System of three non-linear equations

Question

Can anyone help me solve this system of equations? $$a_1 a_3 -a_2 ^2=0$$ $$a_1+a_3-2a_2-16=0$$ $$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$ After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Thank you for your time

Answer 1

By the second equation we get $$a_3=2a_2+16-a_1$$ so we get in (1) $$2a_1a_2+16a_1-a_1^2-a_2^2=0$$ (I)

and in (3)

$$2a_1a_2+16a_1-a_1^2+64a_1-a_2^2-16a_2-64=0$$(II)

with $$2a_1a_2=a_1^2-16a_1+a_2^2$$ we get in (II):

We get $$4a_1-a_2=4$$

Now e can eliminate $a_2$ $$2a_1(4a_1-4)+16a_1-a_1^2-(4a_1-4)^2=0$$ Now you can solve this equation for $a_1$.

Answer 2

Subtract the first row from the last row of the system: then you get $$ \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 64a_1 -16 a_2 -64 &=0 \end{split} $$ Then multiply by 8 the second row and subtract it again from the third row: $$ \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 64a_1 -16 a_2 -64 &=0 \end{split}\implies \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 56a_1 -8a_3+64 &=0 \end{split} $$ Then you can proceed by substitution and do the following steps $$ \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ a_1 &=\frac{1}{7}(a_3-8) \end{split}\implies \begin{split} \frac{1}{7}a_3^2-\frac{8}{7}a_3-a_2^2 &=0\\ a_2&=\frac{4}{7}(a_3-15)\\ a_1&=\frac{1}{7}(a_3-8) \end{split} $$ and finally we have $$ \begin{split} 9a_3^2-424a_3-3600 &=0\\ a_2&=\frac{4}{7}(a_3-15)\\ a_1 &=\frac{1}{7}(a_3-8) \end{split} $$ where the first equation is a quadratic equation respect to the single $a_3$. In sum, a nice way to proceed for such system is to see if you can subtract a multiple of one row from another, in order to simplify the structure and possibly arrive to a system where all the equations are linear except one, which however contains only one of the variables.

Answer 3

As long as $a_1a_3-a_2^2= 0$ we have

$$ a_1+a_3-2a_2= 16\\ 64a_1 -16 a_2 =64 $$

or

$$ a_1+a_3-2a_2=16\\ 4a_1 - a_2 =4 $$

solving for $a_1,a_2$ we have

$$ a_1 = \frac 17(a_3-8)\\ a_2 = \frac 47(a_3-15) $$

substituting now into $a_1a_3-a_2^2= 0$ we have

$$ 9a_3^2-424a_3-3600=0 $$

and solving gives

$$ a_3 = \left\{\frac{100}{9},\ 36\right\} $$

etc.

Answer 4

I'm going to use $x=a_1, \ y=a_2, \ z=a_3$

$$xz - y^2=0 \tag A$$ $$x + z - 2y - 16 = 0 \tag B$$ $$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$

Substitute $xz = y^2$ from (A) into (C)

\begin{align} xz + 64x - y^2 - 16y - 64 &= 0 \\ y^2 + 64x - y^2 - 16y - 64 &= 0 \\ 64x - 16y - 64 &= 0 \\ y &= 4x - 4 \tag D \end{align}

Substitute $y = (4x - 4)$ from (D) into (B)

\begin{align} x + z - 2y - 16 = 0 \\ x + z - 2(4x - 4) - 16 = 0 \\ -7x + z - 8 &= 0 \\ z &= (7x + 8) \tag E \end{align}

Substitute (D) and (E) into (A) \begin{align} xz - y^2 &= 0 \\ x(7x + 8) - (4x - 4)^2 &= 0 \\ 7x^2 + 8x - 16x^2 + 32x - 16 &= 0 \\ -9x^2 + 40x - 16 &= 0 \\ -(9x - 4)(x - 4) &= 0 \\ x &\in \left\{ \dfrac 49, 4 \right\} \end{align}

We get $$ (a_1,a_2,a_3) = \left(\dfrac 49, -\dfrac{20}{9}, \dfrac{100}{9} \right) \ \text{and} \ (a_1,a_2,a_3) = (4, 12, 36)$$