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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

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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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  • $\begingroup$ Thanks for your reply. Is there advantage to this method comparing to general methods same as moore-penrose? $\endgroup$ – M a m a D Oct 30 '19 at 9:01
  • $\begingroup$ I'm not really the right person to ask here... If you use the pseudoinverse, we still need to check wether we found a solution or not by plugging in. In a specific case, I think Gaussian elimination or similar is still the way to go. But I'm a bit out of my depth here. $\endgroup$ – Milten Oct 30 '19 at 9:10

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