BSC_COMPUTER _SCIENCE_UNIT-3_DISCRETE MATHEMATICS

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Class: B.Sc CS.
Subject: Discrete Mathematics
Unit-3 Basics Of Matrix
RAI UNIVERSITY, AHMEDABAD

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BSC_COMPUTER _SCIENCE_UNIT-3_DISCRETE MATHEMATICS

  1. 1. Class: B.Sc CS. Subject: Discrete Mathematics Unit-3 RAI UNIVERSITY, AHMEDABAD
  2. 2. UNIT-III: Basics Of Matrix  Definition: Matrix οƒ˜ A ractangular array of m rows and n columns, enclosed by brackets [ ] is called a matrix of order π‘š Γ— 𝑛. A matrix of order 3 Γ— 3 is expressed as, 𝐴 = [ π‘Ž11 π‘Ž12 π‘Ž13 π‘Ž21 π‘Ž22 π‘Ž23 π‘Ž31 π‘Ž32 π‘Ž33 ] οƒ˜ An element π‘Žπ‘–π‘— denotes, ith row and jth column π‘Ž23 denotes, 2nd row and 3rd column. οƒ˜ Matrices are denoted by capital letters A,B,C,……..etc.  Types of Matrices: 1. Row Matrix: A matrix having only single row is called row matrix. Its order is 1 Γ— 𝑛. For example, 𝐴 = [1 4]1Γ—2 𝐴 = [2 4 βˆ’3]1Γ—3 2. Column Matrix: A matrix having single column is called column matrix. Its order is 𝑛 Γ— 1. For example, 𝐴 = [ 2 1 ] 2Γ—1 𝐴 = [ 4 βˆ’2 6 ] 3Γ—1 3. Square Matrix: A matrx in which the number of rows is equal to number columns is called a square matrix.
  3. 3. ∴ π‘š = 𝑛. ∴ No. of rows = No. of columns For example, 𝐴 = [ 1 3 βˆ’4 2 ] 2Γ—2 𝐴 = [ βˆ’3 2 1 2 3 1 3 1 βˆ’5 ] 3Γ—3 4. Null Matrix: A matrix whose all elements are zero, is called null matrix. For example, 𝐴 = [ 0 0 0 0 ] 𝐴 = [ 0 0 0 0 0 0 0 0 0 ] 5. Unit Matrix or Identity Matrix: A matrix in which all the elements of its principal diagonal are unity(one) and remaining elements are zero is called unit matrix. It is denoted by I. 𝐼 = [ 1 0 0 1 ] 𝐼 = [ 1 0 0 0 1 0 0 0 1 ] 6. DiagonalMatrix: A square matrix in which the elements on the principal diagonal are non zero and all the other elements are zero, is called a diagonal matrix. For example, 𝐴 = [ 1 0 0 βˆ’5 ]
  4. 4. 𝐴 = [ 3 0 0 0 βˆ’1 0 0 0 2 ] 7. ScalarMatrix: A diagonal matrix in which all the elements of its principal diagonal are equal is called scalar matrix. For example, 𝐴 = [ 5 0 0 0 5 0 0 0 5 ] 𝐴 = [ βˆ’2 0 0 βˆ’2 ] 8. Transpose Matrix: For a given matrix A, if rows and column are interchanged ,the new matrix obtained 𝐴’ is called transposeof a matrix. οƒ˜ Transposeof matrix A is denoted by 𝐴′ or 𝐴 𝑇 . For example, 𝐴 = [ 2 1 4 7 6 βˆ’3 4 1 0 ] ∴ 𝐴′ = 𝐴 𝑇 = [ 2 7 4 1 6 1 4 βˆ’3 0 ] οƒ˜ ( 𝐴′)β€² = 𝐴 9. Symmetric Matrix: A square matrix 𝐴 = [π‘Žπ‘–π‘—] is said to be symmetric, if π‘Žπ‘–π‘— = π‘Žπ‘—π‘– of each pair (i, j) . οƒ˜ For Symmetric matrix 𝐴′ = 𝐴 For example, 𝐴 = [ 4 2 0 2 βˆ’3 5 0 5 6 ] ∴ 𝐴′ = [ 4 2 0 2 βˆ’3 5 0 5 6 ]
