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

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

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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}

 

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