Solving an overdetermined system of linear equations partially

linear algebrasystems of equations

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?

Thanks in advance

Best Answer

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.

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