Question

If D = $\left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|$ and D´ =$\left| \begin{matrix} a_{1} + pb_{1} & b_{1} + qc_{1} & c_{1} + ra_{1} \\ a_{2} + pb_{2} & b_{2} + qc_{2} & c_{2} + ra_{2} \\ a_{3} + pb_{3} & b_{3} + qc_{3} & c_{3} + ra_{3} \end{matrix} \right|$ then

• A. D´ = D(1 + pqr)
• B. D´ = D
• C. D´ = D(1 – pqr)
• D. D´ = D(1 + p + q + r)

Answer 1

Answer: (A)

D´ = D(1 + pqr)

Answer 2

D´ = $\left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|$+ pqr $\left| \begin{matrix} b_{1} & c_{1} & a_{1} \\ b_{2} & c_{2} & a_{2} \\ b_{3} & c_{3} & a_{3} \end{matrix} \right|$

(All other determinants vanish)

D´ = (1 + pqr) D