Find a $2 \times 2$ matrix $A$ with each main diagonal entries $0$, and with $A^2 = -I.$ I'm not sure how to tackle this problem. 
I'm not certain what is meant by "each main diagonal entries 0". Does this mean:
$$A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$
I'm also not sure what it means by $A^2 = -I$? 
From Wikipedia, I gathered that the identity matrix is an $n \times n$ square matrix with one's on the main diagonal and zero's elsewhere. What would would be considered the main diagonal? 
 A: The main diagonal runs from the top left to the bottom right, so you're looking for a matrix like
$$A = \left[ \begin{array}{cc} 0 & a \\ b & 0 \end{array}\right]$$
such that $A^2 = -I$. This can be done by trial and error, or by writing a system of a few equations after finding what $A^2$ is.
A: Since $A^2=-I$ the only possibilities for the eigenvalues of $A$ are $\pm i$.
Now, if your matrix is real, the characteristic polynomial is a monic quadratic polynomial with $i$ or $-i$ as a root. The only possibility is $x^2+1$.
Therefore $\det(A)=1$. 
Conversely, if $\det(A)=1$, since $tr(A)=0$ the characteristic polynomial of $A$ is $x^2+1$.  Then $A$ has two eigenvalues $\pm i$ with algebraic multiplicuty $1$. Hence $A$ is diagonalisable and then 
$$A=P \left[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right] P^{-1} \Rightarrow A^2=P(-I)P^{-1}=-I$$
So the problem reduces to finding all such matrices of determinant 1. It is easy to find ALL matrices which satisfy the given condition of determinant 1, and these are all solutions to the question. 
A: From the relation given that $A^2 =-I$
We get an annihilating polynomial $p(x)=x^2+1$
And clearly the minimal polynomial of A divides this and hence the roots are $i$ and $-i$.
Just choose any a,b such that their product is $-1$ and you'll get your matrix..the prod has to be  $-1$ since the product of the eigenvalues is the determinant and for your matrix this comes out to be $-ab$. So $$-ab=1$$
