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#### Please Solve R.D. Sharma class 12 Chapter determinants Exercise 5.2 Question 2 sub question 7 Maths textbook Solution.

Answer:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|=(x+2)(x-1)^{2}$

Hint: We will try to make some elements of the determinant into zero

Given:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|$

Solution:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|$

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

\begin{aligned} &\text { On taking common }(x+2) \text { from } \mathrm{R}_{1}\\ &=(x+2)\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=(x+2)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1-x & x-1 & 1 \\ 1 & 1-x & x \end{array}\right| \end{aligned}

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

Hence $\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|=(x+2)(x-1)^{2}$