X+py+(p^2)z=(p^3); x+qy+(q^2)z=(q^3); x+ry+(r^2)z=(r^3). How to solve these equations without Cramer's rule?

Question

X+py+(p^2)z=(p^3); x+qy+(q^2)z=(q^3); x+ry+(r^2)z=(r^3). How to solve these equations without Cramer's rule?

Answer 1

The problem is a rather tricky one. Well, you may employ the following method:

All the three given equations have the common form x+ty+t²z=t³

(if we replace p, q and r by t)

So we may say that p, q and r are the roots of the cubic equation

x+ty+t²z=t³, that is, t³–t²z–ty–x=0.

(where t is the variable and x, y, z are coefficients)

Now, from the relation between the roots and coefficients of an equation,

x=pqr; y=–(pq+qr+rp) and z=p+q+r.

Don't expect that this method will be applicable to all problems of this type. In fact this is a special case. For other problems follow THE nerd's suggestion carefully.

Answer 2

I'm not sure that I'm understanding the problem correctly, but are the p, q, and r's just arbitrary constants? If so, follow this method:

x + py + p²z = p³ ...In this equation, the p's are just some constant.

x + qy + q²z = q³...In this equation, the q's are just some constant.

x + ry +  r²z = r³ ...In this equation, the r's are just some constant.

We could use subsitution, but it gets too messy after a while. Plus, you probably want to use a method involving a matrix right?

I'd do it like this: set up a matrix of these equations.

Ax = B   , where A, B are matrices, and x is unknowns.

| 1   p   p²  | p³ |

| 1   q   q²  | q³ |

| 1   r    p²  | r³  |      

This is called the augmented matrix. On the left side, you have the constants in front of the x, y, z in the equation.  On the right side, you have the values (constants) to which the equations are equal.

You could use elementary row operations to solve this easily, for x, y, and z

Example)

|1  2  1| 1|

|0  1  0| 2| ---> - R3 + R1

|0  0  1| 4|

|1  2  0| -3|

|0  1  0|  2| ---> - 2R2 + R1

|0  0  1|  4|

|1  0  0| -7|

|0  1  0|  2|

|0  0  1|  4|

This is reduced row echelon form. Now we can see that:

x = -7, y = 2, z = 4

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This is one way, you could also find the inverse, by using the adjoint, There are many ways to solve systems like this.

Hope this helps!

Answer 3

substitution