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  • explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 36 maths

explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 36 maths

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best_answer

Answer: 0,-6

Hint: Here we use basic determinant

Given: \left[\begin{array}{ccc} 2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x \end{array}\right]=0

Solution: Here, firstly

                2-x\left[\begin{array}{ll} 2-x & 2+x \\ 2+x & 2-x \end{array}\right]-2+x\left[\begin{array}{cc} 2+x & 2+x \\ 2+x & 2-x \end{array}\right]+2+x\left[\begin{array}{cc} 2+x & 2-x \\ 2+x & 2+x \end{array}\right]

                \begin{aligned} &0=(2-x)\left[(2-x)^{2}-(2+x)^{2}\right]-[2+x]\left[2^{2}-x^{2}[2+x]^{2}\right]+ &(2+x)\left[(2+x)^{2}-\left[\left(2^{2}-x^{2}\right)\right]\right. \end{aligned}

                \begin{aligned} &0=8 x^{2}+16 x+8 x+8 x^{2}+8 x^{2}+4 x^{3}\end{aligned}

                \begin{aligned} &0=8 x^{2}+8 x^{2}+8 x^{2}+4 x^{3} \end{aligned}

                \begin{aligned} &0=24 x^{2}+4 x^{3} \end{aligned}

Let’s take 4x2 common

                \begin{aligned} &4 x^{2}(6+x)=0 \\ &4 x^{2}=0 \text { and } 6+x=0 \\ &x^{2}=0 \text { and } x=-6 \end{aligned}

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