For a system of n equations in n variables, represented by
A*$\vec x$=$\vec b$ ,
- There exists a unique solution if $\lvert A\rvert$$\neq$O
if$\lvert A\rvert$=O
- There exist infinite solutions if adj(A)*$\vec b$=O
- There exists no solution if adj(A)*$\vec b$$\neq$O
The first is easily justified. I request help for the other two statements.
Attempt: Infinite solutions will exist iff $\vec b$ $\in$ column space of A
Infinite solutions exist when adj(A)*$\vec b$=O , that is when $\vec b$ $\in$ Nullspace of adj(A)
So, i need to show that the column space of A is equal to the null space of adj(A)
and that, is a dead end. Because it isn't true.
I have since come across an exception to this rule, one where adj(A)=O and $\vec b$ doesnt lie on the column space of A. So, When exactly does this rule hold, and how is it justified?
Now, adj(A)*A = $\lvert$A$\rvert$*I This can be shown computationally, for any matrix.For invertible matrices, we have the luxury of avoiding pure computation by observing
adj(A)=$\lvert$A$\rvert$ A$^{-1}$
adj(A)*A = $\lvert$A$\rvert$ (A$^{-1}$ *A)
Nevertheless, for singular matrices, adj(A)*A= O
So, we see that every column in matrix A transforms into a null vector. every column of A as well as their linear combinations, belong to the nullspace of adj(A). From that, we can conclude,
The column space of a singular square matrix is a subset of the nullspace of its adjoint.
Of course, being a subset alone is of little use, because if there are exceptions, the formula cannot be used to find all vectors that lie on the span of the columns of A. We need to now see when these two are identical and when its just a subset.
If the rank of A is $\leq$ n-2 , that is, at least three of the columns are linearly dependent, irrespective of what column or row is deleted, the remaining matrix will have at least 2 dependent columns so Its determinant will be zero. So every element in the adjoint is zero. The null space of the adjoint is all of space if rank < n-1. That is, in most cases, the formula is useless and the adjoint will be the null matrix.
The only other case left for singular matrices is rank n-1.we already see that this entire n-1 dimensional space is in the null space of the adjoint. If it gets any bigger than that, it can only be the whole space and that means its the null matrix.Since the adjoint is definitely not the null matrix, it cannot be larger than the column space of A.So they are equal and the formula works.