Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 23 Maths textbook Solution.

Please Solve R. D. Sharma class 12 Chapter determinants Exercise  5.2  Question 23 Maths textbook Solution.

Answers (1)

Answer:\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)

Hint  We will make some elements of the determinant zero

Given:\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)

Solution:

\text { L.H.S }\left|\begin{array}{ccc} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|

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

\begin{aligned} &\text { Consider } a^{2}+b c-b^{2}-c a \\ &=a^{2}-b^{2}+b c-c a \\ &=(a-b)(a+b)-(a-b) \\ &=(a-b)(a+b-c) \end{aligned}

\begin{aligned} &\text { Similarly }\\ &b^{2}+c a-c^{2}-a b=(b-c)(b+c-a)\\ &\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \end{aligned}

\begin{aligned} &\text { Now from }(1)\\ &=\left|\begin{array}{ccc} 0 & (a-b)(a+b-c) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b-c)(b+c-a) & (b-c)\left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \end{aligned}

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

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

\begin{aligned} &\text { Consider } a^{2}+a b-b c-c^{2} \\ &=a^{2}-c^{2}+a b-b c \\ &=(a+c)(a-c)+b(a-c) \\ &=(a-c)(a+c+b) \end{aligned}

From (2)

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{1}\\ &=-(a-b)(b-c)(c-a)\left[0\left|\begin{array}{ccc} b+c-a & b^{2}+b c+c^{2} \\ c^{2}+a b & c^{3} \end{array}\right|-0\left|\begin{array}{cc} 2 & a+b+c \\ c^{2}+a b & c^{3} \end{array}\right|+1 \mid \begin{array}{cc} 2 & a+b+c \\ b+c+a & b^{2}+b c+c^{2} \end{array}\right]\\ &=-(a-b)(b-c)(c-a)\left[0-0+\left\{2\left(b^{2}+b c+c^{2}\right)-(b+c-a)(b+c-a)\right\}\right]\\ &=-(a-b)(b-c)(c-a)\left[2 b^{2}+2 b c+2 c^{2}-\left\{(b+c)^{2}-a^{2}\right\}\right]\\ &=-(a-b)(b-c)(c-a)\left\{2 b^{2}+2 b c+2 c^{2}-\left(b^{2}+2 b c+c^{2}-a^{2}\right)\right\}\\ &=-(a-b)(b-c)(c-a)\left(2 b^{2}+2 b c+2 c^{2}-b^{2}-2 b c-c^{2}+a^{2}\right)\\ &=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right) \end{aligned}

Hence it is proved that

\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)

 

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question