Why can I solve an impossible equation using linear algebra? I am currently learning matlab and linear algebra side by side and I stumbled upon this example from mathworks
A = [1 2 0; 0 4 3];
b = [8; 18];
x = A\b

x = 3×1

     0
4.0000
0.6667

which in my mind translates to
$$
A = \left[
\begin{matrix}
1 & 2 & 0 \\
0 & 4 & 3
\end{matrix}\right]
B = \left[
\begin{matrix}
8 \\ 18
\end{matrix}\right]
x = \left[
\begin{matrix}
a \\ b \\ c
\end{matrix}\right]
$$
$$
Ax = \left[
\begin{matrix}
1 & 2 & 0 \\
0 & 4 & 3
\end{matrix}\right] \times \left[
\begin{matrix}
a \\ b \\ c
\end{matrix}\right] 
= \left[
\begin{matrix}
a + 2b \\ 4b + 3c
\end{matrix}\right]
$$
which boils down to
$$
\left[
\begin{matrix}
a + 2b \\ 4b + 3c
\end{matrix}\right] = \left[
\begin{matrix}
8\\ 18
\end{matrix}\right] \Rightarrow \begin{matrix}a + 2b = 8 \\4b + 3c = 18\end{matrix}
$$
which is an equation with 3 unknown (a, b and c) with two equations, which is impossible! Yes there is a solution
$$
x = \left[
\begin{matrix}
0 \\ 4 \\ 2/3
\end{matrix}\right]
$$
How can I solve an impossible equation (three unknown and two equations) using linear algebra?
 A: What MATLAB is doing, here, is finding a solution to the underdetermined system of equations that has the fewest possible non-zero elements. That is, it maximises the number of zeros in $x$.
MATLAB can also be applied to actually-unsolvable systems (overdetermined), in which it will give the "solution" with the smallest error (that is, the one that minimises $||Ax-b||_2$).
A: It is not impossible. The problem has a geometric interpretation which may clear things up for you. We know that all points $(x,y,z)$ on a plane in three-dimensional space satisfy equations of the form $Ax+By+Cz=D$. We may therefore interpret the equations $a+2b=8$ and $4b+3c=18$ as planes in three-dimensional space. We know that if two planes in three-dimensional space are not parallel to one another, then they must intersect along a line. Therefore, any point $(a,b,c)$ (like $(0,4,2/3)$, for example) that lies on this line will satisfy the system of equations in your question.
Here is a diagram to illustrate my point. The point $(0,4,2/3)$ is indicated, as well

In order to actually find these points, if the line is not parallel to any of the axes, then you may simply pick a value for $a$, $b$, or $c$, leaving you with only two unknowns, and then solve the resulting system of equations as you would normally.
A: A system of  2 (or $n$) equations in 3 (or $m$) unknowns (with $n<m$) if it has a solution then it has infinite number of solutions.
A system of 3 (or $n$) equations in 2 (or $m$) unknowns (with $n>m$) might not have a solution.
A: If there are fewer equations than unknowns, usually there are many solutions. It is not impossible, but indeterminate.
An extreme example is this:  one unknown, but no equation !
A: The set of solutions $S = \{ x \vert Ax=b \}$ will be one of three cases:


*

*No solution, $S = \emptyset$, $b \not\in \{ Ax \mid x\in V\}$, where $A: V \to W$.

*One solution, $S = \{ y \}$

*Infinite many solutions


The "impossible case" is 1., but your system is of case 3.