  5. 5. 10. Skew Symmetric Matrix: A square matrix 𝐴 = [π‘Žπ‘–π‘—] is said to be skew symmetric if π‘Žπ‘–π‘— = βˆ’π‘Žπ‘—π‘– for each pair (i, j). For Skew symmetric matrix 𝐴′ = βˆ’π΄. For example, 𝐴 = [ 0 3 5 βˆ’3 0 4 βˆ’5 βˆ’4 0 ] 𝐴′ = [ 0 βˆ’3 βˆ’5 3 0 βˆ’4 5 4 0 ] 11. Singular Matrix: For a square matrix, if value of its determinant is zero, it is called singular matrix. For example, 𝐴 = | 6 2 9 3 | ∴ | 𝐴| = (6 Γ— 3) βˆ’ (9 Γ— 2) = 18 βˆ’ 18 = 0 𝐴 = | 1 2 3 4 1 5 3 6 9 | ∴ | 𝐴| = 1(9βˆ’ 30) βˆ’ 2(36βˆ’ 15) + 3(24 βˆ’ 3). | 𝐴| = βˆ’21 βˆ’ 42+ 63 | 𝐴| = 0 If | 𝐴| = 0 is called singular matrix. If | 𝐴| β‰  0 is called Non singular matrix. 12. Equal Matrix: For two matrices, if their correspondingelements are equal, they are called equal matrices. For example, 𝐴 = [ 2 3 βˆ’4 5 ] 𝐡 = [ 2 3 βˆ’4 5 ] ∴ 𝐴 = 𝐡 13. Negative matrix:
  6. 6. For two matrices, if their correspondingelements are equal but opposite, they are called Negative matrix. For example, 𝐴 = [ 1 βˆ’2 3 βˆ’5 ] ∴ (βˆ’π΄) = [ βˆ’1 2 βˆ’3 5 ] 14. Orthogonalmatrix: For square matrix A if Productof matirx A & its transpose matrix A’ (i.e. AA’) is Identity matrix (I). 𝐴 = [ π‘π‘œπ‘ πœƒ βˆ’π‘ π‘–π‘›πœƒ π‘ π‘–π‘›πœƒ π‘π‘œπ‘ πœƒ ] 𝐴′ = [ π‘π‘œπ‘ πœƒ π‘ π‘–π‘›πœƒ βˆ’π‘ π‘–π‘›πœƒ π‘π‘œπ‘ πœƒ ] 𝐴𝐴′ = [ 1 0 0 1 ] = 𝐼 15. Upper Triangular Matrix: For square matrix A if all the elements below the main diagonal are zero then it is called a Upper Triangular Matrix. For example, 𝐴 = [ 1 4 5 0 βˆ’2 3 0 0 3 ] 16. LowerTriangular Matrix: For square matrix A if all the elements above the main diagonal are zero then it is called a Lower Triangular Matrix. For example, 𝐴 = [ 1 0 0 4 2 0 3 βˆ’1 6 ] 17. Trace ofMatrix:
  7. 7. For square matrix A sum of all diagonal elements are called trace of matrix A. For example, 𝐴 = [ βˆ’1 2 7 3 5 βˆ’8 1 2 7 ] π‘‘π‘Ÿ( 𝐴) = βˆ’1 + 5 + 7 = 11 18. Idempotent Matrix: Matrix A is said to be Idempotent matrix if matrix 𝐴 satisfy the equation 𝐴2 = 𝐴. For example 𝐴 = [ 2 βˆ’2 βˆ’4 βˆ’1 3 4 1 βˆ’2 βˆ’3 ] 19. Involuntary Matrix: Matrix A is said to be Involuntary matrix if matrix A satisfy the equation 𝐴2 = 𝐼. Since 𝐼2 = 𝐼 always. Therefore Unit matrix is involuntary. 20. Conjugate Matrix: Let 𝐴 = [ 1 + 𝑖 2 + 3𝑖 4 7 + 2𝑖 βˆ’π‘– 3 βˆ’ 2𝑖 ] Conjugate of matrix A is 𝐴̅ 𝐴̅ = [ 1 βˆ’ 𝑖 2 βˆ’ 3𝑖 4 7 βˆ’ 2𝑖 𝑖 3 + 2𝑖 ] Note: Transposeof the conjugate of a matrix A is denoted by 𝐴 πœƒ . 21.Unitary Matrix: A square matrix A is said to be unitary if 𝐴 πœƒ 𝐴 = 𝐼 For example,
