Solving a system of equations using Cramer's rule.

Mar 2010
49
0
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?
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Cramer's Rule

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

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


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


For z, you replace column 3, \(\displaystyle \begin{bmatrix}
c\\
f\\
i
\end{bmatrix}\), with \(\displaystyle \begin{bmatrix}
4\\
10\\
3
\end{bmatrix}\) and take the determinant.
 
Mar 2010
49
0
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.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Expanding the determinant down column 3 will be the easiest.

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

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

\(\displaystyle x=\frac{21}{7}=3\)

Double click the images to see how they are entered.
 
Mar 2010
49
0
Alrighty! Here is the working I used in its unabridged form:


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

\(\displaystyle
\begin{vmatrix}
2 & -1 & 0\\
3 & 1 & -1\\
0 & 1 & 1
\end{vmatrix}
\)

Then I defined determinants \(\displaystyle D_x\),\(\displaystyle D_y\) and \(\displaystyle D_z\):

\(\displaystyle D_x\):
\(\displaystyle
\begin{vmatrix}
4 & -1 & 0\\
10 & 1 & -1\\
3 & 1 & 1
\end{vmatrix}
\)

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

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

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:
\(\displaystyle
\begin{vmatrix}
+ & - & +\\
- & + & -\\
+ & - & +
\end{vmatrix}
\)

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

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

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

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

(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:

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

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!
 

masters

MHF Helper
Jan 2008
2,550
1,187
Big Stone Gap, Virginia
Alrighty! Here is the working I used in its unabridged form:


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

\(\displaystyle
\begin{vmatrix}
2 & -1 & 0\\
3 & 1 & -1\\
0 & 1 & 1
\end{vmatrix}
\)

Then I defined determinants \(\displaystyle D_x\),\(\displaystyle D_y\) and \(\displaystyle D_z\):

\(\displaystyle D_x\):
\(\displaystyle
\begin{vmatrix}
4 & -1 & 0\\
10 & 1 & -1\\
3 & 1 & 1
\end{vmatrix}
\)

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

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

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:
\(\displaystyle
\begin{vmatrix}
+ & - & +\\
- & + & -\\
+ & - & +
\end{vmatrix}
\)

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

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

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

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

(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:

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

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 \(\displaystyle D_y\) matrix is incorrect. I highlighted the errors in red.

This should be:

\(\displaystyle D_y\):
\(\displaystyle
\begin{vmatrix}
2 & 4 & 0\\
3 & 10 & -1\\
0 & 3 & 1
\end{vmatrix}=14
\)
 
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Reactions: quikwerk
Mar 2010
49
0
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!