Matrix + Number Theory

To make things clear, Det $(A)$ [also can be written as $|A|$ ] refers to the determinant of the square matrix $A$ and Tr $(A)$ refers to the trace of the square matrix $A$ which is the sum of all the elements of the main diagonal of $A$ . Simply put, for $A$ we have,

$|A|=p^2-q^2\quad \textrm{and}\quad \textrm{Tr}(A)=\sum_{n=1}^2 a_{nn}=a_{11}+a_{22}=p+p=2p$

Now, we are given that $p,q,r$ are primes and $|A|=p^2-q^2=r$ , so,

$(p-q)(p+q)=r$

Now, one of the factors of $r$ must be $1,(-1)$ and the other must be $r,(-r)$ respectively. Just by checking the case where $p-q=1$ and $p+q=r$ , we can get the trivial solution $p=3,q=2$ . If you check the other $3$ cases, you'll see that either $p$ or $q$ aren't real or they are negative.

So, we conclude the answer with $\boxed{\textrm{Tr}(A)=2p=2\times 3 =6}$

Too easy Problem !

think P^2-Q^2=(P-Q)(P+Q)=Prime

P-Q=1 and P+Q=prime

P=3 , Q=2 (only set of prime with difference 1)