#### Please Solve R. D. Sharma class 12 Chapter determinants Exercise  5.2  Question 24 Maths textbook Solution.

Answer:$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$

Hint We will convert some elements of the determinant into zero

Given: $\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right| \\ &=\left|\begin{array}{ccc} a^{2} & b c & c(a+c) \\ a(a+b) & b^{2} & a c \\ a b & b(b+c) & c^{2} \end{array}\right| \end{aligned}

\begin{aligned} &\text { Taking out 'a' common from } \mathrm{C}_{1}, \text { and } \mathrm{b} \text { from } \mathrm{C}_{2} \text { and } \mathrm{c} \text { from } \mathrm{C}_{3}\\ &=a b c\left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \\ &=a b c\left|\begin{array}{ccc} a+c+a+c & c & a+c \\ a+b+b+a & b & a \\ b+b+c+c & b+c & c \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} 2 a+2 c & c & a+c \\ 2 a+2 b & b & a \\ 2 b+2 c & b+c & c \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} 2(a+c) & c & a+c \\ 2(a+b) & b & a \\ 2(b+c) & b+c & c \end{array}\right| \end{aligned}

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

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{2}-\mathrm{C}_{1}, C_{3} \rightarrow C_{3}-C_{1} \\ &=2 a b c\left|\begin{array}{ccc} (a+c) & c-a-c & (a+c)-(a+c) \mid \\ (a+b) & b-a-b & a-a-b \\ (b+c) & (b+c)-(b+c) & c-b-c \end{array}\right| \\ &=2 a b c\left|\begin{array}{ccc} (a+c) & -a & 0 \\ (a+b) & -a & -b \\ (b+c) & 0 & -b \end{array}\right| \end{aligned}

$\text { On taking }(-a) \text { common from } \mathrm{C}_{2} \text { and }(-b) \text { from } \mathrm{C}_{3}$

$=2 a b c(-a)(-b)\left|\begin{array}{lll} (a+c) & 1 & 0 \\ (a+b) & 1 & 1 \\ (b+c) & 0 & 1 \end{array}\right|$

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=2 a^{2} b^{2} c\left[(a+c)\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|-1\left|\begin{array}{ll} a+b & 1 \\ b+c & 1 \end{array}\right|+0\left|\begin{array}{ll} a+b & 1 \\ b+c & 0 \end{array}\right|\right]\\ &=2 a^{2} b^{2} c[(a+c)(1-0)-1(a+b-b-c)+0]\\ &=2 a^{2} b^{2} c[a+c-1(a-c)]\\ &=2 a^{2} b^{2} c(a+c-a+c)\\ &=2 a^{2} b^{2} c \times 2 c\\ &=4 a^{2} b^{2} c^{2}\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$