# Relationship between matrix equations and determinants?

What is the relationship between a matrix equation and the determinants of the matrix?

Say if you have the matrix equation $A^3 - 4A = 0$, what can I learn about the matrix $A$'s determinants?

Or even if you have $A - 3B = 0$, does that imply $\det(A) = 3*\det(B)?$

One of the properties of the determinant (more here: http://en.wikipedia.org/wiki/Determinant) is $$\det(AB) = \det(A)\det(B)$$ and from here you can deduce $$\det(A^n) = (\det(A))^n$$ Another property of determinant is $$\det(cA) = c^n\det(A)$$ if $A$ is a $n \times n$ matrix.
If you apply the determinant on your first equation you get $$\det(A^3) = \det(4A)$$ or $$(\det(A))^3 = 4^n \det(A)$$ implying (if $\det(A)$ is not zero) $$(\det(A))^2 = 4^n$$ and finally $$\det(A) = (\pm2)^n$$ From the equation $$A-3B=0$$ you can deduce $$\det(A) = 3^n \det(B)$$ where $A$ and $B$ are square $n \times n$ matrices.
In general, an annihilating polynomial doesn't give you information about the determinant but rather the eigenvalues of the matrix. There is a well known theorem that says if $m$ is the minimal polynomial of $A$ and $p$ is any polynomial for which $p(A)=0$ then $m\mid p$. What this says is that the eigenvalues of the matrix necessarily forms a subset of the roots of any annihilating polynomial $p$.
For simpler cases, we can extract more information. For your first polynomial, we know that $A$ satisfies the polynomial $p(x)=x^3 - 4x$. This implies that the eigenvalues of $A$ are amongst the roots of $p$, being $\{0,\ 2,\ -2\}$. If the matrix has $0$ as an eigenvalue, that necessarily implies that the determinant is $0$. Otherwise, the determinant is of the form $\pm 2^n$ where $n$ is the size of the matrix. This is the most information that we can extract since all of these cases are possible.
For your second equation, we get $A = 3B$. Taking determinants gives $\det A = 3^n \det B$ where $n$ is the size of the matrix.
More generally, the eigenvalues of $p(A)$ are the images of the eigenvalues of $A$ under the polynomial $p$. So if $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ (counted by algebraic multipicity), $p(A)$ has eigenvalues $p(\lambda_1), \ldots, p(\lambda_n)$ (again counted by algebraic multiplicity), and in particular $\det p(A) = p(\lambda_1) \ldots p(\lambda_n)$.