Suppose that we have a system of three equations having three unknowns.
As we know that we can represent them in a matrix form , having a coefficient matrix multiplied by the variable matrix , and on the right side we have the constant matrix, in the form $AX=B$.
Now suppose we have a system of three such equations where the coefficient of a variable in one of the equations is unknown and we need to find the condition on that variable such that the given system has
1)unique solution
2) infinite solutions.
now I understand that for a system $AX=B$ to have unique solutions , the coefficient matrix should be non singular because only then $X=A^{-1}B$ will exist .
However here’s the part that’s confusing. If we solve the coefficient matrix and equate it to zero we get the value of unknown coefficient that makes $|A|=0$. , for which we do get infinite solutions , or in some cases no solutions.
But , I have two questions
1) why does the condition for infinite solutions overlap with that of no solution ? Isn’t the system required to have atleast a unique solution before having infinite solutions?
2) is it necessary , for such a system to have infinite solutions , that the determinant of the coefficient matrix should always be zero ? Or can we get infinite solutions even if the coefficient matrix is non-singular?
Thanks for helping !
If the matrix is singular, that means that it maps at least one vector to zero. Thus, in this case, if you have any solution at all, you already have infinitely many solutions, since you can add arbitrary multiples of the vector that's mapped to zero to the solution. Thus, a linear system of equations with a singular matrix has either zero or infinitely many solutions.
Conversely, if you have two solutions, their difference is mapped to zero, so in this case the matrix is singular. Thus, the answer to your second question is that the determinant of the matrix is indeed necessarily zero if there are infinitely many solutions.