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#### Explain Solution R. D .Sharma Class 12 Chapter  deteminants Exercise 5.2 Question  18 maths Textbook Solution.

Answer:$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$

Hint  We will make column (3) of L.H.S abc

Given:$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$

Solution:

$\text { L.H.S }\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$

$\text { On multiplying by 'a' in } R_{1}=\text { by } b \text { ' in } R_{2} \text { and by 'c' in } R_{m}$

\begin{aligned} &=\frac{1}{a b c}\left|\begin{array}{lll} a & a^{2} & a b c \\ b & b^{2} & a b c \\ c & c^{2} & a b c \end{array}\right|\\ &\text { Taking common }(a b c) \text { from } \mathrm{C}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \end{aligned}

If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.

\begin{aligned} &C_{2} \leftrightarrow C_{3} \\ &=(-)\left|\begin{array}{lll} a & 1 & a^{2} \\ b & 1 & b^{2} \\ c & 1 & c^{2} \end{array}\right| \\ &C_{1} \leftrightarrow C_{2} \\ &=(-)(-)\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$