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

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

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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}

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