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Math Help - Solving a system of equations using Cramer's rule.

  1. #1
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    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?
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  2. #2
    Master Of Puppets
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    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
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  3. #3
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    Go here:

    Cramer's Rule

    it gives an easy to follow method of solving almost the exact problem you have.
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  4. #4
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    Cramer's Rule

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

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


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


    For z, you replace column 3, \begin{bmatrix}<br />
c\\ <br />
f\\ <br />
i<br />
\end{bmatrix}, with \begin{bmatrix}<br />
4\\ <br />
10\\ <br />
3<br />
\end{bmatrix} and take the determinant.
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  5. #5
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    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.
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  6. #6
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    Expanding the determinant down column 3 will be the easiest.

    \begin{vmatrix}<br />
2 & -1 & 0\\ <br />
3 & 1 & -1\\ <br />
0 & 1 & 1<br />
\end{vmatrix}\rightarrow 0*\begin{vmatrix}<br />
3 & 1\\ <br />
0 & 1<br />
\end{vmatrix}-(-1)*\begin{vmatrix}<br />
2 & -1\\ <br />
0 & 1<br />
\end{vmatrix}+1*\begin{vmatrix}<br />
2 & -1\\ <br />
3 & 1<br />
\end{vmatrix} \rightarrow 0+1*(2-0)+1*(2-(-3))=7

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

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

    Double click the images to see how they are entered.
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  7. #7
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    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

    <br />
\begin{vmatrix}<br />
2 & -1 & 0\\<br />
3 & 1 & -1\\<br />
0 & 1 & 1<br />
\end{vmatrix}<br />

    Then I defined determinants D_x, D_y and D_z:

    D_x:
    <br />
\begin{vmatrix}<br />
4 & -1 & 0\\<br />
10 & 1 & -1\\<br />
3 & 1 & 1<br />
\end{vmatrix}<br />

    D_y:
    <br />
\begin{vmatrix}<br />
2 & 4 & -1\\<br />
3 & 10 & 1\\<br />
0 & 3 & 1<br />
\end{vmatrix}<br />

    D_z:
    <br />
\begin{vmatrix}<br />
2 & -1 & 4\\<br />
3 & 1 & 10\\<br />
0 & 1 & 3<br />
\end{vmatrix}<br />

    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:
    <br />
\begin{vmatrix}<br />
+ & - & +\\<br />
- & + & -\\<br />
+ & - & +<br />
\end{vmatrix}<br />

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

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

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

    D:
    <br />
\begin{vmatrix}<br />
2 & -1 & 0\\<br />
3 & 1 & -1\\<br />
0 & 1 & 1<br />
\end{vmatrix}  \rightarrow 2 \begin{vmatrix}1 & -1\\<br />
1 & 1<br />
\end{vmatrix}- (-1) \begin{vmatrix}3 & -1\\<br />
0 & 3<br />
\end{vmatrix}+ 0\begin{vmatrix}3 & 1\\<br />
0 & 1<br />
\end{vmatrix} \rightarrow 2(1-(-1))-(-1)(3-0)+ 0(3-0)  \rightarrow 4-(-3)+0=7<br />
    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!
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  8. #8
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by quikwerk View Post
    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

    <br />
\begin{vmatrix}<br />
2 & -1 & 0\\<br />
3 & 1 & -1\\<br />
0 & 1 & 1<br />
\end{vmatrix}<br />

    Then I defined determinants D_x, D_y and D_z:

    D_x:
    <br />
\begin{vmatrix}<br />
4 & -1 & 0\\<br />
10 & 1 & -1\\<br />
3 & 1 & 1<br />
\end{vmatrix}<br />

    D_y:
    <br />
\begin{vmatrix}<br />
2 & 4 & {\color{red}-1}\\<br />
3 & 10 & {\color{red}1}\\<br />
0 & 3 & 1<br />
\end{vmatrix}<br />

    D_z:
    <br />
\begin{vmatrix}<br />
2 & -1 & 4\\<br />
3 & 1 & 10\\<br />
0 & 1 & 3<br />
\end{vmatrix}<br />

    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:
    <br />
\begin{vmatrix}<br />
+ & - & +\\<br />
- & + & -\\<br />
+ & - & +<br />
\end{vmatrix}<br />

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

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

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

    D:
    <br />
\begin{vmatrix}<br />
2 & -1 & 0\\<br />
3 & 1 & -1\\<br />
0 & 1 & 1<br />
\end{vmatrix}  \rightarrow 2 \begin{vmatrix}1 & -1\\<br />
1 & 1<br />
\end{vmatrix}- (-1) \begin{vmatrix}3 & -1\\<br />
0 & 3<br />
\end{vmatrix}+ 0\begin{vmatrix}3 & 1\\<br />
0 & 1<br />
\end{vmatrix} \rightarrow 2(1-(-1))-(-1)(3-0)+ 0(3-0)  \rightarrow 4-(-3)+0=7<br />
    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:
    <br />
\begin{vmatrix}<br />
2 & 4 & 0\\<br />
3 & 10 & -1\\<br />
0 & 3 & 1<br />
\end{vmatrix}=14<br />
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  9. #9
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    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!
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