  8. 8. 𝐴 = [ 1+𝑖 2 βˆ’1+𝑖 2 βˆ’1βˆ’π‘– 2 1βˆ’π‘– 2 ] , 𝐴 πœƒ = [ 1βˆ’π‘– 2 1βˆ’π‘– 2 βˆ’1βˆ’π‘– 2 1+𝑖 2 ], 𝐴. 𝐴 πœƒ = 𝐼 22. Hermitian Matrix: A square matrix 𝐴 = (π‘Žπ‘–π‘—) is called Hermitian matrix, if every i-jth element of A is equal to conjugate complex j-ith element of A. In other words, π‘Žπ‘–π‘— = π‘Žπ‘—π‘–Μ…Μ…Μ…Μ… For example, [ 1 2 + 3𝑖 3 + 𝑖 2 βˆ’ 3𝑖 2 1 βˆ’ 2𝑖 3 βˆ’ 𝑖 1 + 2𝑖 5 ] 23. Skew Hermitian Matrix: A square matrix 𝐴 = (π‘Žπ‘–π‘—) will be calledd a Skew Hermitian matrix if every i-jth element of A is equal to negative conjugate complex of j-ith element of A. In other words , π‘Žπ‘–π‘— = βˆ’π‘Žπ‘—π‘–Μ…Μ…Μ…Μ… For example, [ 𝑖 2 βˆ’ 3𝑖 4 + 5𝑖 βˆ’(2 + 3𝑖) 0 2𝑖 βˆ’(4 βˆ’ 5𝑖) 2𝑖 βˆ’3𝑖 ] 24. Minor: The minor of an element in a third order determinant is a second order determinant obtained by deletinng the row and column which contain that element. For example, 𝐴 = | 1 2 1 3 4 5 βˆ’1 1 2 | Minor of element 2 = | 3 5 βˆ’1 2 |
  9. 9. Minor of element 3 = | 2 1 1 2 |  Operations On Matrices and it’s Properties:  Addition of Matrices : If A and B be two matrices of the same order, then their sum, A+B is defined as the matrix ,each element of which is the sum of the corresponding elements of A and B. Thus if 𝐴 = [ 4 2 5 1 3 βˆ’6 ] , 𝐡 = [ 1 0 2 3 1 4 ] Then 𝐴 + 𝐡 = [ 4 + 1 2 + 0 5 + 2 1 + 3 3 + 1 βˆ’6 + 4 ] 𝐴 + 𝐡 = [ 5 2 7 4 4 βˆ’2 ]. οƒ˜ If 𝐴 = [π‘Žπ‘–π‘—], 𝐡 = [𝑏𝑖𝑗] then 𝐴 + 𝐡 = [π‘Žπ‘–π‘— + 𝑏𝑖𝑗]  Properties Of Matrix Addition: Only matrices of the same order can be added or subtracted. i. Commutative law: A + B = B + A ii. Associative law: A + (B + C) = (A + B) + C  Subtraction of Matrices: The difference of two matrices is a matrix, each element of which is obtained by subtracting the elements of the second matrix from the Corresponding element of the first. 𝐴 βˆ’ 𝐡 = [π‘Žπ‘–π‘— βˆ’ π‘Žπ‘—π‘–] Thus [ 8 6 4 1 2 0 ] βˆ’ [ 3 5 1 7 6 2 ] = [ 8 βˆ’ 3 6 βˆ’ 5 4 βˆ’ 1 1 βˆ’ 7 2 βˆ’ 6 0 βˆ’ 2 ] = [ 5 1 3 βˆ’6 βˆ’4 βˆ’2 ]  ScalarMultiple of a Matrix:
  10. 10. If a matrix is multiplied by a scalar quantity K, then each element is multiplied by k, i.e. 𝐴 = [ 2 3 4 4 5 6 6 7 9 ] 3𝐴 = 3[ 3 Γ— 2 3 Γ— 3 3 Γ— 4 3 Γ— 4 3 Γ— 5 3 Γ— 6 3 Γ— 6 3 Γ— 7 3 Γ— 9 ] = [ 6 9 12 12 15 18 18 21 27 ]  Multiplication: The productof two matrices A and B is only possible if the number of columns in A is equal to the number of rows in B. Let 𝐴 = [π‘Žπ‘–π‘—] be an π‘š Γ— 𝑛 matrix and 𝐡 = [𝑏𝑖𝑗] be an 𝑛 Γ— 𝑝 matrix. Then the productAB of these matrices is an π‘š Γ— 𝑝 matrix 𝐢 = [𝑐𝑖𝑗] where, 𝑐𝑖𝑗 = π‘Žπ‘–1 𝑏1𝑗 + π‘Žπ‘–2 𝑏2𝑗 + π‘Žπ‘–3 𝑏3𝑗 + β‹―+ π‘Žπ‘–π‘› 𝑏 𝑛𝑗  Properties of Matrix Multiplication: a) Multiplication of matrix is not commutative. 𝑨𝑩 β‰  𝑩𝑨 b) Matrix multiplication is associative , if conformability is assured. 𝑨( 𝑩π‘ͺ) = ( 𝑨𝑩) π‘ͺ c) Matrix multiplication is distributive with respectto addition. 𝑨( 𝑩 + π‘ͺ) = 𝑨𝑩 + 𝑨π‘ͺ d) Multiplcation of matrix A by unit matrix. 𝑨𝑰 = 𝑰𝑨 = 𝑨 e) Multiplicative inverse of a matrix exists if | 𝐴| β‰  0. 𝑨. π‘¨βˆ’πŸ = π‘¨βˆ’πŸ . 𝑨 = 𝑰 f) If A is a square then 𝑨 Γ— 𝑨 = 𝑨 𝟐 , 𝑨 Γ— 𝑨 Γ— 𝑨 = 𝑨 πŸ‘ g) 𝑨 𝟎 = 𝑰 h) 𝑰 𝒏 = 𝑰, where 𝑛 is positive integer.
