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I have an underdetermined system of linear equations of the form $Ax = b$, where:

$$ A = \begin{bmatrix} a_{20} & a_{10} & a_{00} & 0 & 0 \\ 0 & a_{21} & a_{11} & a_{01} & 0 \\ 0 & 0 & a_{22} & a_{12} & a_{02} \\ \end{bmatrix}, \quad x = \begin{bmatrix} x_{0} \\ x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ \end{bmatrix},\quad b = \begin{bmatrix} b_{0} \\ b_{1} \\ b_{2} \\ \end{bmatrix}$$

and

I am interested in solving for $x_2$. Is it possible to get a unique solution for $x_2$ in this underdetermined system?

Additionally, suppose $x_0$ and $x_1$ are known to us. How would this affect the solution for $x_2$? I am interested in both the general solution where $x_0$ and $x_1$ are not known, and the specific solution where $x_0$ and $x_1$ are known.

Any help or guidance would be greatly appreciated!

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1 Answer 1

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Since $A$ is not full column rank, it has nontrivial nullspace, i.e., $\mathcal{N}(A) \neq \{0\}$. Hence, any vector in the affine set $\{x \ |\ x = \bar{x} + w, \ A\bar{x} = b,\ \bar{x}\in \mathcal{R}(A^T),\ \forall w\in\mathcal{N}(A)\}$ is a solution to your system. Here, $\bar{x}$ is the unique solution to the system in the domain of $A$, $\mathcal{R}(A^T)$.

If $a_{00}\neq 0$ and you know $(x_0, x_1)$, then from the first equation $a_{20}x_0 + a_{10}x_1+ a_{00}x_2 = b_0$, you can uniquely determine $x_2$.

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