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.
Best Answer
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.