1. Determinants

2. If is a square matrix of order 1, then |A| = | a 11 | = a 11 If is a square matrix of order 2, then |A| = = a 11 a 22 – a 21 a 12

3. Example

4. Solution If A = is a square matrix of order 3, then [Expanding along first row]

5. Example [Expanding along first row] Solution :

6. Minors

7. M 11 = Minor of a 11 = determinant of the order 2 × 2 square sub-matrix is obtained by leaving first row and first column of A Similarly, M 23 = Minor of a 23 M 32 = Minor of a 32 etc.

8. Cofactors

9. Cofactors (Con.) C 11 = Cofactor of a 11 = (–1) 1 + 1 M 11 = (–1) 1 +1 C 23 = Cofactor of a 23 = (–1) 2 + 3 M 23 = C 32 = Cofactor of a 32 = (–1) 3 + 2 M 32 = etc.

10. Value of Determinant in Terms of Minors and Cofactors

11. Properties 1. If each element of a row (or column) of a determinant is zero, then its value is zero. Proof: Expanding the determinant along the row containing only zeros:

12. Properties 2. If all the elements of a row (or column) is multiplied by a non-zero number k, then the value of the new determinant is k times the value of the original determinant. Proof: expand elong the row multiplied by

13. Properties (Con.) 4.If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two (or more) determinants. Proof: expand along the row containing the sum:

14. Properties of Determinants 2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign. Proof: first we proove it for neighbourh rows:

15. Cont. Interchanging 2 rows (columns) Suppose, that we change the i. and j. rows, and j-i = k. iij i+1ji i+2i+1i+1 ….i+2i+2 I+k=j….… Then we need k neighbourh change to position the j. row right below the i. (see 2nd column) One more change and j stands at the earlier position of i, and i is the row right after j. Again, i must be interchanged by its neighbours k times, so the sum of neighbourgh row changes is: k+k+1=2k+1 – an odd number, so because by each change the deterimant changes it sing, after odd number of changes, its changes its sign.

16. Properties 6.If any two rows (or columns) of a determinant are identical, then its value is zero. Proof: if we interchanged the identical rows, the determinant remains the same, but formally it does changes its sign, so :D = ( - D ), which is possible only if D=0

17. 5.The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column). Properties (Con.) Proof:

18. Properties (Con.) If A is an upper (lower) triangular matrix, then the determinant is equal to the product of the elements of the main diagonal: Proof:

19. 1.The value of a determinant remains unchanged, if its rows and columns are interchanged. (A matrix and its transpose have the same detereminant)

20. Row(Column) Operations Following are the notations to evaluate a determinant: Similar notations can be used to denote column operations by replacing R with C. (i)R i to denote ith row (ii)R i R j to denote the interchange of ith and jth rows. (iii)R i R i + R j to denote the addition of times the elements of jth row to the corresponding elements of ith row. (iv) R i to denote the multiplication of all elements of ith row by.

21. Evaluation of Determinants If a determinant becomes zero on putting is the factor of the determinant., because C 1 and C 2 are identical at x = 2 Hence, (x – 2) is a factor of determinant.

22. Sign System for Expansion of Determinant Sign System for order 2 and order 3 are given by

23. Example-1 Find the value of the following determinants (i)(ii) Solution :

24. Example –1 (ii) (ii)

25. Evaluate the determinant Solution : Example - 2

26. Example - 3 Evaluate the determinant: Solution:

27. Now expanding along C 1, we get (a-b) (b-c) (c-a) [- (c 2 – ab – ac – bc – c 2 )] = (a-b) (b-c) (c-a) (ab + bc + ac) Solution Cont.

28. Without expanding the determinant, prove that Example-4 Solution :

29. Solution Cont.

30. Prove that : = 0, where is cube root of unity. Example -5 Solution :

31. Example-6 Prove that : Solution :

32. Solution cont. Expanding along C 1, we get (x + a + b + c) [1(x 2 )] = x 2 (x + a + b + c) = R.H.S