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
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Misc 17 (Method 1) Choose the correct answer. If a, b, c, are in A.P., then the determinant |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then, b โ€“ a = c โ€“ b b โ€“ a โ€“ c + b = 0 2b โ€“ a โ€“ c = 0 (Common difference is equal) โ€ฆ(1) Solving |โ– 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 |โ– 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 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@๐Ÿ(๐‘ฅ+3)&๐Ÿ(๐‘ฅ+4)&๐Ÿ(๐‘ฅ+2๐‘)@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@2๐‘ฅ+6&2๐‘ฅ+8&2๐‘ฅ+4๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| Applying R2 โ†’ R2 โ€“ R1 โ€“ R3 = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@2๐‘ฅ+6โˆ’(๐‘ฅ+2)โˆ’(๐‘ฅ+4 )&2๐‘ฅ+8โˆ’(๐‘ฅ+3)โˆ’(๐‘ฅ+5)&2๐‘ฅ+4๐‘โˆ’(๐‘ฅ+2๐‘Ž)โˆ’(๐‘ฅ+2๐‘)@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@2๐‘ฅ+6โˆ’๐‘ฅโˆ’2โˆ’๐‘ฅโˆ’4&2๐‘ฅ+8โˆ’๐‘ฅโˆ’3โˆ’๐‘ฅโˆ’5&2๐‘ฅ+4๐‘โˆ’๐‘ฅโˆ’2๐‘Žโˆ’๐‘ฅโˆ’2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@0&0&4๐‘โˆ’2๐‘Žโˆ’2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@0&0&2(๐Ÿ๐’ƒโˆ’๐’‚โˆ’๐’„)@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@0&0&2(๐ŸŽ)@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| (From (1): 2b โ€“ b โ€“ c = 0) = 1/2 |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@0&0&0@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| If any row or column of determinant are zero, then value of determinant is also zero. = 1/2 ร— 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 |โ– 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 โ€“ b = c โ€“ b b + b = c + a 2b = a + c (Common difference is equal) โ€ฆ(1) Consider |โ– 8(๐‘ฅ+2&๐‘ฅ+3&๐‘ฅ+2๐‘Ž@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| Applying R1 โ†’R1 + R3 โ€“ 2R2 = |โ– 8((๐‘ฅ+2)+(๐‘ฅ+4)โˆ’2(๐‘ฅ+3)&(๐‘ฅ+3)+(๐‘ฅ+5)โˆ’2(๐‘ฅ+4)&(๐‘ฅ+2๐‘Ž)+(๐‘ฅ+2๐‘)โˆ’2(๐‘ฅ+2๐‘)@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = |โ– 8(๐‘ฅ+2+๐‘ฅ+4โˆ’2๐‘ฅโˆ’6&๐‘ฅ+3+๐‘ฅ+5โˆ’2๐‘ฅโˆ’8&๐‘ฅ+2๐‘Ž+๐‘ฅ+2๐‘โˆ’2๐‘ฅโˆ’4๐‘@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = |โ– 8(2๐‘ฅโˆ’2๐‘ฅ+6โˆ’6&2๐‘ฅโˆ’2๐‘ฅ+8โˆ’8&2๐‘ฅโˆ’2๐‘ฅ+2๐‘Ž+2๐‘โˆ’4๐‘@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = |โ– 8(0&0&0+2(๐’‚+๐’„โˆ’2๐‘)@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = |โ– 8(0&0&2(๐Ÿ๐’ƒโˆ’2๐‘)@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| = |โ– 8(0&0&0@๐‘ฅ+3&๐‘ฅ+4&๐‘ฅ+2๐‘@๐‘ฅ+4&๐‘ฅ+5&๐‘ฅ+2๐‘)| 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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