Is the sum of singular and nonsingular matrix always a nonsingular matrix? If $A$ and $B$ are singular and nonsingular respectively, where both are square, is $A+B$ always nonsingular?
Suppose that $A$ is a singular matrix and that $B$ is nonsingular, where both are square of the same dimension. It is not hard to see that $AB$ and $BA$ are both singular. It seems natural to ask whether the same is true for addition of matrices instead of product.
For $1\times1$ matrices (i.e., numbers), the only singular matrix is $0$; so if we add it to any nonsingular (invertible) matrix, it remains nonsingular. So to find a counterexample, we have to look at bigger matrices.
 A: Not true even for positive matrices:
$$
\begin{pmatrix}1 & 1\\2 & 2\end{pmatrix}+
\begin{pmatrix}3 & 2\\2 & 1\end{pmatrix}=
\begin{pmatrix}4 & 3\\4 & 3\end{pmatrix}.
$$
A: No. Consider the matrices 
$$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix},
   B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$
Then: 


*

*$A$ is regular.

*$B$ is singular.

*$A + B = \binom{2\,0}{0\,1}$ is regular.

*$A - B = A+(-B) = \binom{0\,0}{0\,1}$ is singular.

A: Let me tell one particular way of generating lots of examples. We will find $A$ such that  $A +I$ will be  singular. You can easily adapt this method  to use with any non-singular matrix instead of identity. We will work backwards to get solutions.
Take a matrix with two identical rows as $A+I$.
This gives the condition that 
$$A+\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr  * & * & *\cr}$$
First two rows of $A$ are forced. To ensure  singularity of $A$ make the last row identical to 2nd row:
$$\pmatrix{a-1 & b& c\cr a &b-1 & c\cr a & b-1 & c} +\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr  * & * & *\cr} $$
Now we can work backwards to get the values to be used in place of stars: 
$$\pmatrix{a-1 & b& c\cr a &b-1 & c\cr a & b-1 & c} +\pmatrix{1 &0&0\cr0&1&0\cr 0&0&1\cr}=\pmatrix{a & b &c \cr a &b & c\cr  a & b-1 & c+1\cr} $$ Now replace $a,b,c$ with your ATM pin number, your friend's age, and your annual salary in Euros respectively, you will get a solution.
