Solving a system of equations using Cramer's rule.

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May 13th 2010, 02:04 PM
quikwerk

Solving a system of equations using Cramer's rule.

OK, I have this question here that says:

Using Cramer's rule, solve

2x-y=4
3x+y-z=10
y+z=3

I have no idea how to show my working on the board, but having gone through the working to the best of my ability I came up with the answer:
x=2.33 y=1.55 and z=0.11

I'm certain that I'm wrong, I checked using a math program, but I can't tell where I went wrong with my working.
Could I get a little help figuring out how to post my working, and then troubleshoot it?
May 13th 2010, 02:16 PM
pickslides

Do you know how to find a determinant of a 3x3 mtrix? If so applying cramer's rule is quite easy, have a read.

Cramer's rule - Wikipedia, the free encyclopedia

Spoiler:

$x=3,y=2,z=1$
May 13th 2010, 02:18 PM
shenanigans87

Go here:

Cramer's Rule

it gives an easy to follow method of solving almost the exact problem you have.
May 13th 2010, 02:23 PM
dwsmith

Cramer's Rule

$x,y,z=\frac{\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}}{\begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix}}$

For x, you replace column 1, $\begin{bmatrix} a\\ d\\ g \end{bmatrix}$ , with $\begin{bmatrix} 4\\ 10\\ 3 \end{bmatrix}$ and take the determinant.

For y, you replace column 2, $\begin{bmatrix} b\\ e\\ h \end{bmatrix}$ , with $\begin{bmatrix} 4\\ 10\\ 3 \end{bmatrix}$ and take the determinant.

For z, you replace column 3, $\begin{bmatrix} c\\ f\\ i \end{bmatrix}$ , with $\begin{bmatrix} 4\\ 10\\ 3 \end{bmatrix}$ and take the determinant.
May 13th 2010, 03:52 PM
quikwerk

I have some familiarity with it but my book has a very poor explanation of the concept--I had to search the internet just to find out that I had to use alternating signs.

I did the working by expanding the determinants by their minors on the first row.
I will do the working again by using the main diagonal(?) method, and report the result.
I would like to try to "master"(to some degree) the minor method and fix my flawed working.
That said, how can I post said working?
I think I can use latex but I'm not sure how to make it look good.
May 13th 2010, 04:14 PM
dwsmith

Expanding the determinant down column 3 will be the easiest.

$\begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix}\rightarrow 0*\begin{vmatrix} 3 & 1\\ 0 & 1 \end{vmatrix}-(-1)*\begin{vmatrix} 2 & -1\\ 0 & 1 \end{vmatrix}+1*\begin{vmatrix} 2 & -1\\ 3 & 1 \end{vmatrix}$ $\rightarrow 0+1*(2-0)+1*(2-(-3))=7$

Again it is best to expand by column 3.
$x=\frac{\begin{vmatrix} 4 & -1 & 0\\ 10 & 1 & -1\\ 3 & 1 & 1 \end{vmatrix}}{7}\rightarrow 0*\begin{vmatrix} 10 & 1\\ 3 & 1 \end{vmatrix}-(-1)*\begin{vmatrix} 4 & -1\\ 3 & 1 \end{vmatrix}+1*\begin{vmatrix} 4 & -1\\ 10 & 1 \end{vmatrix}$
$=0+(4+3)+(4+10)=21$

$x=\frac{21}{7}=3$

Double click the images to see how they are entered.
May 14th 2010, 12:01 PM
quikwerk

Alrighty! Here is the working I used in its unabridged form:

I started my working by defining system of equations and Determinant $D$ :
$2x-y=4$
$3x+y-z=10$
$y+z=3$

$ \begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix} $

Then I defined determinants $D_x$ , $D_y$ and $D_z$ :

