Misc 17 - If a, b, c, are in AP, then the determinant - Miscellaneous

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  1. Chapter 4 Class 12 Determinants
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Misc 17 Choose the correct answer. If a, b, c, are in A.P., then the determinant 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then b – a = c – b b – a – c + b = 0 2b – a – c = 0 Solving x+2﷮x+3﷮x+2a﷮x+3﷮x+4﷮x+2b﷮x+4﷮x+5﷮x+2c﷯﷯ Multiplying and dividing 2 = 2﷮2﷯ x+2﷮x+3﷮x+2a﷮x+3﷮x+4﷮x+2b﷮x+4﷮x+5﷮x+2c﷯﷯ Multiplying R2 by 2 = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮𝟐(𝑥+3)﷮𝟐(𝑥+4)﷮𝟐(𝑥+2𝑏)﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮2𝑥+6﷮2𝑥+8﷮2𝑥+4𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ Applying R2 → R2 – R1 – R3 = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮2𝑥+6− 𝑥+2﷯−(𝑥+4 )﷮2𝑥+8− 𝑥+3﷯−(𝑥+5)﷮2𝑥+4𝑏− 𝑥+2𝑎﷯−(𝑥+2𝑐)﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮2𝑥+6−𝑥−2−𝑥−4﷮2𝑥+8−𝑥−3−𝑥−5﷮2𝑥+4𝑏−𝑥−2𝑎−𝑥−2𝑐﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮0﷮0﷮4𝑏−2𝑎−2𝑐﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮0﷮0﷮2(𝟐𝒃−𝒂−𝒄)﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮0﷮0﷮2(𝟎)﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 1﷮2﷯ 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮0﷮0﷮0﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ If any row as column of determinant are zero, then value of determinant is also zero. = 1﷮2﷯ × 0 = 0 Thus, the value of determinant is 0 Correct Answer is A Misc. 17(Method 2) Choose the correct answer. If a, b, c, are in A.P., then the determinant x+2﷮x+3﷮x+2a﷮x+3﷮x+4﷮x+2b﷮x+4﷮x+5﷮x+2c﷯﷯ is A. 0 B. 1 C. x D. 2x We need to find value of x+2﷮x+3﷮x+2a﷮x+3﷮x+4﷮x+2b﷮x+4﷮x+5﷮x+2c﷯﷯ Since a, b & c are in A.P Then a – b = c – b b + b = c + a 2b = a + c Consider 𝑥+2﷮𝑥+3﷮𝑥+2𝑎﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ Applying R1 →R1 + R3 – 2R2 = 𝑥+2﷯+ 𝑥+4﷯−2(𝑥+3)﷮ 𝑥+3﷯+ 𝑥+5﷯−2(𝑥+4)﷮ 𝑥+2𝑎﷯+ 𝑥+2𝑐﷯−2(𝑥+2𝑏)﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 𝑥+2+𝑥+4−2𝑥−6﷮𝑥+3+𝑥+5−2𝑥−8﷮𝑥+2𝑎+𝑥+2𝑐−2𝑥−4𝑏﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 2𝑥−2𝑥+6−6﷮2𝑥−2𝑥+8−8﷮2𝑥−2𝑥+2𝑎+2𝑐−4𝑏﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 0﷮0﷮0+2(𝒂+𝒄−2𝑏)﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 0﷮0﷮2(𝟐𝒃−2𝑏)﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ = 0﷮0﷮0﷮𝑥+3﷮𝑥+4﷮𝑥+2𝑏﷮𝑥+4﷮𝑥+5﷮𝑥+2𝑐﷯﷯ If any row as column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0 Correct Answer is A

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