# Example 29 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Sept. 21, 2018 by Teachoo

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6 Deleted for CBSE Board 2022 Exams

Example 7 Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Example 9 Important Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Deleted for CBSE Board 2022 Exams

Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Important Deleted for CBSE Board 2022 Exams

Example 16 Important Deleted for CBSE Board 2022 Exams

Example 17

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23

Example 24 Important

Example 25

Example 26 Important

Example 27

Example 28 Important

Example 29 You are here

Example 30 Deleted for CBSE Board 2022 Exams

Example 31 Important Deleted for CBSE Board 2022 Exams

Example 32 Important Deleted for CBSE Board 2022 Exams

Example 33 Important

Example 34 Important Deleted for CBSE Board 2022 Exams

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

Last updated at Sept. 21, 2018 by Teachoo

Example 29 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method. Let the first, second & third number be x, y, z respectively Given, x + y + z = 6 y + 3z = 11 x + z = 2y or x 2y + z = 0 Step 1 Write equation as AX = B 1 1 1 0 1 3 1 2 1 = 6 11 0 Hence A = 1 1 1 0 1 3 1 2 1 , X = & B = 6 11 0 Step 2 Calculate |A| = 1 1 1 0 1 3 1 2 1 = 1 (1 + 6) 0 (1 + 2) + 1 (3 1) = 7 + 2 = 9 So, |A| 0 The system of equation is consistent & has a unique solutions Now, AX = B X = A-1 B Hence A = 1 1 1 0 1 3 1 2 1 , X = & B = 6 11 0 = 1 (1 + 6) 0 (1 + 2) + 1 (3 1) = 7 + 2 = 9 0 Since determinant is not equal to O, A 1 exists Now find adj (A) adj (A) = 11 12 13 21 22 23 31 32 33 = 11 21 31 12 22 32 13 32 33 Now, AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1 |A| adj (A) adj A = A11 A12 A13 A21 A22 A23 A31 A32 A33 = A11 A21 A31 A12 A22 A32 A13 A23 A33 A = 1 1 2 3 4 5 2 1 3 11 = 1 1 3 ( 2) = 1 + 6 = 7 12 = 0 1 3 1 = ( 3) = 3 13 = 0 2 1 1= 1= 21 = 1 1 2 1 = 1+2 = 3 22 = 1 1 1 1 = 1 1 = 0 23 = 1 2 1 1 = 2 1 = 3 = 3 31 = 1 3 1 1 = 3 1 = 2 32 = 1 3 0 1 = 3 0 = 3 33 = 1 1 1 0 = 1 0 = 1 Hence, adj (A) = 7 3 2 3 0 3 1 3 1 Now, A 1 = 1 adj (A) A 1 = 1 9 7 3 2 3 0 3 1 3 1 Solution of given system of equations is X = A 1 B = 1 9 7 3 2 3 0 3 1 3 1 6 11 0 = 1 9 42 33+0 18+0+0 6+33+0 = 1 9 9 18 27 = 1 2 3 x = 1, y = 2, z = 3