#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 29 Maths Textbook Solution.

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

Hint We will try to convert some elements of determinant into zero

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

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$

\begin{aligned} &=\left|\begin{array}{ccc} 1+a^{2} & a b & a c \\ a b & 1+b^{2} & b c \\ c a & c b & 1+c^{2} \end{array}\right|\\ &\text { Multiply } C_{1} \text { by } a, C_{2} \text { by } b \text { and } C_{3} \text { by } c\\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(1+a^{2}\right) & a b^{2} & a c^{2} \\ a^{2} b & b\left(1+b^{2}\right) & b c^{2} \\ c a^{2} & c b^{2} & c\left(1+c^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking } a \text { common from } \mathrm{R}_{1}, \mathrm{~b} \text { from } \mathrm{R}_{2} \text { and } \mathrm{c} \text { from } \mathrm{R}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3}\\ &=\left|\begin{array}{ccc} 1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & 1+b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & b^{2} & 1+c^{2} \end{array}\right| \end{aligned}

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

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

Hence it is proved that

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