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Misc 17 - If a, b, c are in AP, then the determinant | x+2 x+3 x+2a

Misc 17 - Chapter 4 Class 12 Determinants - Part 2
Misc 17 - Chapter 4 Class 12 Determinants - Part 3
Misc 17 - Chapter 4 Class 12 Determinants - Part 4
Misc 17 - Chapter 4 Class 12 Determinants - Part 5
Misc 17 - Chapter 4 Class 12 Determinants - Part 6
Misc 17 - Chapter 4 Class 12 Determinants - Part 7

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Misc 17 (Method 1) Choose the correct answer. If a, b, c, are in A.P., then the determinant |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+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 (Common difference is equal) …(1) Solving |β– 8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying and dividing 2 = 2/2 |β– 8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying R2 by 2 = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@𝟐(π‘₯+3)&𝟐(π‘₯+4)&𝟐(π‘₯+2𝑏)@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@2π‘₯+6&2π‘₯+8&2π‘₯+4𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| Applying R2 β†’ R2 – R1 – R3 = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@2π‘₯+6βˆ’(π‘₯+2)βˆ’(π‘₯+4 )&2π‘₯+8βˆ’(π‘₯+3)βˆ’(π‘₯+5)&2π‘₯+4π‘βˆ’(π‘₯+2π‘Ž)βˆ’(π‘₯+2𝑐)@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@2π‘₯+6βˆ’π‘₯βˆ’2βˆ’π‘₯βˆ’4&2π‘₯+8βˆ’π‘₯βˆ’3βˆ’π‘₯βˆ’5&2π‘₯+4π‘βˆ’π‘₯βˆ’2π‘Žβˆ’π‘₯βˆ’2𝑐@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@0&0&4π‘βˆ’2π‘Žβˆ’2𝑐@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@0&0&2(πŸπ’ƒβˆ’π’‚βˆ’π’„)@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@0&0&2(𝟎)@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| (From (1): 2b – b – c = 0) = 1/2 |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@0&0&0@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| If any row or column of determinant are zero, then value of determinant is also zero. = 1/2 Γ— 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 |β– 8(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 a – b = c – b b + b = c + a 2b = a + c (Common difference is equal) …(1) Consider |β– 8(π‘₯+2&π‘₯+3&π‘₯+2π‘Ž@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| Applying R1 β†’R1 + R3 – 2R2 = |β– 8((π‘₯+2)+(π‘₯+4)βˆ’2(π‘₯+3)&(π‘₯+3)+(π‘₯+5)βˆ’2(π‘₯+4)&(π‘₯+2π‘Ž)+(π‘₯+2𝑐)βˆ’2(π‘₯+2𝑏)@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = |β– 8(π‘₯+2+π‘₯+4βˆ’2π‘₯βˆ’6&π‘₯+3+π‘₯+5βˆ’2π‘₯βˆ’8&π‘₯+2π‘Ž+π‘₯+2π‘βˆ’2π‘₯βˆ’4𝑏@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = |β– 8(2π‘₯βˆ’2π‘₯+6βˆ’6&2π‘₯βˆ’2π‘₯+8βˆ’8&2π‘₯βˆ’2π‘₯+2π‘Ž+2π‘βˆ’4𝑏@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = |β– 8(0&0&0+2(𝒂+π’„βˆ’2𝑏)@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = |β– 8(0&0&2(πŸπ’ƒβˆ’2𝑏)@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| = |β– 8(0&0&0@π‘₯+3&π‘₯+4&π‘₯+2𝑏@π‘₯+4&π‘₯+5&π‘₯+2𝑐)| If any row or column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0 Correct Answer is A (From (1): 2b = a + c)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.