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Need solution for RD Sharma maths class 12 chapter Determinants exercise 5.1 question 12 subquestion (ii)

Answers (1)

Answer:

x=2\; or\; x=-1

Hint:

If matrix is singular, then |A| = 0

Given:

\mathrm{A}=\left[\begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array}\right]

Solution:

\begin{aligned} &|\mathbf{A}|=\mathbf{0} \\ &(\mathrm{x}-1)\left(\mathrm{x}^{2}-2 \mathrm{x}\right)-1(\mathrm{x}-2)+1(2-\mathrm{x})=0 \\ &\mathrm{x}^{3}-2 \mathrm{x}^{2}-\mathrm{x}^{2}+2 \mathrm{x}-\mathrm{x}+2-\mathrm{x}+2=0 \\ &\mathrm{x}^{3}-3 \mathrm{x}^{2}+4=0 \\ &(\mathrm{x}-2)^{2}(\mathrm{x}+1)=0 \\ &\Rightarrow[\mathrm{x}=2] \text { or }[\mathrm{x}=-1] \\ &x=-1,2 \end{aligned}

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Gurleen Kaur

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