#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 2 sub question 4 Maths Textbook Solution.

Answer:-$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=(a-b)(b-c)(c-a)$

Hint: We will try to make some elements of the determinant into zero

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

Solution :$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} 0 & a-b & b c-c a \\ 0 & b-c & c a-a b \\ 1 & c & a b \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & a-b & -c(a-b) \\ 0 & b-c & -a(b-c) \\ 1 & c & a b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking common }(a-b) \text { from } \mathrm{R}_{1} \text { and }(b-c) \text { from } \mathrm{R}_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & 1 & -c \\ 0 & 1 & -a \\ 1 & c & a b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a-b)(b-c)\left[0\left|\begin{array}{ll} 1 & -a \\ c & a b \end{array}\right|-1\left|\begin{array}{cc} 0 & -a \\ 1 & a b \end{array}\right|+(-c)\left|\begin{array}{ll} 0 & 1 \\ 1 & c \end{array}\right|\right]\\ &=(a-b)(b-c)[0-1\{0 \times a b-(-a) \times 1\}-c\{0 \times c-1 \times 1\}]\\ &=(a-b)(b-c)[-1(+a)-c(-1)]\\ &=(a-b)(b-c)(-a+c)\\ &=(a-b)(b-c)(c-a) \end{aligned}

Hence $\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=(a-b)(b-c)(c-a)$

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