# Overdetermined linear system?

What should be the conditions on coefficients $$a_i$$ and $$b_i$$ such that the following overdetermined linear system of equations has unique solution. $$a_i x+y=b_i$$ where $$i=1,2,3...,n$$.

The system represents $$n$$ straight lines and it is possible to make them ins tersect at one point, hence sytem must have a unique solution for some $$a_i,b_i$$.

• Where is $b_i$ in your equation? – induction601 Dec 17 '18 at 6:35
• Typo! Fixed. Sorry – ersh Dec 17 '18 at 6:42

The system is over-determined for $$n>2$$ but here is a general method. We can write the augmented matrix for the system $$A\begin{bmatrix}x\\y\end{bmatrix}=B$$ as under:

$$\begin{bmatrix}a_1&1&\Big|&b_1\\a_2&1&\Big|&b_2\\\vdots&\vdots&\Big|&\vdots\\a_n&1&\Big|&b_n\end{bmatrix}$$

For a unique solution to exist, we should have $$2$$ linearly-independent equations to solve for the $$2$$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $$2$$. Recall that no more than $$2$$ vectors in $$\Bbb R^2$$ can be linearly independent, so the rank of the coefficient matrix $$A, \text{rank}(A)\le2$$. For $$\text{rank}(A)=2$$, we need to ensure at-least two $$a_i$$ are distinct. Say we have distinct $$a_1\ne0,a_2\ne a_1$$.

The point of intersection of $$a_1x+y=b_1,a_2x+y=b_2$$ is given by $$(X,Y)=\displaystyle\Big(\frac{b_1-b_2}{a_1-a_2},\frac{a_1b_2-a_2b_1}{a_1-a_2}\Big)$$.

$$\displaystyle R_i\to R_i-\frac{a_i}{a_1}\cdot R_1,\ i>1$$

$$\displaystyle R_j\to R_j-\frac{1-\frac{a_i}{a_1}}{1-\frac{a_2}{a_1}}\cdot R_2,\ j>2$$

$$\sim\begin{bmatrix}a_1&1&\Big|&b_1\\0&1-\frac{a_2}{a_1}&\Big|&b_2-\frac{a_2}{a_1}\cdot b_1\\0&0&\Big|&b'_3\\\vdots&\vdots&\Big|&\vdots\\0&0&\Big|&b'_n\end{bmatrix}$$

$$\displaystyle b'_i=b_i-\frac{a_i}{a_1}\cdot b_1-\frac{1-\frac{a_i}{a_1}}{1-\frac{a_2}{a_1}}\cdot\Big[b_2-\frac{a_2}{a_1}\cdot b_1\Big] \forall i>2$$

For the rank of the augmented matrix to be $$0$$, we require $$b'_i=0$$

$$\displaystyle\therefore b_i=\frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=\Big[\frac{b_1-b_2}{a_1-a_2}\Big]\cdot a_i+\Big(\frac{a_1b_2-a_2b_1}{a_1-a_2}\Big),\ \forall i>2$$

Therefore, if:

• There are two lines $$L_i,L_j$$ not parallel to each other $$(a_i\ne a_j)$$ intersecting at $$(X,Y)$$;

• $$\displaystyle b_k=\Big[\frac{b_i-b_j}{a_i-a_j}\Big]\cdot a_k+\Big(\frac{a_ib_j-a_jb_i}{a_i-a_j}\Big)=Xa_k+Y,\ \forall k\ne i,j$$; that is, $$(X,Y)$$ lies on all the remaining lines;

Then, the straight lines $$L_i:= a_ix+y=b_i,i\in\{1,2,...,n\}$$ intersect at the point $$\displaystyle(X,Y)=\Big(\frac{b_i-b_j}{a_i-a_j},\frac{a_ib_j-a_jb_i}{a_i-a_j}\Big)$$ uniquely.