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Misc. 16 Important

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Matrices and Determinants - Formula Sheet and Summary Important

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at March 16, 2023 by Teachoo

Misc 17 (Method 1) Choose the correct answer. If a, b, c, are in A.P., then the determinant |■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, b – a = c – b b – a – c + b = 0 2b – a – c = 0 (Common difference is equal) …(1) Solving |■8(x+2&x+3&[email protected]+3&x+4&[email protected]+4&x+5&x+2c)| Multiplying and dividing 2 = 2/2 |■8(x+2&x+3&[email protected]+3&x+4&[email protected]+4&x+5&x+2c)| Multiplying R2 by 2 = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@𝟐(𝑥+3)&𝟐(𝑥+4)&𝟐(𝑥+2𝑏)@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(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 |■8(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 |■8(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 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&4𝑏−2𝑎−2𝑐@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&2(𝟐𝒃−𝒂−𝒄)@𝑥+4&𝑥+5&𝑥+2𝑐)| = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&2(𝟎)@𝑥+4&𝑥+5&𝑥+2𝑐)| (From (1): 2b – b – c = 0) = 1/2 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@0&0&[email protected]𝑥+4&𝑥+5&𝑥+2𝑐)| If any row or 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 |■8(x+2&x+3&[email protected]+3&x+4&[email protected]+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 |■8(𝑥+2&𝑥+3&𝑥+2𝑎@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| Applying R1 →R1 + R3 – 2R2 = |■8((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)| = |■8(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)| = |■8(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)| = |■8(0&0&0+2(𝒂+𝒄−2𝑏)@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(0&0&2(𝟐𝒃−2𝑏)@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐)| = |■8(0&0&[email protected]𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+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)