0 votes

In a non-homogeneous equation Ax = b, x has a unique solution when $A^{-1}$ exists i.e x = $A^{-1}$b

but when det(A) = 0 then we have infinite solution or many solution.

please give a mathematical explanation of how the 2nd statement occurs?

but when det(A) = 0 then we have infinite solution or many solution.

please give a mathematical explanation of how the 2nd statement occurs?

0 votes

$A^{-1}$ exists is same as saying det(A) $\neq$ 0.

What happens when det(A)=0? It simply means that few of the vectors are linearly dependent and therefore one unique solution isn’t possible. WHY? Because few of the vectors are not giving any new information regarding the spaces they span but just lie in the span of other vectors.

For example, $2x+3y=9$ and $4x+6y=18$ are representing just one line, in terms of vector algebra, they are spanning the same line. So there can exist multiple solutions for the same. Like $(0,3),(1,\frac{7}{3}),(2,\frac{5}{3})$ and so on….

You can see $\begin{vmatrix} 2 & 3 \\ 4 & 6 \end{vmatrix}$ = $0$. Therefore had they been spanning different spaces, we would have got some unique intersection point.

Now in case of homogeneous system, all the vectors pass through the origin, and that is what is known as the trivial solution.

Now if you want some non-trivial solution then you must have at least one dependent vector so that it collapses or mixes (non-formal terms) with some other given vector and therefore all the points that lie on that dependent vector(infinite points) will give solutions.