# Find a 2x2 matrix A such that A^2=-I

1. Apr 2, 2017

### seanthinks

This is a problem from Lang's Introduction to Linear Algebra. The problem statement is:

Find a 2 x 2 matrix A such that A2= $\begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}$ = -I

The solution is available in the answer section of the book, but it is not shown how the author comes up with the solution.

My initial attempt at the problem involved multiplying both sides of the equation by the inverse of A in attempt to isolate A, that only produced IA=-A-1 which is really no more clear than what I started with.

I then attempted to express A as $\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$ such that A2= $\begin{pmatrix} a^2+bc & ab+bd \\ ca+dc & cb + d^2 \\ \end{pmatrix}$=-I, but it this didn't lead me anywhere either.

This section hasn't introduced to determinants so I certain that isn't a part of the approach. What am I missing?

Thanks.

2. Apr 2, 2017

### vela

Staff Emeritus

Where'd you get stuck with this approach? It will work.

3. Apr 3, 2017

### seanthinks

I got stuck after setting it up as a system of linear equations. I had $$a^2 + bc = -1 \\ ab + bd = 0 \\ ca + dc = 0 \\ cb + d^2 = -1$$ which when I solved yielded $a = -d$, which means $a^2=d^2$.

Then I get stuck on either $ab + bd$ or $ca + dc$ because that means either $a, b, c, d$ can be zero. But if $b, c = 0$ then that implies $a^2=-1$ which isn't true in $\mathbb {R}$.

4. Apr 3, 2017

### vela

Staff Emeritus
Did you consider the possibility $a=d=0$?

5. Apr 3, 2017

### vela

Staff Emeritus
If you have $a=-d$, the second and third equations are satisfied. This leaves you with $bc=-1-a^2$. You should be able to find three numbers which satisfy the one equation.

6. Apr 3, 2017

### seanthinks

Ohhhh okay, I think the problem I'm having is that when I get there I see the solutions as being either $b = -1/c$ or $c = -1/b$, neither of which seemed particular illuminating. And the solutions 1 and -1, while obvious seemed like I was just blindly guessing rather than reasoning them out. Why not 6 and -1/6 and so forth. But I see now that any solutions x and -1/x would simply be scalar multiples of 1 and -1.

7. Apr 3, 2017

### seanthinks

Actually, scratch that last part, the error in my thinking is glaring to me now lol

8. Apr 3, 2017