Question

Let$\left| \begin{matrix} 6i & - 3i & 1 \\ 4 & 3i & - 1 \\ 20 & 3 & i \end{matrix} \right| = x + iy$, then.

• A. $x = 3,y = 1$
• B. $x = 0,y = 0$
• C. $x = 0,y = 3$
• D. $x = 1,y = 3$

Answer 1

Answer: (B)

$x = 0,y = 0$

Answer 2

$\mathbf{(a + b + c}\mathbf{)}^{\mathbf{2}}\left| \begin{matrix} \mathbf{2bc} & \mathbf{-}\mathbf{2c} & \mathbf{-}\mathbf{2b} \\ \mathbf{b}^{\mathbf{2}} & \mathbf{c + a}\mathbf{-}\mathbf{b} & \mathbf{0} \\ \mathbf{c}^{\mathbf{2}} & \mathbf{0} & \mathbf{a + b}\mathbf{-}\mathbf{c} \end{matrix} \right|$

$\Rightarrow 6i( - 3 + 3) + 3i(4i + 20) + 1(12 - 60i) = x + iy$

$\Rightarrow (C_{1} \rightarrow C_{1} + C_{2} + C_{3})$ $\Rightarrow$ $x = 0,y = 0$.