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First of all I am sorry because I have asked similar kind of question a few days ago.But I still have problem with row reductions when there are letters in a matrix.The question is asking the value of 'a' when the rank of matrix is 1 , 2 , 3 and 4. I am not good at row reductions.In each row operations the matrix became more confusing.If someone help me with the row reductions , I will be very happy. $$ \left( \begin{array}{ccc} 1&1&2&0\\ 2&a+1&3&a-1\\ -3&a-2&a-5&a+1\\ a+2&2&a+4&-2a \end{array} \right) $$

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As I mentioned last time, it's best if a question has a tag in addition to the [homework] tag, indicating the area of mathematics in question. –  Arturo Magidin Jun 21 '11 at 21:24
    
@ Arturo Pardon me.Is it okey now ? –  Piril Jun 21 '11 at 21:27
    
Why are you reverting the linear-algebra tag? It's been added twice by two different people, you even responded positively to the comment suggesting that it be added; why have you twice removed it? –  joriki Jun 21 '11 at 21:31
    
Not the title, the tags at the bottom. –  Arturo Magidin Jun 21 '11 at 21:36
    
@Joriki :( I am sorry. I am confused with tags.It must be homework + linear algebra tag.Right ? As Arturo Magidin tried to explain me million times. –  Piril Jun 21 '11 at 21:37

2 Answers 2

up vote 7 down vote accepted

You just need to work the algebra, just like you do with regular algebra.

Subtract twice the first row from the second row to get $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ -3 & a-2 & a-5 & a+1\\ a+2 & 2 & a+4 & -2a \end{array}\right).$$ This is just algebra (e.g., the (2,2) entry is $a-1$ because $(a+1)-2 = a-1$).

Add three times the first row to the second row. The (3,2) entry will be $(a-2) + 3(1) = a+1$; the (3,3) entry will be $(a-5)+3(2) = a+1$. Etc. You get: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & a+1 & a+1 & a+1\\ a+2 & 2 & a+4 & -2a \end{array}\right).$$ Subtract $a+2$ times the first row from the fourth row; that means subtracting $1(a+2)$ from the first entry; $1(a+2)$ from the second entry; $2(a+2)$ from the third entry; and $0$ from the fourth entry. You get: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & a+1 & a+1 & a+1\\ 0 & -a & -a & -2a \end{array}\right).$$ Add the fourth row to the third row to simplify things: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & 1 & 1 & -a+1\\ 0 & -a & -a & -2a \end{array}\right).$$ Exchange second and third row: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & 1 & 1 & -a+1\\ 0 & a-1 & -1 & a-1\\ 0 & -a & -a & -2a \end{array}\right).$$ Add $1-a$ times the second row to the third row: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & 1 & 1 & -a+1\\ 0 & 0 & -a & -1+2a-a^2\\ 0 & -a & -a & -2a \end{array}\right).$$

We are just doing algebra, only in several columns at the same time. Keep going until you get a matrix for which the rank will be easy to figure out, depending on the values of $a$. I've gotten you almost all of the way there.

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Okey.Thank you very much.I will keep on simplifying this matrix.thanks for your patience. –  Piril Jun 21 '11 at 21:59

here is an answer for rank=1:

if rank=1, that means all the rows are dependent. The first row is given, so every other row should be a multiple of the first row. Let's start with the second row. since the first element on the second row is 2, and the first element on the first row is 1, then (a+1) should also be 2*1 and (a-1) should be 2*0. so $a$ should be equal to 1. However if you check the third row, substituting $a$ with 1, we have [-3, -1, -4, 2] which is not a multiple of the first row. hence rank=1 has no solution.

You can do row manipulation to make the matrix lower triangular. then the determinant would be the product of the elements on the diagonal.

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Thank you very much. –  Piril Jun 21 '11 at 22:00

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