Question

The determinant $\Delta=\begin{vmatrix} a l+a^{\prime} l^{\prime} & a m+a^{\prime} m^{\prime} & a n+a^{\prime} n^{\prime} \\ b l+b^{\prime} l^{\prime} & b m+b^{\prime} m^{\prime} & b n+b^{\prime} n^{\prime} \\ c l+c^{\prime} l^{\prime} & c m+c^{\prime} m^{\prime} & c n+c^{\prime} n^{\prime} \end{vmatrix}$ is equal to

• A. $\left(a b c+a^{\prime} b^{\prime} c^{\prime}\right)\left(l m n+l^{\prime} m^{\prime} n^{\prime}\right)$
• B. $a b c l m n+a^{\prime} b^{\prime} c^{\prime} l^{\prime} m^{\prime} n^{\prime}$
• C. $\left(a^2+b^2+c^2\right)\left(l^2+m^2+n^2\right) +\left(a^{\prime 2}+b^{\prime 2}+c^{\prime 2}\right)\left(l^{\prime 2}+m^{\prime 2}+n^{\prime 2}\right)$
• D. 0

Answer 1

Answer: (D)

Write $\Delta=l \Delta_1+l^{\prime} \Delta_2$, where
$\Delta_1 =\begin{vmatrix}a & a m+a^{\prime} m^{\prime} & a n+a^{\prime} n^{\prime} \\b & b m+b^{\prime} m^{\prime} & b n+b^{\prime} n^{\prime} \\c & c m+c^{\prime} m^{\prime} & c n+c^{\prime} n^{\prime}\end{vmatrix} \text { and } $
$\Delta_2 =\begin{vmatrix}a^{\prime} & a m+a^{\prime} m^{\prime} & a n+a^{\prime} n^{\prime} \\b^{\prime} & b m+b^{\prime} m^{\prime} & b n+b^{\prime} n^{\prime} \\c^{\prime} & c m+c^{\prime} m^{\prime} & c n+c^{\prime} n^{\prime}\end{vmatrix}$
In $\Delta_1$ apply $C_2 \rightarrow C_2-m C_1, C_3 \rightarrow C_3-n C_1$ to obtain
$\Delta_1=\begin{vmatrix}a & a^{\prime} m^{\prime} & a^{\prime} n^{\prime} \\b & b^{\prime} m^{\prime} & b^{\prime} n^{\prime} \\c & c^{\prime} m^{\prime} & c^{\prime} n^{\prime}\end{vmatrix}=0$
Similarly $\Delta_2=0$. Thus, $\Delta=0$