anonymous
  • anonymous
How do you write this system of equations in matrix form? x²+y²=17 x+y=5
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I tried this, but that can't be right: [ 1 1 | 17 ] [ 1 1 | 5 ]
Owlcoffee
  • Owlcoffee
I might be wrong on this one but: \[x^2+y^2=17\] \[x+y=5\] Then: \[x.x+y.y=17\] \[x+y=5\] \[\iff \left[\begin{matrix}x & y & 17 \\ 1 & 1 & 5\end{matrix}\right]\]
Owlcoffee
  • Owlcoffee
Are we speaking about Cramers method of reduction?

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anonymous
  • anonymous
Hmm, no, not Cramers. Gauss-Jordan's elimination actually.
TrojanPoem
  • TrojanPoem
Gauss elimination works with linear equations and the second one isn't , so somehow we have to turn it into one
TrojanPoem
  • TrojanPoem
x²+y²=17 <<< if it was x^2 - y^2 It would have been easier.
anonymous
  • anonymous
Certain. I solved it using regular algebra, the results were 1 and 4. Was just wondering if it's possible using linear algebra too.
TrojanPoem
  • TrojanPoem
HOw about this ? (x+y) ^2 = 25 x^2 + y^2 + 2xy = 25 let x^2 = t , y^2 = k , xy = c t + k + 2c = 25 -> first eqn x^2+y^2 = 17 t + k = 17 -> 2nd
TrojanPoem
  • TrojanPoem
Now we can put it into a matrix.
TrojanPoem
  • TrojanPoem
|dw:1450012536227:dw|
anonymous
  • anonymous
Seem to be stuck eventually at t+k=42. I think I'll just stick to the algebraic solution but thank you both very much for the interesting ideas to solve this.
TrojanPoem
  • TrojanPoem
@Ankh Nope, It's not stuck like that. It has solution |dw:1450017373188:dw| -2c = -8 c = 4 (remember c = xy) xy = 4 y = 4/x t+k = 17 x^2 + y^2 = 17 x^2 + 16/x^2 = 17 x^4 + 16 = 17x^2 x^4 - 17x^2 +16 = 0 (x^2 - 16) (x^2-1) = 0 x^2 = 16 x= - or + 4 , x = + or - 1 SS = -1 , 1 , -4 , 4 You didn't said how to solve this, you said how to put it in Gauss elimination matrix.
TrojanPoem
  • TrojanPoem
typo: you didn't say*
TrojanPoem
  • TrojanPoem
To get y put in the values of x -4, 4, -1, 1 (corresponding to -1, 1, -4, 4)
TrojanPoem
  • TrojanPoem
You can confirm that this values are true by substituting once again in the main function.

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