  11. 11. i) ( 𝑨𝑩)β€² = 𝑩′ 𝑨′ j) ( 𝑨𝑩)βˆ’πŸ = π‘©βˆ’πŸ π‘¨βˆ’πŸ  Example-1. If 𝑨 = [ 𝟐 𝟐 𝟐 𝟐 𝟏 βˆ’πŸ‘ 𝟏 𝟎 πŸ’ ], 𝑩 = [ πŸ‘ πŸ‘ πŸ‘ πŸ‘ 𝟎 πŸ“ πŸ— πŸ— βˆ’πŸ ], π‘ͺ = [ πŸ’ πŸ’ πŸ’ πŸ“ βˆ’πŸ πŸ“ βˆ’πŸ• πŸ– βˆ’πŸ ] Find πŸπ‘¨ βˆ’ πŸ‘π‘© + π‘ͺ. Solution: 2𝐴 βˆ’ 3𝐡 + 𝐢 = 2 [ 2 2 2 2 1 βˆ’3 1 0 4 ] βˆ’ 3[ 3 3 3 3 0 5 9 9 βˆ’1 ] + [ 4 4 4 5 βˆ’1 5 βˆ’7 8 βˆ’1 ] 2𝐴 βˆ’ 3𝐡 + 𝐢 = [ 4 4 4 4 2 βˆ’6 2 0 8 ] + [ βˆ’9 βˆ’9 βˆ’9 βˆ’9 0 βˆ’15 βˆ’27 βˆ’27 3 ] + [ 4 4 4 5 βˆ’1 5 βˆ’7 8 βˆ’1 ]. 2𝐴 βˆ’ 3𝐡 + 𝐢 = [ 4 βˆ’ 9 + 4 4 βˆ’ 9 + 4 4 βˆ’ 9 + 4 4 βˆ’ 9 + 5 2 + 0 βˆ’ 1 βˆ’6 βˆ’ 15 + 5 2 βˆ’ 27 βˆ’ 7 0 βˆ’ 27 + 8 8 + 3 βˆ’ 1 ]. 2𝐴 βˆ’ 3𝐡 + 𝐢 = [ βˆ’1 βˆ’1 βˆ’1 0 1 βˆ’16 βˆ’32 βˆ’19 10 ].  Example-2. If 𝑨 = [ 𝟏 𝟐 𝟏 πŸ‘ πŸ’ 𝟐 ] , 𝑩 = [ πŸ‘ βˆ’πŸ πŸ’ 𝟏 πŸ“ 𝟎 ] find matrix𝑿 from 𝑿 + 𝑨 + 𝑩 = 𝟎 Solution: 𝑋 + 𝐴 + 𝐡 = 0 οƒ˜ ∴ 𝑋 + [ 1 2 1 3 4 2 ] + [ 3 βˆ’2 4 1 5 0 ] = 0. οƒ˜ ∴ 𝑋 + [ 4 0 5 4 9 2 ] = [ 0 0 0 0 0 0 ] οƒ˜ ∴ 𝑋 = [ βˆ’4 0 βˆ’5 βˆ’4 βˆ’9 βˆ’2 ]  Example-3: show that any square matrix can be expressedas the sum of two matrices, one symmetric and the other anti-symmetric. Solution: Let A be a given square matrix.
