#### YehiaMedhat

##### Junior Member

- Joined
- Oct 9, 2022

- Messages
- 74

- No solution.
- Many solutions.
- Unique solution.

Reducing the augmented matrix, and applying the condition for no solution case for the first one, which is [imath]\text{Rank}(A)\neq \text{Rank}(A_{\text{augmented}})[/imath].

[math]\begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ a & 0 & 4 &\bigm| 1\\ a & a & 4 &\bigm| 5\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ 0 & -a & 4-a &\bigm| 1\\ 0 & 0 & 4-a &\bigm| 5\\ \end{bmatrix}[/math]And then taking the determinant of that matrix to get some expression from which I can use the condition of no solution case.

[math]\text{det} \left( \begin{bmatrix} 1 & 1 & 1\\ 0 & -a & 4-a\\ 0 & 0 & 4-a\\ \end{bmatrix} \right) = -a(4-a) = a^2-4a[/math]And get the determinant of the [imath]\text{Rank}(A_{augmented})[/imath]:

[math]\text{det} \left( \begin{bmatrix} 1 & 1 & 0\\ -a & 4-a & 1\\ 0 & 4-a & 5\\ \end{bmatrix} \right) = a-4 + 5(4-a - (-a)) = a + 16[/math]Applying the condition that: [imath]a^2-4a \neq a + 16,\ \therefore a^2-5a-16 \neq 0[/imath]

The problem is that that kind of solution doesn't yield any integer solutions, and the solution that was done by the TA was too much trivial than mine, actually it was better because it was just trivial.

The solution of my TA was:

[math]\begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ a & 0 & 4 &\bigm| 1\\ a & a & 4 &\bigm| 5\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ 0 & -a & 4-a &\bigm| 1\\ 0 & 0 & 4-a &\bigm| 5\\ \end{bmatrix}[/math]Hence by inspection, in case of no solution for the system, [imath]a[/imath] is equal to [imath]4, 0[/imath]