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- 1. DETERMINANT
- 2. DEFINITION
- 3. EXPANSION OF DETERMINANT
- 4. MINORS
- 5. COFACTORCofactor of the element aij is Cij = (–1 )i+j. Mij ; where i & jdenotes the row & column in which the particular elementlies.Note that the value of a determinant of order three in termsof ‘Minor’ & ‘Cofactor’ can be written as: D = a11M11 – a12M12 + a13M13 or D = a11C11 + a12C12 + a13C13 & so on.
- 6. TRANSPOSE OF A DETERMINANT
- 7. SYMMETRIC, SKEW-SYMMETRIC, ASYMMETRICDETERMINANTSI. A determinant is symmetric if it is identical to its transpose. Its ith row is identical to its ith column i.e. aij = aji for all values of i and j II. A determinant is skew-symmetric if it is identical to its transpose having sign of each element inverted i.e. aij = – aji for all values of i and j . A skew-symmetric determinant has all elements zero in its principal diagonal.III. A determinant is asymmetric if it is neither symmetric nor skew-symmetric.
- 8. PROPERTIES OF DETERMINANTS
- 9.
- 10.
- 11.
- 12. EXAMPLES
- 13. USE OF FACTOR THEOREM TO FIND THE VALUE OFDETERMINANTIf by putting x = a the value of a determinant vanishes then(x – a) is a factor of the determinant.
- 14. EXAMPLES
- 15. MULTIPLICATION OF TWO DETERMINANTS
- 16. EXAMPLES
- 17. SUMMATION OF DETERMINANTS
- 18. EXAMPLES
- 19. INTEGRATION OF A DETERMINANT
- 20. EXAMPLES
- 21. DIFFERENTIATION OF DETERMINANT
- 22. EXAMPLES
- 23. CRAMERS RULE: SYSTEM OF LINEAR EQUATIONS
- 24.
- 25. Consistency of a system of Equations i. If D 0 and alteast one of D1, D2, D3 0, then the given system of equations are consistent and have unique non trivial solution. ii. If D 0 & D1 = D2 = D3 = 0, then the given system of equations are consistent and have trivial solution only. iii. If D = D1 = D2 = D3 = 0, then the given system of equations have either infinite solutions or no solution.
- 26.
- 27. EXAMPLES Find the nature of solution for the given system of equations. x + 2y + 3z = 1 2x + 3y + 4z = 3 3x + 4y + 5z = 0 Ans. No Solution Solve the following system of equations x + 2y + 3z = 1 2x + 3y + 4z = 2 3x + 4y + 5z = 3 Ans. x = 1 + t y = –2t z = t where t R
- 28. EXAMPLES Solve the following system of equations x + 2y + 3z = 0 2x + 3y + 4z = 0 x–y–z=0 Ans. x = 0, y = 0, z = 0 Let 2x + 3y + 4 = 0 ; 3x + 5y + 6 = 0, 2x2 + 6xy + 5y2 + 8x + 12y + 1 + t = 0, if the system of equations in x and y are consistent then find the value of t. Ans. t = 7
- 29. APPLICATION OF DETERMINANTS
- 30.

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