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  • Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 12 maths

Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 12 maths

Answers (1)

Answer:

Correct option (b)

Hint:

First solve the given determinant, then conclude answer.

Given:

        \left | a\, b\, 2a\alpha +3b\: b\: c\: 2b\alpha +3c\: 2a\alpha +3b\: 2b\alpha +3c\: 0 \right |=0

We have to find the conclusion of determinant

Solution:

        \begin{vmatrix} a &b &2a\alpha +3b \\ b &c &2b\alpha +3c \\ 2a\alpha +3b &2b\alpha +3c &0 \end{vmatrix}=0

Expanding along R1, we have

a\left[-(2 b \alpha+3 c)^{2}\right]-b[-(2 b \alpha+3 c)(2 a \alpha+3 b)]+(2 a \alpha+3 b)[b(2 b \alpha+3 c)-c(2 a \alpha+3 b)]=0

        \Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2b^{2}\alpha +3bc-2ac\alpha -3bc)=0

        \Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2\alpha (b^{2}-ac))=0

        \Rightarrow (2b\alpha +3c)[-2ab\alpha -3ac+2ab\alpha +3b^{2}]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0

        \Rightarrow (2b\alpha +3c)[3b^{2}-3ac]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0

        \Rightarrow (b^{2}-ac)[6b\alpha +9c+4a\alpha ^{2}+6b\alpha ]=0

        \Rightarrow (b^{2}-ac)[4a\alpha ^{2}+12b\alpha +9c ]=0

        \Rightarrow (b^{2}-ac)=0\; \; or\; \;[4a\alpha ^{2}+12b\alpha +9c ]=0

        \Rightarrow a,b,c\: are in\; G.P\; or\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; whose\; root\; is\; \alpha .

        Hence\; \alpha \; is\;a\; root\; o\! f\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; or\; a,b,c\; are\; in\; G.P .

Posted by

Gurleen Kaur

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