# Example 31 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

Transcript

Example 31 (Method 1) If a, b, c, are in A.P, find value of |■8(2y+4&5y+7&8y+a@3y+5&6y+8&9y+b@4y+6&7y+9&10y+c)| Given a, b & c are in A.P Then, b – a = c – b b – a – c + b = 0 2b – a – c = 0 Solving (Common difference is equal) |■8(2y+4&5y+7&8y+a@3y+5&6y+8&9y+b@4y+6&7y+9&10y+c)| Multiply & Divide by 2 = 2/2 |■8(2y+4&5y+7&8y+a@3y+5&6y+8&9y+b@4y+6&7y+9&10y+c)| Multiplying 2 to R2 = 1/2 |■8(2y+4&5y+7&8y+a@𝟐(3y+5)&𝟐(6y+8)&𝟐(9y+b)@4y+6&7y+9&10y+c)| = 1/2 |■8(2y+4&5y+7&8y+a@6y+10&12y+16&18y+2b@4y+6&7y+9&10y+c)| Applying R2 →R2 – R1 – R3 = 1/2 |■8(2y+4&5y+7&8y+a@6y+10−(2𝑦+4)−(4𝑦+6)&12y+16−(5𝑦+7)−(7𝑦+9)&18y+2b−(8y+a)−(10y+c)@4y+6&7y+9&10y+c)| = 1/2 |■8(2y+4&5y+7&8y+a@6y+10−2𝑦−4−4𝑦−6&12y+16−5𝑦−7−7𝑦−9&18y+2b−2𝑦−𝑎−10𝑦−𝑐@4y+6&7y+9&10y+c)| = 1/2 |■8(2𝑦+4&5𝑦+7&8𝑦+𝑎@0&0&𝟐𝒃−𝒂−𝒄@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = 1/2 |■8(2𝑦+4&5𝑦+7&8𝑦+𝑎@0&0&𝟎@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = 1/2 × 0 = 0 Thus, the value of determinant is 0 (From (1): 2b – a – c = 0) If any row or column of determinant are zero, then value of determinant is also zero. Example 31 (Method 2) If a, b, c, are in A.P, find value of |■8(2𝑦+4&5𝑦+7&8𝑦+𝑎@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| Given a, b & C are in A.P Then b – a = c – b b + b = a + c 2b = a + c (Common difference is equal) Consider |■8(2𝑦+4&5𝑦+7&8𝑦+𝑎@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| Applying R1 → R1 + R3 – 2R2 = |■8(2𝑦+4+(4𝑦+6)−2(3𝑦+5)&5𝑦+7+(7𝑦+9)−2(6𝑦+8)&8𝑦+𝑎+(10𝑦+𝑐)−2(9𝑦+𝑏)@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = |■8(2𝑦+4+4𝑦+6−6𝑦−10&5𝑦+7+7𝑦+9−12𝑦−16&8𝑦+𝑎+10𝑦+𝑐−18𝑦−2𝑏@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = |■8(6𝑦−6𝑦+10−10&12𝑦−12𝑦+16−16&18𝑦−18𝑦+𝑎+𝑐−2𝑏@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = |■8(0&0&𝒂+𝒄−2𝑏@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = |■8(0&0&𝟐𝒃−2𝑏@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| = |■8(0&0&0@3𝑦+5&6𝑦+8&9𝑦+𝑏@4𝑦+6&7𝑦+9&10𝑦+𝑐)| If any row or column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0 (From (1): 2b = a + c)

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Chapter 4 Class 12 Determinants (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.