Show that $|A|$ is an integer multiple of 125 Let A=$(a_{ij})\in M_{4\times4}(\mathbb Q)$ be a matrix each entry of which is either -2 or 3. Show that $|A|$ is an integer multiple of 125.
Would it be best to look at the $2\times 2$, minor determinants, or is there some better way?
 A: Hint: Try proving the analogous statement for $n\times n$ matrices (that the determinant is divisible by $5^{n-1}$) by induction on $n$.  You can start with a specific example of $A$ where you know the determinant is $0$, and then consider what happens when you change one entry of $A$ at a time.
A full proof is hidden below:

 We prove by induction on $n\geq 1$ that if $A\in M_{n\times n}(\mathbb{Z})$ has all entries either $3$ or $-2$, then $\det(A)$ is divisible by $5^{n-1}$.  The case $n=1$ is trivial.  Let $n>1$, and suppose the statement is true for $n-1$.  First, note that the statement is trivially true for the matrix $A_0$ all of whose entries are $-2$, since $\det(A_0)=0$.  Now I claim that if $A$ is such a matrix and $\det(A)$ is divisible by $5^{n-1}$ and $A'$ is obtained by changing one entry of $A$ from $-2$ to $3$, then $\det(A')$ is also divisible by $5^{n-1}$.  Indeed, computing $\det(A')$ using minors, the difference between $\det(A')$ and $\det(A)$ is just $\pm 5$ (the difference between the values in the changed entry) times the $(n-1)\times(n-1)$ minor corresponding to the changed entry.  By the induction hypothesis, this minor is divisible by $5^{n-2}$, and so the product is divisible by $5^{n-1}$.  Thus $\det(A')$ must also be divisible by $5^{n-1}$.  Now every matrix $A$ whose entries are all $-2$ or $3$ can be obtained from $A_0$ by changing one entry at a time from $-2$ to $3$, so this shows $\det(A)$ is divisible by $5^{n-1}$ for any such matrix.

A: $A$ can be written as
$$A = B + 5C,$$
where $B$ is the matrix with only $-2$ entries and $C$ has $1$ or $0$ entries. $B$ can be writen as 
$$B = \pmatrix{1\\1\\ \vdots \\1} (-2 -2 \dots -2) = a b^t.$$
we notice that $A$ is invertible iff $C$ is invertible and the case where $A$ is not invertible is trivial, so we may assume that $C$ is invertible.
$$\det(A) = \det(5 C)( 1 + b^t \frac{C^{-1} }{5}a)=5^n \det(C)(1 + b^t \frac{C^{-1}}{5}a)  = 5^n \det(C) + 5^{n - 1}b^t det(C)C^{-1}a $$
The first term is clearly a multiple of $5^{n-1}$ as $\det(C)$ is an integer. For the second, we notice that $\det(C) C^{-1} = adj(C)$ only has integer entries so $b^t det(C)C^{-1}a$ is an integer.
A: Reduce all entries modulo 5 and you will see every entry is now $3 \pmod 5$. That means the $2\times 2$ determinant of this type is $0 \pmod 5$ i.e. it is divisible by 5. That in turns means the $3 \times 3$ determinant of this type is divisible by $25$ (expanding from Row 1 for instance) so that in general the determinant of order $n$ is divisible by $5^{n-1}$