  12. 12. οƒ˜ Then 𝐴 = 1 2 ( 𝐴 + 𝐴′) + 1 2 (𝐴 βˆ’ 𝐴′ ) οƒ˜ Now, ( 𝐴 + 𝐴′)β€² = 𝐴′ + ( 𝐴′)β€² = 𝐴′ + 𝐴 = 𝐴 + 𝐴′ οƒ˜ ∴ 𝐴 + 𝐴′ is a symmetric matrix. οƒ˜ Also, ( 𝐴 βˆ’ 𝐴′)β€² = 𝐴′ βˆ’ ( 𝐴′)β€² = 𝐴′ βˆ’ 𝐴 = βˆ’(𝐴 βˆ’ 𝐴′ ) οƒ˜ ∴ ( 𝐴 βˆ’ 𝐴′) π‘œπ‘Ÿ 1 2 (𝐴 βˆ’ 𝐴′ ) is an anti symmetric matrix. οƒ˜ ∴ 𝑨 = 𝟏 𝟐 ( 𝑨 + 𝑨′) + 𝟏 𝟐 (𝑨 βˆ’ 𝑨′ )  Example-4.Express 𝑨 = [ 𝟏 βˆ’πŸ βˆ’πŸ‘ πŸ‘ 𝟎 πŸ“ πŸ“ πŸ” 𝟏 ] as the sum of a lower triangular matrix and upper triangular matrix. Solution: Let 𝐴 = 𝐿 + π‘ˆ οƒ˜ [ 1 βˆ’2 βˆ’3 3 0 5 5 6 1 ] = [ π‘Ž 0 0 𝑏 𝑐 0 𝑑 𝑒 𝑓 ] + [ 1 𝑝 π‘ž 0 1 π‘Ÿ 0 0 1 ] οƒ˜ [ 1 βˆ’2 βˆ’3 3 0 5 5 6 1 ] = [ π‘Ž + 1 0 0 𝑏 + 0 𝑐 + 1 0 + π‘Ÿ 𝑑 + 0 𝑒 + 0 𝑓 + 1 ] οƒ˜ Equating the correspondingelements on both the sides, we get π‘Ž + 1 = 1 𝑝 = βˆ’2 π‘ž = βˆ’3 𝑏 = 3 𝑐 + 1 = 0 π‘Ÿ = 5 𝑑 = 5 𝑒 = 6 𝑓 + 1 = 1 οƒ˜ On Solving these equations, we get π‘Ž = 0 𝑝 = βˆ’2 π‘ž = βˆ’3 𝑏 = 3 𝑐 = βˆ’1 π‘Ÿ = 5 𝑑 = 5 𝑒 = 6 𝑓 = 0 οƒ˜ Hence 𝐿 = [ 0 0 0 3 βˆ’1 0 5 6 0 ] & π‘ˆ = [ 1 βˆ’2 βˆ’3 0 1 5 0 0 1 ]
  13. 13.  Example-5 If 𝑨 = [ 𝟏 𝟐 πŸ‘ πŸ’ πŸ“ πŸ” ] and 𝑩 = [ 𝟏 𝟐 𝟐 𝟏 𝟏 𝟐 ] find 𝑨𝑩 and 𝑩𝑨. Solution: 𝐴𝐡 = [ 1 2 3 4 5 6 ] [ 1 2 2 1 1 2 ] οƒ˜ 𝐴𝐡 = [ (1 Γ— 1) + (2 Γ— 2) + (3 Γ— 1) (1 Γ— 2) + (2 Γ— 1) + (3 Γ— 2) (4 Γ— 1) + (5 Γ— 2) + (6 Γ— 1) (4 Γ— 2) + (5 Γ— 1) + (6 Γ— 2) ] οƒ˜ 𝐴𝐡 = [ (1 + 4 + 3) (2 + 2 + 6) (4 + 10 + 6) (8 + 5 + 12) ] οƒ˜ 𝐴𝐡 = [ 8 10 20 25 ] Now 𝐡𝐴 = [ 1 2 2 1 1 2 ][ 1 2 3 4 5 6 ] οƒ˜ 𝐴 = [ (1 Γ— 1) + (2 Γ— 4) (1 Γ— 2) + (2 Γ— 5) (1 Γ— 3) + (2 Γ— 6) (2 Γ— 1) + (1 Γ— 4) (2 Γ— 2) + (1 Γ— 5) (2 Γ— 3) + (1 Γ— 6) (1 Γ— 1) + (2 Γ— 4) (1 Γ— 2) + (2 Γ— 5) (1 Γ— 3) + (2 Γ— 6) ] οƒ˜ 𝐴 = [ (1 + 8) (2 + 10) (3 + 12) (2 + 4) (4 + 5) (6 + 6) (1 + 8) (2 + 10) (3 + 12) ] οƒ˜ 𝐴 = [ 9 12 15 6 9 12 9 12 15 ]  Example-6. If 𝑨 = [ 𝟏 𝟐 βˆ’πŸ πŸ‘ ] , 𝑩 = [ 𝟐 𝟏 𝟐 πŸ‘ ] and π‘ͺ = [ βˆ’πŸ‘ 𝟏 𝟐 𝟎 ]. Verify that ( 𝑨𝑩) π‘ͺ = 𝑨(𝑩π‘ͺ) and 𝑨( 𝑩 + π‘ͺ) = 𝑨𝑩 + 𝑨π‘ͺ. Solution: 𝐴𝐡 = [ 1 2 βˆ’2 3 ] Γ— [ 2 1 2 3 ] οƒ˜ 𝐴𝐡 = [ (1)(2) + (2)(2) (1)(1) + (2)(3) (βˆ’2)(2) + (3)(2) (βˆ’2)(1) + (3)(3) ] οƒ˜ 𝐴𝐡 = [ 6 7 2 7 ]
