Example 31 - Chapter 4 Class 12 Determinants

Transcript

Example 31 (Method 1) If a, b, c, are in A.P, find value of 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 2y+4 5y+7 8y+a 3y+5 6y+8 9y+b 4y+6 7y+9 10y+c Multiply & Divide by 2 = 2 2 2y+4 5y+7 8y+a 3y+5 6y+8 9y+b 4y+6 7y+9 10y+c Multiplying 2 to R2 = 1 2 2y+4 5y+7 8y+a (3y+5) (6y+8) (9y+b) 4y+6 7y+9 10y+c = 1 2 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 2y+4 5y+7 8y+a 6y+10 2 +4 (4 +6) 12y+16 5 +7 (7 +9) 18y+2b 8y+a (10y+c) 4y+6 7y+9 10y+c = 1 2 2y+4 5y+7 8y+a 6y+10 2 4 4 6 12y+16 5 7 7 9 18y+2b 2 10 4y+6 7y+9 10y+c = 1 2 2 +4 5 +7 8 + 0 0 4 +6 7 +9 10 + = 1 2 2 +4 5 +7 8 + 0 0 4 +6 7 +9 10 + 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 Example 31(Method 2) If a, b, c, are in A.P, find value of 2 +4 5 +7 8 + 3 +5 6 +8 9 + 4 +6 7 +9 10 + We need to find value of 2 +4 5 +7 8 + 3 +5 6 +8 9 + 4 +6 7 +9 10 + Given a, b & C are in A.P Then b a = c b b + b = a + c 2b = a + c Consider 2 +4 5 +7 8 + 3 +5 6 +8 9 + 4 +6 7 +9 10 + Applying R1 R1 + R3 2R2 = 2 +4+ 4 +6 2(3 +5) 5 +7+ 7 +9 2(6 +8) 8 + + 10 + 2(9 + ) 3 +5 6 +8 9 + 4 +6 7 +9 10 + = 2 +4+4 +6 6 10 5 +7+7 +9 12 16 8 + +10 + 18 2 3 +5 6 +8 9 + 4 +6 7 +9 10 + = 6 6 +10 10 12 12 +16 16 18 18 + + 2 3 +5 6 +8 9 + 4 +6 7 +9 10 + = 0 0 + 2 3 +5 6 +8 9 + 4 +6 7 +9 10 + = 0 0 2 3 +5 6 +8 9 + 4 +6 7 +9 10 + = 0 0 0 3 +5 6 +8 9 + 4 +6 7 +9 10 + If any row as column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0