# Linear Algebra - Identity matrices

1. Oct 1, 2008

### GTRockstar24

I am having some difficulty with identity matrices in linear algebra at the moment. I am sure it is fairly simple to solve, but I just cannot follow the logic behind this particular problem.

I need to come up with a matrix B (2x2), such that B =/= I but B2 = I

Since
Code (Text):
I = (1 0)
(0 1)
If I am attempting this correctly, I have to find a number that isnt 1, but if you square it, it equals 1. So I came up with this.
Code (Text):
B = (-1 0)
(0 -1)

2. Oct 1, 2008

### HallsofIvy

Staff Emeritus
Well? Have you tried it? What is
$$\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$$

3. Oct 1, 2008

### sutupidmath

well, dang, i don't know how to type matirices in latex, so i'll just try to give some hints.

let the matrix $$B=[B_1,B_2]$$ where $$B_1,B_2$$ are the columns of B, so

$$B_1=[a.... b]^T, and, B_2=[c ... d]^T$$ now multiply $$BB=[BB_1 .... BB_2]$$ now we want the following to hold

$$BB_1=e_1,BB_2=e_2$$ where $$I=[e_1 .... e_2]$$, $$e_1=[1 .... 0]^T, e_2=[0 .... 1]^T$$

now you will end up with some eq. involving in it a, b, c ,d so you can figure out the conditions that a, b, c, and d have to satisfy in order for BB=I. THis way you can find more than one such matrix.

I hope i was of any help