What's wrong with my solution for this usual linear algebra problem?

[YehiaMedhat]

For the system $x+y+z = 0,\ ax + 4z = 5,\ ax +ay + 4z = 5$, find all the values of such $a$ that the system has:

1. No solution.
2. Many solutions.
3. Unique solution.

In this problem I tried to solve it the following way:
Reducing the augmented matrix, and applying the condition for no solution case for the first one, which is $\text{Rank}(A)\neq \text{Rank}(A_{\text{augmented}})$.
$$\begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ a & 0 & 4 &\bigm| 1\\ a & a & 4 &\bigm| 5\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ 0 & -a & 4-a &\bigm| 1\\ 0 & 0 & 4-a &\bigm| 5\\ \end{bmatrix}$$And then taking the determinant of that matrix to get some expression from which I can use the condition of no solution case.
$$\text{det} \left( \begin{bmatrix} 1 & 1 & 1\\ 0 & -a & 4-a\\ 0 & 0 & 4-a\\ \end{bmatrix} \right) = -a(4-a) = a^2-4a$$And get the determinant of the $\text{Rank}(A_{augmented})$:
$$\text{det} \left( \begin{bmatrix} 1 & 1 & 0\\ -a & 4-a & 1\\ 0 & 4-a & 5\\ \end{bmatrix} \right) = a-4 + 5(4-a - (-a)) = a + 16$$Applying the condition that: $a^2-4a \neq a + 16,\ \therefore a^2-5a-16 \neq 0$
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:
$$\begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ a & 0 & 4 &\bigm| 1\\ a & a & 4 &\bigm| 5\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ 0 & -a & 4-a &\bigm| 1\\ 0 & 0 & 4-a &\bigm| 5\\ \end{bmatrix}$$Hence by inspection, in case of no solution for the system, $a$ is equal to $4, 0$

 

[blamocur]

Where does the last column (0,1,5) in the augmented matrix come from?

 

[YehiaMedhat]

Where does the last column (0,1,5) in the augmented matrix come from?

It's from question header its self. Look at the first equations and you'll see that the absolute values are the 0,1,5

 

[blamocur]

It's from question header its self. Look at the first equations and you'll see that the absolute values are the 0,1,5

Do you mean (0,5,5) ?

 

[YehiaMedhat]

Do you mean (0,5,5) ?

No, they were exactly (0,1,5)

 

[blamocur]

No, they were exactly (0,1,5)

Then I don't understand how you constructed you matrix

 

[Dr.Peterson]

For the system $x+y+z = {\color{green}0},\ ax + 4z = {\color{red}5},\ ax +ay + 4z = {\color{green}5}$, find all the values of such $a$ that the system has:
...
In this problem I tried to solve it the following way:
Reducing the augmented matrix, and applying the condition for no solution case for the first one, which is $\text{Rank}(A)\neq \text{Rank}(A_{\text{augmented}})$.
$$\begin{bmatrix} 1 & 1 & 1 &\bigm| {\color{green}0}\\ a & 0 & 4 &\bigm| {\color{red}1}\\ a & a & 4 &\bigm| {\color{green}5}\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 1 &\bigm| 0\\ 0 & -a & 4-a &\bigm| 1\\ 0 & 0 & 4-a &\bigm| 5\\ \end{bmatrix}$$

Do you not see the difference between the first line and your matrix?

I suspect you just typed the first line wrong, since it's 1 everywhere else.

Once we know what the actual problem is, you can explain your work in detail.

 

[YehiaMedhat]

Do you not see the difference between the first line and your matrix?

I suspect you just typed the first line wrong, since it's 1 everywhere else.

Once we know what the actual problem is, you can explain your work in detail.

Yes, I mistyped it, it was supposed to be $ax + 4z = 1$, sorry for this confusion
It's just really sarcastic when someone misses something that's too obvious

 

[YehiaMedhat]

Where does the last column (0,1,5) in the augmented matrix come from?

Sorry Mr. @blamocur

 

[khansaheb]

Yes, I mistyped it, i

I mistyped it, it was supposed to be ax+4z=1ax + 4z = 1

$ax + 4z = 1$, sorry for this confusion
It's just really sarcastic when someone misses something that's too obvious

Please post the COMPLETE & CORRECTED PROBLEM - again along with your CORRECTED WORK - so that "helpers" don't have to flip through screens !

 

[blamocur]

Sorry Mr. @blamocur

It's OK, no penalties for mistakes on this forum

Regarding your op: in the first formula, you seem to be applying Gaussian elimination to the first 3 columns but not to the last one.

And get the determinant of the Rank(Aaugmented)\text{Rank}(A_{augmented})Rank(Aaugmented):

"determinant of the Rank" ? Moreover, where the inequality of the determinants (as opposed to the inequality of the ranks) come from ?

 

[YehiaMedhat]

It's OK, no penalties for mistakes on this forum

Regarding your op: in the first formula, you seem to be applying Gaussian elimination to the first 3 columns but not to the last one.


"determinant of the Rank" ? Moreover, where the inequality of the determinants (as opposed to the inequality of the ranks) come from ?

No, I have done it to the third column, but since the absolute value is just 0 in the first row, It will not add or subtract any thing from the other below absolute values.

 

[YehiaMedhat]

"determinant of the Rank" ? Moreover, where the inequality of the determinants (as opposed to the inequality of the ranks) come from ?

Please guys, I surely have mistyped this also, I was typing in my break in work
To be honest, just forget about it, I feel I couldn't get to the main point, may I had a really horrible way to get explain my approach, but I don't think it's worth it discussing this question any more.

 

[blamocur]

No, I have done it to the third column, but since the absolute value is just 0 in the first row, It will not add or subtract any thing from the other below absolute values.

You are right: I missed that fact -- sorry.

 

[blamocur]

Please guys, I surely have mistyped this also, I was typing in my break in work

But then you are looking at the equality of the two determinants. Moreover, the augmented matrix isn't square and does not have a determinant -- instead you are using "reduced" version to make it square.

To be honest, just forget about it, I feel I couldn't get to the main point, may I had a really horrible way to get explain my approach, but I don't think it's worth it discussing this question any more.

This is completely up to you, but we'll do our best to help if you change your mind.
Best of luck with your studies!

 

[YehiaMedhat]

But then you are looking at the equality of the two determinants. Moreover, the augmented matrix isn't square and does not have a determinant -- instead you are using "reduced" version to make it square.


This is completely up to you, but we'll do our best to help if you change your mind.
Best of luck with your studies!

I really appreciate it Mr. @blamocur