# Solve this

Question:

If $\Delta_{1}=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 & b c & a \\ 1 & c a & b \\ 1 & a b & c\end{array}\right|$, then

(a) $\Delta_{1}+\Delta_{2}=0$

(b) $\Delta_{1}+2 \Delta_{2}=0$

(c) $\Delta_{1}=\Delta_{2}$

(d) none of these

Solution:

(a) $\Delta_{1}+\Delta_{2}=0$

$\Delta_{2}=\mid \begin{array}{lll}1 & b c & a\end{array}$

$1 \quad c a \quad b$

$\begin{array}{lll}1 & a b & c\end{array}$

$=\frac{1}{a b c} \mid \begin{array}{lll}a & a b c & a^{2}\end{array}$

$\begin{array}{lcc}c & c a b & c^{2} \mid\end{array}\left[R_{1}, R_{2}, R_{3}\right.$ are multiplied by a, b and c respectively, therefore we divide by abc]

$=\frac{a b c}{a b c} \mid a \quad 1 \quad a^{2}$

$b \quad 1 \quad b^{2}$

$c \quad 1 \quad c^{2}$       [Taking $a b c$ common from $C_{2}$ ]

$=-\mid \begin{array}{lll}1 & a & a^{2}\end{array}$

$1 \quad b \quad b^{2}$

$\begin{array}{llll}1 & c & c^{2} & C_{1} \leftrightarrow C_{2}\end{array}$

We know that the value of a determinant remains unchanged if its rows and columns are interchanged. So,

$\Delta_{2}=-\mid \begin{array}{lll}1 & 1 & 1\end{array}$

$a \quad b \quad c$

$\begin{array}{lll}a^{2} & b^{2} & c^{2}\end{array}$

$=-\Delta_{1}$

$\Rightarrow \Delta_{1}+\Delta_{2}=0$