being {-1,3,2} I don't get it?
• May 30th 2006, 12:41 PM
TD!
I had overwritten a part of my own post, I added it again. The post should make more sense now :)
• May 30th 2006, 01:49 PM
CaptainBlack
Quote:

Originally Posted by kbryant05
2x +2y+3z=10
4x-y+z=-5
5x-2y+6z=1

This is a problem using cramer's rule and I am lost trying to get this one to work out.

Re write the equations in matrix form:

$
\left[\begin{array}{ccc}
2&+2&3\\
4&-1&1\\
5&-2&6
\end{array}\right]
\left[\begin{array}{c}
x\\
y\\
z
\end{array}\right]=
\left[\begin{array}{c}
10\\
-5\\
1
\end{array}\right]
$

Then use Cramers rule to invert the matrix, and then use:

$\left[ \begin{array}{c}
x\\
y\\
z
\end{array}\right]=
\left[\begin{array}{ccc}
2&+2&3\\
4&-1&1\\
5&-2&6
\end{array}\right]^{-1}
\left[\begin{array}{c}
10\\
-5\\
1
\end{array}\right]
$

to solve for $x,\ y$ and $z$.

RonL
• May 30th 2006, 01:51 PM
earboth
Quote:

Originally Posted by kbryant05
I have problems with all the above on the determinants, arithmatic and getting the equation to work out properly. I have managed to do several other ones but maybe my dilexia is kicking in.

Hello,

with 3X3-determinants there exist another method to calculate their values. Maybe this is of some help:

Because I don't know to write determinants in Latex I've attached an image to sho you what you have to calculate:

Copy the first two columns of the determinant and add them to the determinant. Then multiply always three numbers as I've indicated by black and red lines. You have to add the products which are produced by the black lines and then you have to subtract the products which are produced by the red lines.

Greetings

EB
• May 30th 2006, 02:02 PM
ThePerfectHacker
I have a question about a method of computing determinants.

In earboth's post he used the diagnol multiplication method. I believe it only works for 3x3 correct?
• May 30th 2006, 02:09 PM
TD!
It's more of a mnemonic named after Sarrus. It comes down to the same thing as expanding the determinant.
• May 30th 2006, 02:25 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
It's more of a mnemonic named after Sarrus. It comes down to the same thing as expanding the determinant.

But does it work for higher dimesnions?
• May 30th 2006, 09:24 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
I have a question about a method of computing determinants.
In earboth's post he used the diagnol multiplication method. I believe it only works for 3x3 correct?

Hello,

you are right. It is a kind of condensed Sarrus' rule.

But in school determinants are not very often bigger than 3X3. So this rule (if you are used to it and trained to use it) is very helpful. If you use integers only (like here) you can even solve such a problem much faster than those poor kids who only learned to hit keys on their calculators.

Greetings

EB
• May 31st 2006, 12:40 AM
TD!
Quote:

Originally Posted by ThePerfectHacker
But does it work for higher dimesnions?

Not with the same trick, there may be other tricks though :)