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  • 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)

 

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