Question

1 & 1 + ac & 1 + bc \\ 1 & 1 + ad & 1 + bd \\ 1 & 1 + ae & 1 + be \end{matrix} \right| =$$

• A. 1
• B. 0
• C. 3
• D. $a + b + c$

Answer 1

Answer: (B)

0

Answer 2

Applying $C_{3} \rightarrow C_{3} - C_{1}$and $C_{2} \rightarrow C_{2} - C_{1}$, we get

$\left| \begin{matrix} 1 & ac & bc \\ 1 & ad & bd \\ 1 & ae & be \end{matrix} \right| = ab\left| \begin{matrix} 1 & c & c \\ 1 & d & d \\ 1 & e & e \end{matrix} \right| = 0$, $\{\because C_{2} \equiv C_{3}\}$.