**Propositional Logic – Discrete Mathematics**

p = (begin{vmatrix} x & 3 & 0 \ 2 & y & 3\ 4 & 2 & 4 end{vmatrix})

Q = (begin{vmatrix} x & 2 & z \ 3 & y & 2\ 0 & 3 & z end{vmatrix})

PQ is equivalent to

**A.**PP^{T}**B.**pp^{-1}**C.**qp**D.**pp

##### Correct Answer: Option A

##### Explanation

p = (begin{vmatrix} 0 & 3 & 0 \ 2 & 1 & 3\ 4 & 2 & 2 end{vmatrix})

Q = (begin{vmatrix} 0 & 2 & 4 \ 3 & 1 & 2\ 0 & 3 & 2 end{vmatrix}) = p^{T}

pq = pp^{T}