# Matrix

• Mar 6th 2011, 02:28 AM
Matrix
I have a revision question in my book about matrices but because I learnt it a year ago I can't quite remember all the details so i'll write what I know and hopefully some of you can help me

I have a matrix which is (a,b ; c,d) as in ab on top and cd below and I need to find the conditions on abcd so that it is row equivalent to the identity

For one I know the identity is (1,0 ; 0,1) and im pretty sure the determinant can't be zero in order for it to be row equivalent and equal to the identity but apart from that I don't know the other conditions for it..

The second question is just to check my answer

I have (lamda - 2)x + y = 0

x + (lamda - 2)y = 0

and I have to find the non trivial solutions

In order to do this I find the determinant ( lamda squared + - 4 lamda + 4) then make that equal to zero find the solution for lamda then the non trival solutions are all those except where lamda makes the determinant zero? is that correct?
• Mar 6th 2011, 04:02 AM
FernandoRevilla
Quote:

For one I know the identity is (1,0 ; 0,1) and im pretty sure the determinant can't be zero in order for it to be row equivalent and equal to the identity but apart from that I don't know the other conditions for it..

The condition $\det A\neq 0$ is necessary and sufficient.
• Mar 6th 2011, 04:13 AM
FernandoRevilla
Quote:

I have (lamda - 2)x + y = 0

x + (lamda - 2)y = 0

and I have to find the non trivial solutions

In order to do this I find the determinant ( lamda squared + - 4 lamda + 4) then make that equal to zero find the solution for lamda then the non trival solutions are all those except where lamda makes the determinant zero? is that correct?

$\det A=(\lambda-2)^2-1=0\Leftrightarrow \lambda=3\;\vee\;\lambda=1$

If $\lambda \neq 3\;\wedge\; \lambda \neq 1$ there are no trivial solutions. For $\lambda =3\;\vee\; \lambda \neq 1$ the system has more than one solution so, it has non trivial solutions.