  14. 14. οƒ˜ 𝐡𝐢 = [ 2 1 2 3 ] Γ— [ βˆ’3 1 2 0 ] οƒ˜ 𝐡𝐢 = [ βˆ’6 + 2 2 + 0 βˆ’6 + 6 2 + 0 ] οƒ˜ 𝐡𝐢 = [ βˆ’4 2 0 2 ] οƒ˜ 𝐴𝐢 = [ 1 2 βˆ’2 3 ] Γ— [ βˆ’3 1 2 0 ] οƒ˜ 𝐴𝐢 = [ βˆ’3 + 4 1 + 0 6 + 6 βˆ’2 + 0 ] οƒ˜ 𝐴𝐢 = [ 1 1 12 βˆ’2 ] οƒ˜ 𝐡 + 𝐢 = [ 2 1 2 3 ] + [ βˆ’3 1 2 0 ] οƒ˜ 𝐡 + 𝐢 = [ 2 + (βˆ’3) 1 + 1 2 + 2 3 + 0 ] οƒ˜ 𝐡 + 𝐢 = [ βˆ’1 2 4 3 ] i. ( 𝐴𝐡) 𝐢 = [ 6 7 2 7 ] Γ— [ βˆ’3 1 2 0 ] οƒ˜ ( 𝐴𝐡) 𝐢 = [ βˆ’18 + 14 6 + 0 βˆ’6 + 14 2 + 0 ] οƒ˜ ( 𝐴𝐡) 𝐢 = [ βˆ’4 6 8 2 ] .……………………………………………..(1) οƒ˜ & 𝐴( 𝐡𝐢) = [ 1 2 βˆ’2 3 ] Γ— [ βˆ’4 2 0 2 ] οƒ˜ 𝐴( 𝐡𝐢) = [ βˆ’4 + 0 2 + 4 8 + 0 βˆ’4 + 6 ] οƒ˜ 𝐴( 𝐡𝐢) = [ βˆ’4 6 8 2 ] ……………………………………………….(2) Thus from (1) and (2), we get οƒ˜ ( 𝐴𝐡) 𝐢 = 𝐴(𝐡𝐢) ii. 𝐴( 𝐡 + 𝐢) = [ 1 2 βˆ’2 3 ][ βˆ’1 2 4 3 ]
  15. 15. οƒ˜ 𝐴( 𝐡 + 𝐢) = [ βˆ’1 + 8 2 + 6 2 + 12 βˆ’4 + 9 ] οƒ˜ 𝐴( 𝐡 + 𝐢) = [ 7 8 14 5 ] ………………………………………….(3) οƒ˜ 𝐴𝐡 + 𝐴𝐢 = [ 6 + 1 7 + 1 2 + 12 7 βˆ’ 2 ] οƒ˜ 𝐴𝐡 + 𝐴𝐢 = [ 7 8 24 5 ] …………………………………………..(4) Thus from (3) and (4), we get 𝐴( 𝐡 + 𝐢) = 𝐴𝐡 + 𝐴𝐢  Example-7. If 𝑨 = [ 𝟏 𝟐 𝟐 𝟐 𝟏 𝟐 𝟐 𝟐 𝟏 ] show that 𝑨 𝟐 βˆ’ πŸ’π‘¨ βˆ’ πŸ“π‘° = 𝟎 where I, 0 are the unit matrix and the null matrix of order 3 respectively. Use this result to find π‘¨βˆ’πŸ . Solution: Here, we have 𝐴 = [ 1 2 2 2 1 2 2 2 1 ] οƒ˜ 𝐴2 = [ 1 2 2 2 1 2 2 2 1 ][ 1 2 2 2 1 2 2 2 1 ] οƒ˜ 𝐴2 = [ 9 8 8 8 9 8 8 8 9 ] οƒ˜ 𝐴2 βˆ’ 4𝐴 βˆ’ 5𝐼 = [ 9 8 8 8 9 8 8 8 9 ] βˆ’ 4[ 1 2 2 2 1 2 2 2 1 ] βˆ’ 5[ 1 0 0 0 1 0 0 0 1 ] οƒ˜ 𝐴2 βˆ’ 4𝐴 βˆ’ 5𝐼 = [ 9 βˆ’ 4 βˆ’ 5 8 βˆ’ 8 + 0 8 βˆ’ 8 + 0 8 βˆ’ 8 + 0 9 βˆ’ 4 βˆ’ 5 8 βˆ’ 8 + 0 8 βˆ’ 8 + 0 8 βˆ’ 8 + 0 9 βˆ’ 4 βˆ’ 5 ] οƒ˜ 𝐴2 βˆ’ 4𝐴 βˆ’ 5𝐼 = [ 0 0 0 0 0 0 0 0 0 ]
