I have an induced form of Augmented Matrix of the system
$$A = \left[\begin{array}{ccc|c}1&2&1&1 \\ 0&1&1&0 \\ 0&0&a^2-4&a^2 + 2\end{array}\right].$$
If $a^2-4 \neq 0$, we have unique solution. (For each a value, there is only one solution for this. I think this is ok.)
If $a^2-4 = 0$, there exists no solutions.
- Is this approach valid?
- When does this system have a infinitely many solutions? When $a^2-4=a^2+2=0$, there is no satisfying $a$ for this solution. Does it mean there is not infinitely many solutions for this system?
Thanks in advance.
Best Answer
If $a^2 - 4 \neq 0$, then yes, there is a unique solution. Dividing by the non-zero number $a^2 - 4$ (whatever it is equal to) will put a pivot in each column, and finding the solution from there involves back-substitution.
If $a^2 - 4 = 0$, then we don't have a pivot in each (non-augmented) column. This doesn't mean there is no solution necessarily; there might be infinitely many. To have any solutions at all, we must verify that, once in row-echelon form, every $0$ row extends to the augmented column. That is, we have no rows of the form
$$[\begin{array}{cccc|c} 0 & 0 & \cdots & 0 & 2\end{array}]$$
for example. Anything non-zero in the augmented column will mean there's no solutions.
In this case, when $a^2 - 4 = 0$, then we get $a^2 + 2 = 6 \neq 0$ (in fact, $a^2 + 2$ is never equal to $0$ for $a \in \Bbb{R}$). So, indeed, there will be no solutions when $a^2 - 4 = 0$. Please note that this wasn't guaranteed by the zero row; we needed to check the augmented column was non-zero.
Infinitely many solutions occur when the system is consistent (i.e. at least one solution exists), and there is a column without a pivot. When considering a square matrix of coefficients, this must be accompanied by a zero row (extending into the augmented column). This can never happen in the given matrix, as we have just shown.
Hope that helps!