# Thread: Writing system of linear equations.

1. ## Writing system of linear equations.

Could someone help me with the folowing question please

Using the following data:

Write down the system of linear equations with
a1....... a11 as unknowns such that
f(xj) = yj , 1<=j <=11. Please write all powers as z^b.

Solution:
Would it be written as
f(0.540)= a1+a2(0.467)+a3(0.467)^2..... ?
And do that for all values of X and y?
Im just confused on what the question is exactly asking and how to contruct the system with z^b

2. ## System of linear equations solution check Please!

ANY help or TIPS would be GREATLY Appriciated!!

Using the following data

Write down the system of linear equations with a1+....... a11 as unknowns such that f(xj)= yj , 1<=j <=11. Please write all powers as z^b.

Solution: ???
Would it be written as
0.540= a1+ a2(0.467)+ a3(0.467)2+a4 (0.467)3+ a5(0.467)4+ a6(0.467)5+a7 (0.467)6+ a8(0.467)7+ a9(0.467)8+a10 (0.467)9+a11 (0.467)10 ???

How would you use this then to make coefficient matrix T where T stands for power ten?

3. Originally Posted by kensington
ANY help or TIPS would be GREATLY Appriciated!!

Using the following data

Write down the system of linear equations with a1+....... a11 as unknowns such that f(xj)= yj , 1<=j <=11. Please write all powers as z^b.

Solution: ???
Would it be written as
0.540= a1+ a2(0.467)+ a3(0.467)2+a4 (0.467)3+ a5(0.467)4+ a6(0.467)5+a7 (0.467)6+ a8(0.467)7+ a9(0.467)8+a10 (0.467)9+a11 (0.467)10 ???

How would you use this then to make coefficient matrix T where T stands for power ten?

That is a complete mystery, Try posting the question with the exact wording from the source, and more contextual background would also help.

CB

4. It looks to me like you are trying to find the 10th degree polynomial that will pass through the given 10 points.
Any such polynomial is of the form $\displaystyle y= a_1+ a_2x+ a_3x^2+ a_4x^3+ a_5x^4+ a_6x^5+ a_7x^6+ a_8x^7+ a_9x^8+ a_{10}x^9+ a_{11}x^{10}$.

Yes, from x= 0.467, y= 0.540, your first equation will be $\displaystyle a_1+ a_2(0.467)+ a_3(0.467)^2+ a_4(0.467)^3+ a_5(0.467)^4+ a_6(0.467)^5+ a_7(0.467)^6+ a_8(0.467)^7+ a_9(0.467)^8+ a_{10}(0.467)^9+ a_{11}(0.467)^{10}= 0.540$

The first row of your coefficient matrix will be those powers of 0.467:
$\displaystyle \begin{bmatrix}1 & 0.540 & (0.467)^2 & (0.467)^3 & (0.467)^4 & (0.467)^4 & (0.467)^5 & (0.467)^6 & (0.467)^7 & (0.467)^8 (0.467)^9 & (0.467)^{10}\end{bmatrix}$,
the second row those same powers of 1.443, etc. I still have no idea why that matrix would "stand for power 10"! I suspect that is a translation error.

5. Originally Posted by HallsofIvy
It looks to me like you are trying to find the 10th degree polynomial that will pass through the given 10 points.
Any such polynomial is of the form $\displaystyle y= a_1+ a_2x+ a_3x^2+ a_4x^3+ a_5x^4+ a_6x^5+ a_7x^6+ a_8x^7+ a_9x^8+ a_{10}x^9+ a_{11}x^{10}$.

Yes, from x= 0.467, y= 0.540, your first equation will be $\displaystyle a_1+ a_2(0.467)+ a_3(0.467)^2+ a_4(0.467)^3+ a_5(0.467)^4+ a_6(0.467)^5+ a_7(0.467)^6+ a_8(0.467)^7+ a_9(0.467)^8+ a_{10}(0.467)^9+ a_{11}(0.467)^{10}= 0.540$

The first row of your coefficient matrix will be those powers of 0.467:
$\displaystyle \begin{bmatrix}1 & 0.540 & (0.467)^2 & (0.467)^3 & (0.467)^4 & (0.467)^4 & (0.467)^5 & (0.467)^6 & (0.467)^7 & (0.467)^8 (0.467)^9 & (0.467)^{10}\end{bmatrix}$,
the second row those same powers of 1.443, etc. I still have no idea why that matrix would "stand for power 10"! I suspect that is a translation error.
Thank you sooo soo soo much! This is exactly what i thought. I was also really confused on the "strand for power 10" which is why i kept questioning the equation.

PS. Also why did u include a one in the first column? Is it because of the a1? If so, then why is the 0.540 included if we are just looking at the coefficents of the "a" ?
Below is the copied document of what is asked of us:

Thanks again Appriciate it