  16. 16. οƒ˜ 𝐴2 βˆ’ 4𝐴 βˆ’ 5𝐼 = 0 ⟹ 5𝐼 = 𝐴2 βˆ’ 4𝐴 οƒ˜ Both the side multiplying by π΄βˆ’1 , we get οƒ˜ 5π΄βˆ’1 = 𝐴 βˆ’ 4𝐼 οƒ˜ 5π΄βˆ’1 = [ 1 2 2 2 1 2 2 2 1 ] βˆ’ 4[ 1 0 0 0 1 0 0 0 1 ] οƒ˜ 5π΄βˆ’1 = [ βˆ’3 2 2 2 βˆ’3 2 2 2 βˆ’3 ] οƒ˜ π΄βˆ’1 = 1 5 [ βˆ’3 2 2 2 βˆ’3 2 2 2 βˆ’3 ]  Adjoint of a square Matrix: Let the determinant of the square matrix 𝐴 be | 𝐴| = [ π‘Ž1 π‘Ž2 π‘Ž3 𝑏1 𝑏2 𝑏3 𝑐1 𝑐2 𝑐3 ] οƒ˜ The matrix formed by the co-factors ofthe elements in | 𝐴| 𝑖𝑠 [ 𝐴1 𝐴2 𝐴3 𝐡1 𝐡2 𝐡3 𝐢1 𝐢2 𝐢3 ] οƒ˜ Where 𝐴1 = | 𝑏2 𝑏3 𝑐2 𝑐3 |= 𝑏2 𝑐3 βˆ’ 𝑏3 𝑐2, οƒ˜ 𝐴2 = βˆ’| 𝑏1 𝑏3 𝑐1 𝑐3 | = βˆ’π‘1 𝑐3 βˆ’ 𝑏3 𝑐1, οƒ˜ 𝐴3 = | 𝑏1 𝑏2 𝑐1 𝑐2 | = 𝑏1 𝑐2 βˆ’ 𝑏2 𝑐1, οƒ˜ 𝐡1 = βˆ’ | π‘Ž2 π‘Ž3 𝑐2 𝑐3 | = βˆ’π‘Ž2 𝑐3 + π‘Ž3 𝑐2, οƒ˜ 𝐡2 = | π‘Ž1 π‘Ž3 𝑐1 𝑐3 | = π‘Ž1 𝑐3 βˆ’ π‘Ž3 𝑐1, οƒ˜ 𝐡3 = βˆ’| π‘Ž1 π‘Ž2 𝑐1 𝑐2 | = βˆ’π‘Ž1 𝑐2 + π‘Ž2 𝑐1, οƒ˜ 𝐢1 = | π‘Ž2 π‘Ž3 𝑏2 𝑏3 | = π‘Ž2 𝑏3 βˆ’ π‘Ž3 𝑏2,
  17. 17. οƒ˜ 𝐢2 = βˆ’| π‘Ž1 π‘Ž3 𝑏1 𝑏3 | = βˆ’π‘Ž1 𝑏3 + π‘Ž3 𝑏1, οƒ˜ 𝐢3 = | π‘Ž1 π‘Ž2 𝑏1 𝑏2 | = π‘Ž1 𝑏2 βˆ’ π‘Ž2 𝑏1, οƒ˜ Then the transposeof the matrix of co-factors [ 𝐴1 𝐡1 𝐢1 𝐴2 𝐡2 𝐢2 𝐴3 𝐡3 𝐢3 ] Is called the adjoint of the matrix 𝐴 and is written as π‘Žπ‘‘π‘— 𝐴. Note:For 𝟐 Γ— 𝟐 order matrix Adj A is defined as: If 𝐴 = [ π‘Ž1 π‘Ž2 𝑏1 𝑏2 ] then Adj A = [ 𝑏2 βˆ’π‘Ž2 𝑏1 π‘Ž1 ] i.e. Change location of elelment of principal diagonal and change sign of elements of subsidary diagonal.  Property of Adjoint Matrix: The productof a matrix 𝐴 and its adjoint is equal to unit matrix multiplied by the determinant 𝐴.  Example-1: If 𝑨 = [ πŸ“ 𝟐 πŸ• πŸ‘ ] then find π‘¨βˆ’πŸ . Solution: Since here, given matrix 𝐴 is of the Order 2 Γ— 2. οƒ˜ ∴ π‘Žπ‘‘π‘— 𝐴 = [ 3 βˆ’2 βˆ’7 5 ]  Example-2: For matrix 𝑨 = [ 𝟐 𝟏 πŸ“ 𝟎 πŸ‘ βˆ’πŸ 𝟐 πŸ“ 𝟎 ] find co-factormatrix 𝒂𝒅𝒋 𝑨. Solution: Let 𝐴 = [π‘Žπ‘–π‘—] οƒ˜ First we will find co-factors ofeach element. οƒ˜ 𝐴11 = | 3 βˆ’1 5 0 | = 0 βˆ’ (βˆ’5) = 5 οƒ˜ 𝐴12 = βˆ’| 0 βˆ’1 2 0 | = βˆ’[0 βˆ’ (βˆ’2)] = βˆ’2
