Given a matrix, how to build an equation or inequality using variables that gives unique solution?

Question

I want to solve a system of equations of the form $$ Ax = b $$ where $A$ is a $6\times 15$ matrix, $x$ is $15\times 1$, and $b$ is $6\times 1$. In the absence of other constraints, this system of equations does not have a unique solution.

Suppose that I have the following composite matrix, $C = [A\,|\,b]$: $$ \left[ \begin{array}{ccccccccccccccc|l} 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2A& 0& 0& e_{1}^{2} - A^2\\ 0& 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2B& 0& e_{2}^{2} - B^2\\ 0& 0& 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2C& e_{3}^{2} - C^2\\ 1& 1& 0& 1& 1& 0& -2& 0& 0& -2& 0& 0& 0& 0& 0& P^2\\ 1& 0& 1& 1& 0& 1& 0& -2& 0& 0& -2& 0& 0& 0& 0& Q^2\\ 0& 1& 1& 0& 1& 1& 0& 0& -2& 0& 0& -2& 0& 0& 0& R^2 \end{array} \right] $$

Can I build an equation or inequality using terms $A, B, C, P, Q, R, e_1, e_2, e_3$ That gives me only one solution? e.g. $A \neq B$; $P+Q+R = C$ and $e_1, e_2, e_3$ are not significant.

Moreover, can I say something like "If $A = B = C$, then the solution for space is between $-2$ and $5$ etc.?

I have used Gauss-Jordan Elimination. I have only used the coefficients of terms ($2$ for $2A,2B,2C$ and $1$ for all the elemenst of $b$). The result was the following:

$x_1 = 2x_8 + 2x_{11} + 2x_{15}$

$x_2 = 1 - x_5 - 2x_{14}$

$x_3 = 1 - x_6 - 2x_{15}$

$x_4 = 1 - 2x_8 - 2x_{11} - 2x_{13} - 2x_{15}$

$x_7 = 1/2 - x_10 - x_{13} - x_{14}$

$x_9 = 1/2 - x_{12} - x_{14} - x_{15}$

$x_5, x_6, x_8, x_10, x_{11}, x_{12}, x_{13}, x_{14}, x_{15}$ - free

But still, my question remains.

Answer 1

The matrix that represents the row operations that brings A to row echelon form is: $$E = \begin{bmatrix} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ -1&-1&0&1&0&0\\ -1&0&-1&0&1&0\\ 0&-1&-1&0&0&1\\ \end{bmatrix}$$

Multiplying on the left by $E$ brings $A$ to row echelon form. That is, $ref(A) = EA$. I get $$EA= \begin{array}{ccccccccccccccc} x_1&x_2&x_3&x_4&x_5&x_6&x_7&x_8&x_9&x_{10}&x_{11}&x_{12}&x_{13}&x_{14}&x_{15}\\ \hline 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2A& 0& 0\\ 0& 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2B& 0\\ 0& 0& 1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 2C\\ 0& 0& 0& 0& 0& 0&-2& 0& 0&-2& 0& 0& -2A& -2B& 0\\ 0& 0& 0& 0& 0& 0& 0&-2& 0& 0&-2& 0& -2A& 0& -2C\\ 0& 0& 0& 0& 0& 0& 0& 0&-2& 0& 0&-2& 0& -2B& -2C \end{array} $$

And performing the exact same row operations on $b$ I get,

$$Eb = \begin{bmatrix} e_{1}^{2} - A^2\\ e_{2}^{2} - B^2\\ e_{3}^{2} - C^2\\ A^2+B^2+P^2-e_1^2-e_2^2\\ A^2+C^2+Q^2-e_1^2-e_3^2\\ R^2+B^2+C^2-e_1^2-e_3^2 \end{bmatrix}$$

So, $x_1, x_2, x_3, x_7, x_8\,\text{and} \, x_9$ are the basic variables, and $x_4, x_5, x_6, x_{10}, x_{11}, x_{12}, x_{13}, x_{14}, x_{15}$ are free variables. Let's write equations for each of the basic variables in terms of the free variables... $$\begin{align} x_1 &= -x_4 -2Ax_{13} + e_{1}^{2} - A^2\\ x_2 &= -x_5 -2Bx_{14} + e_{2}^{2} - B^2\\ x_3 &= -x_6 -2Cx_{15} + e_{3}^{2} - C^2\\ -2x_7 &= 2x_{10}+ 2Ax_{13} + 2Bx_{14} + A^2+B^2+P^2-e_1^2-e_2^2\\ -2x_8 &= 2x_{11}+ 2Ax_{13} + 2Cx_{15} + A^2+C^2+Q^2-e_1^2-e_3^2\\ -2x_9 &= 2x_{12}+ 2Bx_{14} + 2Cx_{15} + R^2+B^2+C^2-e_1^2-e_3^2 \end{align}$$

Now, you can choose whatever value seems convenient for each of the free variables.

Answer 2

You can use the method of linear least squares to find the solution to $\text{min}_x ||Ax-b||_2$ where $||\cdot ||_2$ represent the vector 2-norm. This will give a solution to the equation $Ax=b$ if one exists, and it will still give a "closest fit" solution even when an actual solution does not exist. The solution is only unique when $A$ is full rank. To enforce a kind of uniqueness you can restrict $x$ such that $x$ has smallest norm and satisfies the minimization problem. I forget off the top of my head but I believe using the pseudo inverse from the singular value decomposition will automatically give the $x$ of smallest 2-norm.