$D_x$ :
$ \begin{vmatrix} 4 & -1 & 0\\ 10 & 1 & -1\\ 3 & 1 & 1 \end{vmatrix} $

$D_y$ :
$ \begin{vmatrix} 2 & 4 & -1\\ 3 & 10 & 1\\ 0 & 3 & 1 \end{vmatrix} $

$D_z$ :
$ \begin{vmatrix} 2 & -1 & 4\\ 3 & 1 & 10\\ 0 & 1 & 3 \end{vmatrix} $

I expanded the determinants by their minor elements, in this instance I expanded them all about their first row elements.
I used the following template to help me remember the signs I had to use:
$ \begin{vmatrix} + & - & +\\ - & + & -\\ + & - & + \end{vmatrix} $

$D_x$ :
$ \begin{vmatrix} 4 & -1 & 0\\ 10 & 1 & -1\\ 3 & 1 & 1 \end{vmatrix}$ $\rightarrow 4 \begin{vmatrix}1 & -1\\ 1 & 1 \end{vmatrix}-(-1)\begin{vmatrix}10 & -1\\ 3 & 1 \end{vmatrix}+0\begin{vmatrix}10 & 1\\ 3 & 1 \end{vmatrix}\rightarrow 4(1-(-1))-(-1)(10-(-3))+0(10-3)$ $\rightarrow 8-(-13)+0=21 $
Final value of $D_x = 21$

$D_y$ :
$ \begin{vmatrix} 2 & 4 & -1\\ 3 & 10 & 1\\ 0 & 3 & 1 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}10 & 1\\ 3 & 1 \end{vmatrix}- 4 \begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix}+(-1)\begin{vmatrix}3 & 10\\ 0 & 3 \end{vmatrix} \rightarrow 2(10-3)-4(3-0)+(-1)(9-0)$ $\rightarrow 14-12+(-9)=-7 $
Final value of $D_y = -7$

$D_z$ :
$ \begin{vmatrix} 2 & -1 & 4\\ 3 & 1 & 10\\ 0 & 1 & 3 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}3 & 10\\ 1 & 3 \end{vmatrix}- (-1) \begin{vmatrix}3 & 10\\ 0 & 3 \end{vmatrix}+ 4\begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix} \rightarrow 2(3-10)-(-1)(9-0)+ 4(3-0)$ $\rightarrow -14-(-9)+12=-7 $
Final value of $D_z=-7$

$D$ :
$ \begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}1 & -1\\ 1 & 1 \end{vmatrix}- (-1) \begin{vmatrix}3 & -1\\ 0 & 3 \end{vmatrix}+ 0\begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix} \rightarrow 2(1-(-1))-(-1)(3-0)+ 0(3-0)$ $\rightarrow 4-(-3)+0=7 $
Final value of $D=7$

(I can already see where this might be going wrong.
Interestingly enough, I find myself getting a totally different answer from the first one I worked out.)

In order to find out what the values of x, y and z are, I divided the values of their determinants by D:

$x=\frac{D_x}{D}\rightarrow\frac{21}{7}=3$
$y=\frac{D_y}{D}\rightarrow\frac{-7}{7}=-1$
$z=\frac{D_z}{D}\rightarrow\frac{7}{7}=1$

Threrefore: x=3, y=-1, z=1.

I know that the solution for y is wrong but I can't see what I did wrong.

Thank you for your help so far!
May 14th 2010, 12:20 PM
masters

Quote:

Originally Posted by quikwerk

Alrighty! Here is the working I used in its unabridged form:

I started my working by defining system of equations and Determinant $D$ :
$2x-y=4$
$3x+y-z=10$
$y+z=3$

$ \begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix} $

Then I defined determinants $D_x$ , $D_y$ and $D_z$ :

$D_x$ :
$ \begin{vmatrix} 4 & -1 & 0\\ 10 & 1 & -1\\ 3 & 1 & 1 \end{vmatrix} $

$D_y$ :
$ \begin{vmatrix} 2 & 4 & {\color{red}-1}\\ 3 & 10 & {\color{red}1}\\ 0 & 3 & 1 \end{vmatrix} $

