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  • Need Solution for R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 30 Maths Textbook Solution.

Need Solution for R. D. Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 30 Maths Textbook Solution.

Answers (1)

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

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

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

Solution:

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

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 1+a^{2}+a & a+1+a^{2} & a^{2}+a+1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right| \end{aligned}

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

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=\left(a^{2}+a+1\right)(a-1)^{2}\left[0\left|\begin{array}{cc} -1 & a \\ a+1 & 1 \end{array}\right|-0\left|\begin{array}{cc} a+1 & a \\ -a & 1 \end{array}\right|+1\left|\begin{array}{cc} a+1 & -1 \\ -a & a+1 \end{array}\right|\right]\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left[0-0+1\left\{(a+1)^{2}-(-a)(-1)\right\}\right]\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left(a^{2}+2 a+1-a\right)\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left(a^{2}+a+1\right)\\ &=\left\{(a-1)\left(a^{2}+a+1\right)\right\}^{2}\\ &=\left(a^{3}-1\right)^{2}\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}

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