Using Cramer's rule to solve a system?

Question

i need some help solving this system using cramer's rule

xcosø - ysinø = 1
xsinø - ycosø = 1

for the unknowns x and y as a function of ø.

find (x^2) + (y^2) and show that the solution point (x,y) is always a constant distance from the origin.

Please include the steps as i am quite confused... would mean alot to me if it was easy to follow...

thx is advance..

Answer 1

You do realise this is going to be a pain in the **** to type in ASCII, don't you?

You can rewrite the system as a matrix equation

|cos ø, - sin ø| |x|
|sin ø, - cos ø| |y|

=

|1|
|1|.

Cramer's rule tells us that x is equal to the determinant of the matrix

|1, - sin ø |
|1, - cos ø|

divided by the determinant of

|cos ø, - sin ø|
|sin ø, - cos ø|.

This works out as

(-cos ø + sin ø) / (-cos² ø + sin² ø) = 1 / (cos ø + sin ø).

Similarly, Cramer's rule says that y is equal to the determinant of

|cos ø, 1|
|sin ø, 1|

divided by that of

|cos ø, - sin ø|
|sin ø, - cos ø|,

which works out as

(cos ø - sin ø) / (-cos² ø + sin² ø) = - 1 / (cos ø + sin ø).

Thus

x = 1 / (cos ø + sin ø),
y = - 1 / (cos ø + sin ø).

Now observe that the square of the distance of this solution from the origin is given by

x² + y² = 2 / (cos ø + sin ø)².

This *cannot* be a constant distance from the origin, so there's a mistake in the statement of the question. I think you'll find the original question should be

x cos ø - y sin ø = 1,
x sin ø + y cos ø = 1.

Then if you work through the problem using Cramer's rule as above, you'll find that

x = cos ø + sin ø
y = cos ø - sin ø,

and then the square of the distance from the origin becomes

(cos ø + sin ø)² + (cos ø - sin ø)²

= cos² ø + 2cos ø sin ø + sin² ø + cos² ø - 2cos ø sin ø + sin² ø

= 2(cos² ø + sin² ø) = 2,

constant, as required.

Answer 2

geez's solution is correct and complete, i.e. no mistakes.

the 2nd part of you problem however isn't clear.

x^2 + y^2 will be the equation of a circle IF it is equal to a constant. we know that x^2 + y^2 = 2/(1 + sin(2θ)), IF θ is a constant then you are done, x^2 + y^2 is a circle with radius = sqrt(2/(1 + sin(2θ)) ) in the xy-plane. HOWEVER, if θ is allowed to vary ," ...as a function of ø" as you said earlier, then the graph cannot be constant distance from the origin of the xy-plane. it would be a 3D graph and would look like a spiral closing in on itself.

Or if you want to restrict your observations to only cross-sections in the xy-plane then for a given θ the solution x^2 + y^2 = constant.

Answer 3

you opt for to be sure a set of determinants to unravel through Cramer's approach... to unravel for x, divide the determinant: 5 7 sixty 5 -9 by technique of the determinant 8 7 4 -9 which grants: (-40 5-455)/(-seventy 2-28)=-500/-100=5 to unravel for y, divide 8 5 4 sixty 5 by technique of 8 7 4 -9 and get (520-20)/(-100)=-5 and in case you verify you'll discover (5,-5) be sure this set of equations

Answer 4

Cramer's rule states that if A is the original matrix, and A_i is the matrix formed by replacing the ith column with the right-most (vector) column, then x_i = det(A_i)/det(A).

So A is this:
[ cosθ -sinθ ]
[ sinθ -cosθ ]

det(A) = (cosθ)(-cosθ) - (sinθ)(-sinθ) =
-cos²(θ) + sin²(θ) = sin²(θ) - cos²(θ)

A1 is this:
[ 1 -sinθ ]
[1 -cosθ ]

So det(A1) = -cos(θ) - (-sin(θ)) = sin(θ) - cos(θ).
Likewise, det(A2) = cos(θ) - sin(θ)

So x = [sin(θ) - cos(θ)] / [sin²(θ) - cos²(θ)] =
(sin(θ) - cos(θ)) / (sin(θ)+cos(θ))(sin(θ)-cos(θ)) =
1 / (sin(θ)+cos(θ)).

And y = [cos(θ) - sin(θ)] / [sin²(θ) - cos²(θ)] =
-[sin(θ) - cos(θ)] / [sin(θ)+cos(θ)][sin(θ)-cos(θ)] =
-1 / [sin(θ) + cos(θ)]

You can plug these back into the original equations to verify they're solutions.

Showing that (x,y) is a constant distance from the origin is the same as showing that √[(x-0)² + (y-0)²] is a constant.
x² + y² = 1²/(sin(θ)+cos(θ))² + (-1)²/(sin(θ)+cos(θ))² =
2 / (sin²(θ) + 2sin(θ)cos(θ) + cos²(θ)) =
2 / (1 + 2sin(θ)cos(θ) ) = 2 / (1 + sin(2θ) )
I don't see how the square root of this would be a constant. I might have made a mistake somewhere, but at the moment I can't find it.