Show that the system of equations $$\begin{cases} 3x + y - z+u^2=0 \\ x - y + 2z+u=0 \\ 2x + 2y - 3z+2u=0 \end{cases}$$ can be solved for $x,y,u$ in terms of $z$; for $x,z,u$ in terms of $y$; for $y,z,u$ in terms of $x$; but not for $x,y,z$ in terms of $u$.
Proof: Let $\mathbf{f}:\mathbb{R}^4\to \mathbb{R}^3$ defined by $$\mathbf{f}(x,y,z,u)=(3x + y - z+u^2, x - y + 2z+u, 2x + 2y - 3z+2u).$$ Also we see that $\mathbf{f}(0,0,0,0)=(0,0,0)$ and $\mathbf{f}'(0,0,0,0)=A$ where $[A]$ has the following form (relative to the standard basis) $$[A]=\begin{bmatrix} 3 & 1 & -1 & 0 \\ 1 & -1 & 2 & 1 \\ 2 & 2 & -3 & 2 \end{bmatrix}$$ Our linear operator $A\in L(\mathbb{R}^4,\mathbb{R}^3)$ can be written as: $A(x,y,z,u)=A_1(x,y,u)+A_2(z)$ where $A_1(x,y,u)=A(x,y,0,u)$ and $A_2(z)=A(0,0,z,0)$. And $$[A_1]=\begin{bmatrix} 3 & 1 & 0 \\ 1 & -1 & 1 \\ 2 & 2 & 2 \end{bmatrix}$$ Since $\det[A_1]=-12\neq 0$ then $A_1$ is invertible. Then by implicit function theorem exists open neighborhood $W\in \mathbb{R}$ and $U\in \mathbb{R}^4$ of $0$ and $(0,0,0,0)$ respectively. Also to every $z\in W$ exists a unique $(x(z),y(z), u(z))$ such that $$(x(z),y(z), u(z),z)\in U \quad\text{and}\quad \mathbf{f}(x(z),y(z), u(z),z)=0.$$ Hence the system of equations can be solved for $x,y,u$ in terms of $z$. Analogous reasoning can be applied for another cases.
How to show rigorously that above system can not be solved for $x,y,z$ in terms of $u$?
Can anyone give the full answer? I would be very grateful for help!
$\endgroup$3 Answers
$\begingroup$Take the $(a_k,b_k,c_k)$ to mean the 3 vectors of the coefficients before (x,y,z) from the 3 equations.
The system is not solvable because these $3$ vectors are not linearly independent. Multiply the 2nd equation's vector by $-1$ and add it to the 1st one's vector. You get the vector of the 3rd equation.
This means the main determinant is $0$.
See "undetermined system". OK, if you do that operation I mentioned, you get that:
$u^2 - u = 2u$
i.e.
$u=0$ or $u=3$
So:
1) when $u=0$ or $u=3$ the system has infinitely many solutions (because it's practically a system of 2 equations with 3 unknowns, one of the equations just follows from the other 2);
2) when neither $u=0$ nor $u=3$ the system has no solutions (since it leads to a contradiction).
In case 2) it's all clear. In case 1) you cannot solve $x,y,z$ with $u$ because you have to ALSO make one of $x,y,z$ a parameter i.e. you cannot uniquely determine $x,y,z$ as functions of $u$ and $u$ only.
$\endgroup$ 3 $\begingroup$You made an error in your matrix $[A]$. The coefficient $A_{1,3}$ is $-1$ and not $+1$.
Fixing this error, you'll verify that the determinant of the $3 \times 3$ extracted matrix from $[A]$ is equal to $0$. This is why $x,y,z$ cannot be solved in term of $u$.
$\endgroup$ $\begingroup$Solving for $x, y, z$ in terms of $u$ presents us with three linear equations in three unknowns. From familiar linear algebra, the equations have a unique solution if the determinant of the matrix is nonzero. The matrix is $$\begin{bmatrix} 3 & 1 & -1 \\ 1 & -1 & 2 \\ 2 & 2 & -3 \end{bmatrix}$$ It is easy to see that the second and third rows add to the first row. The determinant is therefore zero.
Edited to add discussion of nonlinear aspect. While you seem to have proven, modulo a small error, that the nonlinear equations are solvable in a neighborhood of $(0,0,0)$, their form allows more to be inferred. Suppose $z$ is the independent variable. The matrix for $x,y$ obtained from the last two equations is $$\begin{bmatrix} 1 & -1 \\ 2 & 2 \end{bmatrix}$$ The determinant is easily seen to be nonzero. Thus, we can write $x$ and $y$ as linear functions of $u$ and $z$ based on those two equations. We then substitute those expressions for $x$ and $y$ in the first equation. It becomes a quadratic equation in $u$ with a coefficient dependent on $z$. The leading term is $u^2$ since everything else is linear in $u$ and $z$. Now, all quadratic equations with nonzero leading coefficient have roots, which however may be complex. As a result, if we allow $u$ to be complex, the nonlinear equations are solvable for all $z$, not just in a neighborhood of $z=0$. It takes some calculation to see what $z$ would allow $u$ to be real.
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