Rank of a Matrix with multiple variables

[diogomgf]

I'm struggling with this exercise:
Find the possible ranks of the matrix D, depending on the values of a and b.

$\displaystyle D_{a,b} = \begin{bmatrix}a & -1 & 1 & b \\0 & 1 & a & 1 \\1 & 0 & 2 & 2 \\0 & 1 & 1 & a\end{bmatrix}$

After reducing D to row echelon form I got:

$\displaystyle \begin{bmatrix}1 & 0& 2& 2 \\0 & 1 & a & 1\\0& 0 & 1-a& b-2a+1\\0 & 0& 0& b-a\end{bmatrix}$

Based on this I concluded that if $\displaystyle a = 1$ and $\displaystyle b = 1$ then $\displaystyle r(D) = 2$; If $\displaystyle a = b$ and $\displaystyle a,b \neq 1$ then $\displaystyle r(D) = 3$.

Apparently this is incorrect according to the solutions in the book, and I can't really understand why.

 

[diogomgf]

and r.e.f:

 

[Steven G]

Keep reducing

 

[Romsek]

The condition for rank 3 is $\displaystyle b = 3a-2,~a \neq 1$

 

[Steven G]

The condition for rank 3 is $\displaystyle b = 3a-2,~a \neq 1$

Are you assuming that the OP has the correct reduced matrix? If you are, then why is rank =3 not obtained when b=a, a is not 1?

 

[Romsek]

Are you assuming that the OP has the correct reduced matrix? If you are, then why is rank =3 not obtained when b=a, a is not 1?

I started from the original matrix. You can compute the determinant as

[MATH]D = 3 a^2-a b-5 a+b+2 = 3 (a-1) \left(a-\dfrac{b+2}{3}\right)[/MATH]
If we let $\displaystyle b=a \neq 1$ we have

[MATH]D= \begin{pmatrix} a & -1 & 1 & a \\ 0 & 1 & a & 1 \\ 1 & 0 & 2 & 2 \\ 0 & 1 & 1 & a \\ \end{pmatrix}[/MATH]
which row reduces to $\displaystyle I_4$

If $\displaystyle b = 3a-2,~a\neq 1$ we have

[MATH]D = \begin{pmatrix} a & -1 & 1 & 3 a-2 \\ 0 & 1 & a & 1 \\ 1 & 0 & 2 & 2 \\ 0 & 1 & 1 & a \\ \end{pmatrix}[/MATH]
which row reduces to

[MATH]\left( \begin{array}{cccc} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & a+1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)[/MATH]
Finally if $\displaystyle a=1, b=3a-2 = 1$ we have

[MATH]D= \left( \begin{array}{cccc} 1 & -1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ \end{array} \right)[/MATH]
which row reduces to

[MATH]\left( \begin{array}{cccc} 1 & 0 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)[/MATH]

 

[Steven G]

I'm struggling with this exercise:
Find the possible ranks of the matrix D, depending on the values of a and b.

$\displaystyle D_{a,b} = \begin{bmatrix}a & -1 & 1 & b \\0 & 1 & a & 1 \\1 & 0 & 2 & 2 \\0 & 1 & 1 & a\end{bmatrix}$

After reducing D to row echelon form I got:

$\displaystyle \begin{bmatrix}1 & 0& 2& 2 \\0 & 1 & a & 1\\0& 0 & 1-a& b-2a+1\\0 & 0& 0& b-a\end{bmatrix}$

Based on this I concluded that if $\displaystyle a = 1$ and $\displaystyle b = 1$ then $\displaystyle r(D) = 2$; If $\displaystyle a = b$ and $\displaystyle a,b \neq 1$ then $\displaystyle r(D) = 3$.

Apparently this is incorrect according to the solutions in the book, and I can't really understand why.

Based on your last matrix shown, there is nothing wrong with your conclusion.
Assuming that the book has the correct solution then you did not reduce your matrix correctly. If you think that I am going to reduce the matrix for you (and not make my own mistakes) you are mistaken. Just try again to reduce the matrix. I am sure that if you post all the matrices to the reduced one someone here (including myself) will try to find your error.

 

[diogomgf]

Based on your last matrix shown, there is nothing wrong with your conclusion.
Assuming that the book has the correct solution then you did not reduce your matrix correctly. If you think that I am going to reduce the matrix for you (and not make my own mistakes) you are mistaken. Just try again to reduce the matrix. I am sure that if you post all the matrices to the reduced one someone here (including myself) will try to find your error.

The answer @Romsek gave is the correct one though I don't know what a matrix determinant is so I didn't go that way.
I just reduced to row echelon form (the last matrix I presented) and made my conclusions. Thats the way I have been doing for most of the similar problems and I've been getting the correct results. Don't see why I would need to keep reducing all the way to reduced row echelon form @Jomo ?

Maybe I'm missing something...

EDIT: When I posted this I was already on my third time reducing the matrix to row echelon form always getting the same matrix, so I'm confident I did it right...

 

[Dr.Peterson]

Have you redone your work to check for an error? I find that everything is correct in your REF except for the b-a in the lower right.

 

[Steven G]

I was mistaken when I suggested that you continue reducing--sorry
Look, you have a mistake in your reducing. You have three choices from here.
1) You can post all the matrices that led to your last one so we can check your work.
2) You can not ask for more help regarding this problem.
3) You can keep asking for help but not receive any help (unless you show us your work).

I am not trying to be harsh by my response above. I am just trying to tell you that we can't find your mistake unless we see your work. And if for what ever reason you refuse to show us your work then we can't help you.

 

[diogomgf]

@Dr.Peterson there goes my confidence in being right about it .

@Jomo I will re-do it and in case something goes wrong I will post the step-by-step process.

 

[Steven G]

About confidence.
It is quite hard to do many computation without making a mistake. Laugh it off instead of letting it affect your confidence.
My linear algebra teacher who was a PhD mathematician college professor could not reduce a matrix correctly to save his life. It was funny because we all knew that he was brilliant.
Einstein could not count change to go on the bus.
I used to be terrible in arithmetic up to graduate school. In adding something like 7+2, the job is to add 2. Well I saw 9 so quickly that I added 2 to it and got 11. A fellow student pointed out to be that I was adding the 2nd number twice and I never made that mistake again.

 

[diogomgf]

This time around I did it diferently.

Here is the process for r.e.f and the conclusions, wich are still wrong I guess...

 

[diogomgf]

@Jomo thank you for the kind words.

 

[Steven G]

3rd matrix has an error.

 

[diogomgf]

3rd matrix has an error.

Oh god... Will re-post, now corrected ??

 

[diogomgf]

@Jomo I did it!

The conditions for rank 2, rank 3 and rank 4 now match with the results in the book!
??

 

[Steven G]

Excellent. Good for you!