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I understand that for the matrix to have a unique solution the determinant of matrix A must not be equal to $0$.

Hence, after finding the determinant $(12b-24),$ I found out that b must not be equal to $2$.

Can someone please explain the other conditions for me please? (infinitely many solutions and no solutions). I have tried solving the system by using Gauss-Jordan elimination but to no avail. Any advise or solution to this would be greatly appreciated.

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You are close. If the determinant of the matrix is zero, then the matrix does not have full rank which means it is not invertible. If the determinant is non-zero, then $A^{-1}$ exists and some $x = A^{-1}y$ is the unique solution. Thus it admits a unique solution when $b \neq 2$, assuming you computed the determinant correctly.

On the other hand, consider the augmented matrix when $b=2$, $$ \left( \begin{array}{cccc} 1 & 2 & -3 & | & a \\ 4 & 2 & 0 & | & a+4 \\ -1 & -2 & 3 & | & a-1 \end{array} \right) $$ If you did everything correctly, when you row-reduce this matrix using Gauss-Jordan elimination, you will end up with a row of all zeroes. The corresponding vector on the right-hand side of $|$ must have a zero in that component in order for the system to be consistent. Choosing $a$ so that this is the case will give you a system that has infinitely many solutions. You have too many variables (3) and too few equations (2). On the other hand, every other value $a$ will mean the system has no solutions, since you will have a requirement that one equation be equal to two different constants, which can never hold.

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    Does this mean that the system has: infinitely many solutions when b = 2 and a = 1/2 and no solutions when b = 2 but a ≠ 1/2 – 2017-01-11
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    I haven't actually worked out the numbers, but it will have infinitely many solutions when $b = b^*$ and $a = a^*$, and no solutions when $b=b^*$ and $a \neq a^*$ and have a unique solution when $b \neq b^*$. Assuming you got $b^* = 2$ and $a^* = \frac12$ doing what I described, then yes. – 2017-01-11
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    This is true for this particular example, assuming you computed the determinant correctly. It is very possible in other examples that there could be multiple $b^*$ if the determinant is non-linear in $b$. For instance $$ \left(\begin{array}{cc} b & 1 \\ 1 & b \end{array}\right)$$ is not invertible for $b = 1$ and $b = -1$. – 2017-01-11