1. ## Gauss-Elimination Problems

I stuck with these 2 problems for a long time not sure that it is defined or not

Question 1: Give all solutions, if any, of

x + y =1
sin(x - y) =1/2

with 0 less than or equal to X and X is less than or equal to 3

Question 2: Give all solutions, if any, of

sinx +siny = -1
sinx - siny + 4cosz =1
sinx + siny + 2cosz =2

For this problem; after I did Gauss Ellimination, I got

[1 1 0 -1]
[0 1 -2 -1]
[0 0 1 3/2] Then z=3/2, y=2, x=-3

but siny=sin2=undefined and sinx=sin(-3)=undefined

Thank you

3. ## Re: Gauss-Elimination Problems

In Question 2, I believe you found the supposed values of sin(x), sin(y) and cos(z) correctly. Since sin(x) = -3 has no solutions, the original systems has no solutions, either.

For Question 1, express y through x from the first equation, substitute it into the second equation and use the fact that $\displaystyle \sin\alpha=1/2$ iff $\displaystyle \alpha=\pi/6+2\pi k$, $\displaystyle k\in\mathbb{Z}$ or $\displaystyle \alpha=5\pi/6+2\pi k$, $\displaystyle k\in\mathbb{Z}$. Select suitable values of x between 0 and 3.

4. ## Re: Gauss-Elimination Problems

Thank you emakorov,

I still dont get the question 1.

as you told me the next step should be

x + y = 1
sin(2x-1) =1/2 --> substitute eq1 in eq2

but I don't know what to do next?

[1 1 1]
[2 0 1/2] and so on?

5. ## Re: Gauss-Elimination Problems

Originally Posted by angelme
sin(2x-1) =1/2 --> substitute eq1 in eq2

but I don't know what to do next?

[1 1 1]
[2 0 1/2] and so on?
Of course not. Gauss elimination procedure applies only to a system of linear equations, where unknown variables are multiplied by constants and added. The equation sin(2x-1) =1/2 is nonlinear because it involves a sine applied to an expression with x. I gave the relevant fact in post #3 (in this case, $\displaystyle \alpha=2x-1$).

6. ## Re: Gauss-Elimination Problems

Hello, angelme!

The second problem has no real solutions.

Question 2: Give all solutions, if any, of: .$\displaystyle \begin{Bmatrix}\sin x +\sin y \:=\: -1 & [1] \\ \sin x - \sin y + 4\cos z \:=\:1 & [2] \\ \sin x + \sin y + 2\cos z \:=\:2 & [3] \end{Bmatrix}$

Substitute [1] into [3]: .$\displaystyle -1 + 2\cos z \:=\:2 \quad\Rightarrow\quad 2\cos z \:=\:3$

. . Hence: .$\displaystyle \cos z \:=\:\tfrac{3}{2}$ . . . no real roots.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way, Gaussian elimination could be used
. . with an appropriate substitution.

Let: .$\displaystyle \begin{Bmatrix}X &=& \sin x \\ Y &=& \sin y \\ Z &=& \cos z\end{Bmatrix}$

Then we have: .$\displaystyle \begin{bmatrix}X + Y \:=\:-1 \\ X - Y + 4Z \:=\:1 \\ X + Y + 2Z \:=\:2 \end{bmatrix} \quad\Rightarrow\quad \left|\begin{array}{ccc|c}1&1&0&\text{-1} \\ 1&\text{-}1&4&1 \\ 1&1&2&2 \end{array}\right|$

Gaussian elimination results in: .$\displaystyle \left|\begin{array}{ccc|c}1&0&0&\text{-}3 \\ 0&1&0&2 \\ 0&0&1&\frac{3}{2}\end{array}\right|$

And we have: .$\displaystyle \begin{Bmatrix}X \:=\:\text{-}3 & \Longrightarrow & \sin x \:=\:\text{-}3 \\ Y \:=\:2 & \Longrightarrow & \sin y \:=\:2 \\ Z \:=\:\frac{3}{2} & \Longrightarrow & \cos z \:=\:\frac{3}{2} \end{Bmatrix}$

. . None of the equations has a real solution.

7. ## Re: Gauss-Elimination Problems

Thank you Soroban

8. ## Re: Gauss-Elimination Problems

Hello again, angelme!

$\displaystyle \text{1: Give all solutions, if any, of: }\:\begin{Bmatrix}x + y &=&1 & [1] \\ \sin(x - y) &=& \frac{1}{2} & [2] \end{Bmatrix}\quad 0 \le x \le 3$

From [2], we have: .$\displaystyle \sin(x-y) \:=\:\tfrac{1}{2} \quad\Rightarrow\quad x-y \:=\:\arcsin(\tfrac{1}{2})$

$\displaystyle \begin{array}{ccccccc}\text{We have:} & x-y &=& \arcsin(\frac{1}{2}) \\ \text{Add [1]:} & x+y &=& 1 \end{array}$

. . . . . . . . . . $\displaystyle 2x \:=\:1 + \arcsin(\tfrac{1}{2})$

. . . . . . . . . . $\displaystyle 2x \:=\:\begin{Bmatrix}1 + \frac{\pi}{6} + 2\pi n \\ \\[-3mm]1 + \frac{5\pi}{6} + 2\pi n\end{Bmatrix}$

n . . . . . . . . . $\displaystyle x \:=\:\begin{Bmatrix}\frac{1}{2} + \frac{\pi}{12} + \pi n \\ \\[-3mm] \frac{1}{2} + \frac{5\pi}{12} + \pi n \end{Bmatrix}$

$\displaystyle \text{To satisfy the domain of }x,\:n = 0.$

$\displaystyle \text{Therefore: }\:(x,y) \;=\;\begin{Bmatrix}\left(\frac{1}{2}\!+\!\frac {\pi}{12},\:\frac{1}{2}\!-\!\frac{\pi}{12}\right) \\ \\[-3mm] \left(\frac{1}{2}\!+\!\frac{5\pi}{12},\:\frac{1}{2 }\!-\!\frac{5\pi}{12}\right) \end{Bmatrix}$