$D_z$ :
$ \begin{vmatrix} 2 & -1 & 4\\ 3 & 1 & 10\\ 0 & 1 & 3 \end{vmatrix} $

I expanded the determinants by their minor elements, in this instance I expanded them all about their first row elements.
I used the following template to help me remember the signs I had to use:
$ \begin{vmatrix} + & - & +\\ - & + & -\\ + & - & + \end{vmatrix} $

$D_x$ :
$ \begin{vmatrix} 4 & -1 & 0\\ 10 & 1 & -1\\ 3 & 1 & 1 \end{vmatrix}$ $\rightarrow 4 \begin{vmatrix}1 & -1\\ 1 & 1 \end{vmatrix}-(-1)\begin{vmatrix}10 & -1\\ 3 & 1 \end{vmatrix}+0\begin{vmatrix}10 & 1\\ 3 & 1 \end{vmatrix}\rightarrow 4(1-(-1))-(-1)(10-(-3))+0(10-3)$ $\rightarrow 8-(-13)+0=21 $
Final value of $D_x = 21$

$D_y$ :
$ \begin{vmatrix} 2 & 4 & {\color{red}-1}\\ 3 & 10 & {\color{red}1}\\ 0 & 3 & 1 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}10 & 1\\ 3 & 1 \end{vmatrix}- 4 \begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix}+(-1)\begin{vmatrix}3 & 10\\ 0 & 3 \end{vmatrix} \rightarrow 2(10-3)-4(3-0)+(-1)(9-0)$ $\rightarrow 14-12+(-9)=-7 $
Final value of $D_y = -7$

$D_z$ :
$ \begin{vmatrix} 2 & -1 & 4\\ 3 & 1 & 10\\ 0 & 1 & 3 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}3 & 10\\ 1 & 3 \end{vmatrix}- (-1) \begin{vmatrix}3 & 10\\ 0 & 3 \end{vmatrix}+ 4\begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix} \rightarrow 2(3-10)-(-1)(9-0)+ 4(3-0)$ $\rightarrow -14-(-9)+12=-7 $
Final value of $D_z=-7$

$D$ :
$ \begin{vmatrix} 2 & -1 & 0\\ 3 & 1 & -1\\ 0 & 1 & 1 \end{vmatrix}$ $\rightarrow 2 \begin{vmatrix}1 & -1\\ 1 & 1 \end{vmatrix}- (-1) \begin{vmatrix}3 & -1\\ 0 & 3 \end{vmatrix}+ 0\begin{vmatrix}3 & 1\\ 0 & 1 \end{vmatrix} \rightarrow 2(1-(-1))-(-1)(3-0)+ 0(3-0)$ $\rightarrow 4-(-3)+0=7 $
Final value of $D=7$

(I can already see where this might be going wrong.
Interestingly enough, I find myself getting a totally different answer from the first one I worked out.)

In order to find out what the values of x, y and z are, I divided the values of their determinants by D:

$x=\frac{D_x}{D}\rightarrow\frac{21}{7}=3$
$y=\frac{D_y}{D}\rightarrow\frac{-7}{7}=-1$
$z=\frac{D_z}{D}\rightarrow\frac{7}{7}=1$

Threrefore: x=3, y=-1, z=1.

I know that the solution for y is wrong but I can't see what I did wrong.

Thank you for your help so far!

Hi quikwerk,

The 3rd column in your $D_y$ matrix is incorrect. I highlighted the errors in red.

This should be:

$D_y$ :
$ \begin{vmatrix} 2 & 4 & 0\\ 3 & 10 & -1\\ 0 & 3 & 1 \end{vmatrix}=14 $
May 14th 2010, 12:44 PM
quikwerk

Thank you so much Masters!
This stuff confuzzles the bejesus out of me, but I think I'm getting better!

Thank you all again for your help!