# Homework Help: Matrix problem

1. Nov 29, 2012

### icesalmon

1. The problem statement, all variables and given/known data

Show that the following nonlinear system has 18 solutions if:

0 ≤ α ≤ 2∏
0 ≤ β ≤ 2∏
0 ≤ γ ≤ 2∏

sin(α) + 2cos(β) + 3tan(γ) = 0
2sin(α) + 5cos(β) + 3tan(γ) = 0
-sin(α) -5cos(β) + 5tan(γ) = 0

using the substitutions x = sin(α) y = cos(β) z = tan(γ)
3. The attempt at a solution
I went ahead and substituted and got:

x + 2y + 3z = 0
2x + 5y + 3z = 0
-x -5y + 5z = 0

and put it into an augmented matrix with the coefficients on one side and the constants on the other.
I also tried computing the values at 0 and 2∏ using the substitution, giving me:

[ 0 2 0 | 0 ]
[ 0 5 0 | 0 ]
[ 0 -5 0 | 0 ]

for both values, and I got 16 more matrices for all the other values that I use with these functions, I seem to be missing one. What i'm thinking about, though: don't they all contain the trivial solution? How are they each different in their own way?

Last edited: Nov 29, 2012
2. Nov 29, 2012

### HallsofIvy

Re: matrix

Okay, and what did you do with it then? What are x, y, and z?

3. Nov 29, 2012

### WillemBouwer

Re: matrix

I think that you should't be substituting the values:
0 ≤ α ≤ 2∏
0 ≤ β ≤ 2∏
0 ≤ γ ≤ 2∏

This merely says that the graphs of sin cos and tan is from 0-360 degrees, so for the sin and cos grapghs it should be ne full cycle and for the tan graph it is 2 cycles.

I would think you should find the answers of the variables in the matrix and see how many time that value can occur on the specific graph. Say you get x to be 0.5, how many times can sin(α) on the graph be equal to 0.5... Well that is how I would approach this. Might be wrong but thought you would appreciate any feedback...

4. Nov 29, 2012

### icesalmon

Re: matrix

x, y, z are variables that make the system linear, so ones I can use to row reduce the matrix after the substitution.

What I did was plug the common angles into the functions sin(alpha) cos(beta) tan(gamma), 0pi, pi/4, pi/2, 3pi/4, pi, 5pi/4 3pi/2, 7pi/4, 2pi, pi/6, pi/3, 2pi/3, 5pi/6, , 7pi/6, 4pi/3, 5pi/3, 11pi/6 but these aren't even solutions, they are values to the trig functions so i'm not sure what I did makes any sense. Let me try row reducing this, after all that's how I find solutions to any other matrix I work with.

and as expected, the trivial solution. I guess this is either solution #1 or solution #18

Last edited: Nov 29, 2012
5. Nov 30, 2012

### Staff: Mentor

Re: matrix

No one else seems to have noticed this, so I'll mention it. Your work above is incorrect.

When you have a homogeneous system (all constants on the right side are zero), you can omit that column in your augmented matrix. After all no matter what multiple of any row you add to any other row, the far right column will never have any values other than zero.

The matrix you started with --
$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 5 & -5\end{bmatrix}$$
-- row reduces to the identity matrix. (Note that I multiplied the 3rd row of your matrix by -1 to get the matrix above.

Given that the matrix reduces to I3, what does that say about your solution in terms of x, y, and z?

What does it say about the solution in terms of sin(α), cos(β), and tan(γ), given the constraints on α, β, and γ?

Finally, what are the solutions for α, β, and γ?

6. Nov 30, 2012

### icesalmon

Re: matrix

the values for x, y, and z are the same, 0
that the solution for sin(alpha) cos(beta) and tan(gamma) are the values where these functions are zero in the interval from 0 to 2pi
I have the values, alpha = gamma = pi, 2pi and beta = pi/2 and 3pi/2. :\

7. Nov 30, 2012

### Staff: Mentor

Re: matrix

Yes, and so you also have
sin(α) = 0
cos(β) = 0
tan(γ) = 0

Now find the solutions for these equations for α, β, and γ in the interval [0, $2\pi$]. A sketch of the graph of each function will be helpful.

All told, you should get the 18 solutions that are mentioned in the first post.

8. Dec 1, 2012

### icesalmon

α = sin-1(0)
β = cos-1(0)
γ = tan-1(0)

α = γ = 0∏,∏,2∏

β = ∏/2, 3∏/2

I graphed them on my calculator, the zeros I see are in accordance with my answers for alpha, beta, and gamma.

What am I not seeing here.

9. Dec 1, 2012

### Staff: Mentor

You have 3 choices for α, 2 choices for β, and 3 choices for γ. All together, you have 3 * 2 * 3 solutions.

10. Dec 1, 2012

### icesalmon

okay, why do we multiply these solutions together?
i've been counting them up individually.

11. Dec 1, 2012

### Staff: Mentor

A solution will be the triple (α, β, γ). Since there are three choices for α, two choices for β, and three choices for γ, there will be 3 * 2 * 3 possible solutions.

Suppose you're in the market for a new car, which can come in three possible colors - red, white, black, with either a four-cylinder or V6 engine, and can come in one of three different body styles: convertible, 2-door sedan, or 4-door sedan.

Red, 4, convertible
Red, 4, 2Dr
Red, 4, 4Dr
Red, V6, convertible
Red, V6, 2Dr
Red, V6, 4Dr

These are all of the combinations with a red car. There are 6 more combinations for a white car, and 6 more for a black car, making 18 different possible choices all together.

12. Dec 2, 2012

thank you.