# Solving an overdetermined system of linear equations partially

I have this over-determined system of $$6$$ equations in $$4$$ unknowns.

$$\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\\ a_{51}&a_{52}&a_{53}&a_{54}\\ a_{61}&a_{62}&a_{63}&a_{64} \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} b_1\\ b_2\\ b_3\\ b_4\\ b_5\\ b_6 \end{bmatrix}$$

There are some methods same as Moore-Penrose inverse or QR decomposition to solve such systems. I have an idea I would like to know if it is true. Since the rows are independent of each other, we can solve the first $$4 \times 4$$ system

$$\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} b_1\\ b_2\\ b_3\\ b_4 \end{bmatrix}$$

If determinant of coefficients matrix is nonzero it will have a unique solution. Now we are left with an under-determined system as follows

$$\begin{bmatrix} a_{51}&a_{52}&a_{53}&a_{54}\\ a_{61}&a_{62}&a_{63}&a_{64} \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} b_5\\ b_6 \end{bmatrix}$$

that can be solved in parameterized manner. Is this a true method to solve such systems?

If the $$4$$x$$4$$-system has a unique solution, then there is no more freedom for $$x$$. We should check whether the solution also satisfies the last $$2$$x$$4$$-system. If it does, then we found the unique solution. If not, there is no solution.
You could find the general solution to the $$2$$x$$4$$-system by parameterising, but it is easier to just plug in the known $$4$$x$$4$$-solution.