Non-singularity of a square matrix The diagonal elements of a square matrix $M$ are odd integers while the off diagonals are even integers. Then can we say that $M$ must be non-singular?
We can easily comment on the singularity of the matrix $M$. Let us take the identity matrix of order $2$ which is non-singular.So can I say the $M$ must be non-singular for every such case?
please help me to clear this idea.
 A: Yes, this is true. Consider the matrix $A$ and its determinant modulo $2$. Then $A$ is congruent to $I_n$ modulo $2$, whose determinant is $1$. Hence $A$ has nonzero determinant, and so is non-singular.
A: We can prove by induction that any such matrix had an odd integer determinant. It would follow from this that the determinant is nonzero, and so the matrix is non-singular.
Let $A$ be an $n\times n$ matrix whose diagonal entries are odd integers, and whose non-diagonal entries are even integers. Then $\det A\in 1+2\mathbb Z$.
If $n=1$, then $A = [m]$ for some odd integer $m$, and so $A$ has determinant $m$, which is odd. 
Now suppose the theorem holds for $n\times n$ matrices, and let $A$ be a $(n+1)\times (n+1)$ matrix whose diagonal entries are odd integers, and whose non-diagonal entries are even integers. Let $a_{ij}$ be the $ij$ minor of $A$. Then
$$\det A = \sum_{i=1}^{n+1}(-1)^{1+i}A_{1i}a_{1i}$$
If $i=1$, then $a_{1i}$ is odd by the inductive hyptohesis. $A_{1i}$ also odd due to the properties of $A$. So, the first term in this sum is odd. And since each $A_{1i}$ is even for $i>1$, every other term is even, and so the sum is odd. This completes the proof. 
