#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 22 maths Textbook Solution.

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

Hint $\text { We will try to convert some elements of the determinant }(a-b),(b-c) \text { and }(c-a) \text { to get the final answer }$

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

Solution:

$\text { L.H.S }\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|$

\begin{aligned} &\text { On applying } C_{1} \rightarrow C_{2}-2 C_{1}-2 C_{3}\\ &=\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2}-2 a^{2}-2 b c & b c \\ b^{2} & b^{2}-(c-a)^{2}-2 b^{2}-2 c a & c a \\ c^{2} & c^{2}-(a-b)^{2}-2 c^{2}-2 a b & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -a^{2}-(b-c)^{2}-2 b c & b c \\ b^{2} & -b^{2}-(c-a)^{2}-2 c a & c a \\ c^{2} & -c^{2}-(a-b)^{2}-2 a b & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -\left\{a^{2}+(b-c)^{2}+2 b c\right\} & b c \\ b^{2} & -\left\{b^{2}+(c-a)^{2}+2 c a\right\} & c a \\ c^{2} & -\left\{c^{2}+(a-b)^{2}+2 a b\right\} & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -\left(a^{2}+b^{2}+c^{2}-2 b c+2 b c\right) & b c \\ b^{2} & -\left(b^{2}+c^{2}+a^{2}-2 c a+2 c a\right) & c a \\ c^{2} & -\left(c^{2}+a^{2}+b^{2}-2 a b+2 a b\right) & a b \end{array}\right| \end{aligned}

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

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1} \\ &\left.=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[a^{2}\left|\begin{array}{cc} 0 & c \\ 0 & -b \end{array}\right|-1\left|\begin{array}{cc} -(a+b) & c \\ c+a & -b \end{array}\right|+b c \mid \begin{array}{cc} -(a+b) & 0 \\ c+a & 0 \end{array}\right]\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[a^{2}(0-0)-1\{(-b)(-(a+b)-c(c+a))+b c(0-0)\}\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[0-1\left\{b(a+b)-c^{2}-a c\right\}\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left\{-1\left(a b+b^{2}-c^{2}-a c\right)\right\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left(-a b-b^{2}+c^{2}+a c\right) \end{aligned}

\begin{aligned} &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left(a c-a b-b^{2}+c^{2}\right) \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left\{-a(b-c)-\left(b^{2}-c^{2}\right)\right\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\{-a(b-c)-(b+c)(b-c)\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)(b-c)(-a-b-c) \\ &=(-)(-)\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)(b-c)(a+b+c) \\ &=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\ &=R H . S \end{aligned}

Hence it is proved that

$\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$