1. ## 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?

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 $\displaystyle \det A\neq 0$ is necessary and sufficient.

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?

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

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