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cramer's rule
- Thread starter cloudgirl
- Start date
Cramer's rule uses deteminants.
Given the system:
\(\displaystyle a_{1}x+b_{1}y=k_{1}\)
\(\displaystyle a_{2}x+b_{2}y=k_{2}\)
\(\displaystyle x=\frac{\begin{vmatrix}k_{1}&b_{1}\\k_{2}&b_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}\)
\(\displaystyle y=\frac{\begin{vmatrix}a_{1}&k_{1}\\a_{2}&k_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}\)
Can you do a 2 by 2 determinant?.
Given the system:
\(\displaystyle a_{1}x+b_{1}y=k_{1}\)
\(\displaystyle a_{2}x+b_{2}y=k_{2}\)
\(\displaystyle x=\frac{\begin{vmatrix}k_{1}&b_{1}\\k_{2}&b_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}\)
\(\displaystyle y=\frac{\begin{vmatrix}a_{1}&k_{1}\\a_{2}&k_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{vmatrix}}\)
Can you do a 2 by 2 determinant?.
I'm pretty sure that is a negative since I do not know what that is.
I'm just trying to make this math course to finish my degree. I am so lost it isn't even funny.
I do understand how to do basic matrices...add/sub and even a little of the multiplication. But I have two problems that I am lost on...this is one of them and the other involves something about transformations.
I'm just trying to make this math course to finish my degree. I am so lost it isn't even funny.
I do understand how to do basic matrices...add/sub and even a little of the multiplication. But I have two problems that I am lost on...this is one of them and the other involves something about transformations.
5x + 6y =-1
5x + y = 4
Okay so it will be
5 6 -1
5 1 4
Never mind I do not understand at all. I mean AT all...so, I will just leave it blank. I'm already flunking this assignment, what is another 5 points LOL.
Thanks anyway! I will be back no doubt, but hopefully on something I at least half-way understand.
5x + y = 4
Okay so it will be
5 6 -1
5 1 4
Never mind I do not understand at all. I mean AT all...so, I will just leave it blank. I'm already flunking this assignment, what is another 5 points LOL.
Thanks anyway! I will be back no doubt, but hopefully on something I at least half-way understand.
I'll help you out. It is a matter of plugging in the coefficients. A lot of calculators do determinants.
Look at your coefficients that correspond to the general forms I gave you.
For the one in the numerator, you have \(\displaystyle \begin{vmatrix}-1&6\\4&1\end{vmatrix}=(-1)(1)-(6)(4)=-1-24=-25\)
For the one in the denominator, you have \(\displaystyle \begin{vmatrix}5&6\\5&1\end{vmatrix}=(5)(1)-(6)(5)=-25\)
\(\displaystyle \frac{-25}{-25}=1\)
That is your x value.
Now, you try the y. Let me know what you get. You should get y=1.
Look at your coefficients that correspond to the general forms I gave you.
For the one in the numerator, you have \(\displaystyle \begin{vmatrix}-1&6\\4&1\end{vmatrix}=(-1)(1)-(6)(4)=-1-24=-25\)
For the one in the denominator, you have \(\displaystyle \begin{vmatrix}5&6\\5&1\end{vmatrix}=(5)(1)-(6)(5)=-25\)
\(\displaystyle \frac{-25}{-25}=1\)
That is your x value.
Now, you try the y. Let me know what you get. You should get y=1.