Why can't a linear system of equations $AX= B$ have a unique solution when $\det(A)= 0$? We know that if $\det(A)$ is not equal to 0, then $AX= B$ surely has a unique solution. My question is if its converse is also true? That is to say, if $AX= B $ has a unique solution, then prove that $\det(A)$ is not equal to $0$.
We can also frame the question as: If $\det(A)= 0$, then why are there only two possibilities, namely no solution or infinite solutions?
Let's say that in $AX= B$, and I know that $X=X_1$ is a solution. Further $\det(A)= 0$. Then can you find me a second solution $X=X_2$ for $AX= B$?
 A: Because $$AX=B\iff AX=AX_1\iff A(X-X_1)=0\iff X-X_1\in\text{Ker} A$$
Since $\det A=0$, $\dim\text{Ker} A\geqslant 1$ and thus there is an infinite number of solutions.
A: Det(A)=0 is equivalent to ker(A) is not trivial (by Dimensionformula) so if you have a solution $ x_1 $ of your linear system, then $ x_1 + y $ will be a solution for all $ y\in \ker(A) $.
A: Of course. In fact, I can find infinitely many solutions! The fact that $\det(A) = 0$ tells us that $A$ has a nontrivial kernel, which is defined as
$$\ker(A) : = \{ x : Ax = 0\}$$
If the kernel is nonempty, then it contains infinite (if the field $k$ on which you're working is infinite) elements: in fact, $A(\lambda x) = \lambda (Ax) = 0$ for any $\lambda \in k$.
So if $X=X_1$ is a solution to $AX=B$, just take $K \in \ker(A)$ such that $K \neq 0$. Then for each $\lambda \in k$, $X+\lambda K$ is also a solution, since
$$A(X + \lambda K) = AX + A(\lambda K) = B + \lambda (AK) = B + 0 = B$$
A: The reason for that is geometrically pretty simple:
to solve a linear system $AX=B$, the standard method consists in doing Gauß-Jordan elimination on the augmented matrix $AB$ until the ‘$A$’ part is is in row-reduced echelon form.
Now the linear system has a solution if and only if the matrix $A$ and the augmented matrix $AB$ have the same rank, which is at most $n$ for a $n\times n$ linear system. 
If they don't, there is no solution.
If they have, this common rank is the codimension of the set of solutions (which is an affine subspace). So, if $\det A\ne 0$, the codimension is $n$, i.e. the set of solutions has dimension $0$ – in other words, it consists of a single point. 
If $A$ has rank $\,n-d\;$ ($0<d<n$), the set of solutions has dimension $d$, hence it is infinite – at least if the the base field is not a finite field. Furthermore, all solutions are congruent modulo the kernel, which has dimension $d$.
A: I'll assume $A$ is an $n \times n$ matrix. Let $I$ be the $n \times n$ identity matrix. If $A$ has an inverse, then $\det I = \det(A A^{-1}) = \det(A) \det(A^{-1}) = 1$, which implies that $\det A \neq 0$. Thus, if $\det A = 0$, then $A$ does not have an inverse. 
This means that the $\text{dim}(R(A)) < n$. By the rank-nullity theorem, $\text{dim} N(A) \geq 1$, so $A$ has a nontrivial null vector $v$. If $Ax = b$, then $A(x + \lambda v) = b$ for any scalar value $\lambda$. 
A: The determinant is the product of all the eigenvalues.  If the determinant is zero, then 0 is an eigenvalue.  Take an associated eigenvector $x_0$.  If $x$ is a solution of $Ax=B$, then $x+x_0$ is a solution also, since $A(x+x_0)=Ax+Ax_0=B+0=B$.
A: I’m assuming that you have a square system, over the real numbers. 
If the determinant of $A$ is $0$, then the system has either no solutions, or infinitely many solutions.
The determinant of $A$ is $0$ if and only if the columns of $A$ are linearly dependent as (column) vectors in $\mathbb{R}^n$. That means that if $A_1,\ldots,A_n$ are the columns, then there exist real numbers $a_1,\ldots,a_n$, not all equal to $0$, such that
$$a_1A_1+\cdots + a_nA_n = \left(\begin{array}{c}0\\0\\\vdots\\0\end{array}\right).$$
Say you have a system $A\mathbf{X}= B$. It’s possible the system has no solutions at all. For example, 
$$A=\left(\begin{array}{cc}
1&1\\
1&1\end{array}\right),\qquad B=\left(\begin{array}{c}1\\2\end{array}\right).$$
It should be easy to verify that there is no solution to $A\mathbf{X}=B$ in this example, so “no solutions” is certainly a possibility.
But assume the system has at least one solution, $\mathbf{X}_0$. If we let 
$$\mathbf{Y}=\left(\begin{array}{c}a_1\\a_2\\\vdots\\a_n\end{array}\right),$$
where $a_1,\ldots,a_n$ are the scalars we mentioned above, then for $\mathbf{Y}\neq\mathbf{0}$, and for every scalar $\alpha$, we have
$$A(\alpha \mathbf{Y}) = \alpha(A\mathbf{Y})= \mathbf{0}.$$
That means that since $A\mathbf{X}_0=B$, we also have
$$A(\mathbf{X}_0 + \alpha\mathbf{Y}) = A\mathbf{X}_0 + A(\alpha\mathbf{Y})  = B + \mathbf{0} = B,$$
so $\mathbf{X}_0+\alpha\mathbf{Y}$ is also a solution for every real number $\alpha$. Because $\mathbf{Y}\neq\mathbf{0}$, each value of $\alpha$ gives you a different solution, so you have as many solutions as there are real numbers, infinitely many.
So in summary, for an $n\times n$ system $AX=B$:


*

*As you note, if $\det(A)\neq 0$, then the system has a unique solution.

*If $\det(A)=0$, then either the system has no solutions, or else it has infinitely many solutions. 
