Answer
See proof below.
Work Step by Step
Step 1.
Let $A_{mn}=[a_{ij}]$ and the zero matrix be denoted by $O_{mn}=[o_{ij}]$
where $i=1,2,3,...m$ and $j=1,2,3,...n$ and $o_{ij}=zero $ for all $i=1,2,3,..m $ and $ j=1,2,3,...n$
$A_{mn}+O_{mn}=[a_{ij}]+[o_{ij}]=[a_{ij}+o_{ij}]=[a_{ij}+0]=[a_{ij}]=A_{mn}$
Step 2.
Since
$-A=(-1)A=(-1)[a_{ij}]=[(-1)a_{ij}]$
then
$A+(-A)=A+(-1)A=[a_{ij}]+[(-1)a_{ij}]=[a_{ij}+(-1)a_{ij}]=[(1+(-1))*a_{ij}]=[0*a_{ij}]=O$
Step 3.
To keep that the following statement holds $cA=O_{mn}$ and $c$ is a real number, then it must be:
The first case that $c=0$
or
For $c\ne 0, $ we see that $c$ has an inverse $1/c$, so
$A=(1/c)(cA)=(1/c)*O_{mn}=O_{mn}$