  18. 18. οƒ˜ 𝐴13 = | 0 3 2 5 | = (0 βˆ’ 6) = βˆ’6 οƒ˜ 𝐴21 = βˆ’| 1 5 5 0 | = βˆ’(0 βˆ’ 25) = 25 οƒ˜ 𝐴22 = | 2 5 2 0 | = (0 βˆ’ 10) = βˆ’10 οƒ˜ 𝐴23 = βˆ’| 2 1 2 5 | = βˆ’(10 βˆ’ 2) = βˆ’8 οƒ˜ 𝐴31 = | 1 5 3 βˆ’1 | = (βˆ’1 βˆ’ 15) = βˆ’16 οƒ˜ 𝐴32 = βˆ’| 2 5 0 βˆ’1 | = βˆ’(βˆ’2 βˆ’ 0) = 2 οƒ˜ 𝐴33 = | 2 1 0 3 | = (6 βˆ’ 0) = 6 ∴ π‘Žπ‘‘π‘— 𝐴 = [ 𝐴11 𝐴21 𝐴31 𝐴12 𝐴22 𝐴32 𝐴13 𝐴23 𝐴33 ] = [ 5 25 βˆ’16 βˆ’2 βˆ’10 2 βˆ’6 βˆ’8 6 ]  Inverse of a Matrix: If A and B are two square matrices of the same order, such that 𝐴𝐡 = 𝐡𝐴 = 𝐼 Then 𝐡 is called the inverse of 𝐴 i.e. 𝐡 = π΄βˆ’1 and 𝐴 is the invese of B. οƒ˜ Condition for a square matrix 𝐴 to possess aninverse is that matrix 𝐴 is non-singular. i.e. | 𝑨| β‰  𝟎 οƒ˜ If 𝐴 is square matrix and 𝐡 its inverse, then 𝐴𝐡 = 𝐼. Taking determinant of both sides, we get | 𝐴𝐡| = | 𝐼| π‘œπ‘Ÿ | 𝐴|| 𝐡| = 𝐼 From this relation it is clear that | 𝐴| β‰  0 i.e. the matrix 𝐴 is non singular. To find the inverse matrix with the help of adjoint matrix:
  19. 19. οƒ˜ We know that 𝐴. ( 𝐴𝑑𝑗𝐴) = | 𝐴| 𝐼 οƒ˜ 𝐴. 1 | 𝐴| ( 𝐴𝑑𝑗 𝐴) = 𝐼 (Provided | 𝐴| β‰  0) ………………………..(1) οƒ˜ Since 𝐴. π΄βˆ’1 = 𝐼 …………………………………………….(2) οƒ˜ From (1) and (2), we have ∴ π‘¨βˆ’πŸ = 𝟏 | 𝑨| ( 𝑨𝒅𝒋 𝑨) = 𝑰  Example-1: If 𝑨 = [ 𝟏 𝟐 βˆ’πŸ πŸ‘ ] find π‘¨βˆ’πŸ . Solution: | 𝐴| = | 1 2 βˆ’1 3 | = 3 βˆ’ (βˆ’2) = 5 β‰  0 οƒ˜ ∴ π΄βˆ’1 is possible. οƒ˜ 𝐴𝑑𝑗 𝐴 = [ 3 βˆ’2 1 1 ] οƒ˜ ∴ π΄βˆ’1 = π‘Žπ‘‘π‘— 𝐴 | 𝐴| οƒ˜ π΄βˆ’1 = 1 5 [ 3 βˆ’2 1 1 ] οƒ˜ π΄βˆ’1 = [ 3 5 βˆ’ 2 5 1 5 1 5 ]  Example-2: If 𝑨 = [ 𝟏 βˆ’πŸ 𝟏 𝟐 βˆ’πŸ 𝟎 𝟏 𝟎 𝟏 ] then find π‘¨βˆ’πŸ . Solution: Let given matrix is 𝐴. οƒ˜ ∴ | 𝐴| = | 1 βˆ’1 1 2 βˆ’1 0 1 0 1 | οƒ˜ ∴ | 𝐴| = 1(βˆ’1 βˆ’ 0) + 1(2 βˆ’ 0) + 1[0βˆ’ (βˆ’1)] οƒ˜ ∴ | 𝐴| = βˆ’1 + 2 + 1 = 2 β‰  0 οƒ˜ ∴ π΄βˆ’1 is possible. οƒ˜ First we will calculate, co-factors ofeach element.
  20. 20. οƒ˜ 𝐴11 = | βˆ’1 0 0 1 | = (βˆ’1 βˆ’ 0) = βˆ’1 οƒ˜ 𝐴12 = βˆ’| 2 0 1 1 | = [βˆ’(2βˆ’ 0)] = βˆ’2 οƒ˜ 𝐴13 = | 2 βˆ’1 1 0 | = [0 βˆ’ (βˆ’1)] = 1 οƒ˜ 𝐴21 = βˆ’| βˆ’1 1 0 1 | = [βˆ’(βˆ’1 βˆ’ 0)] = 1 οƒ˜ 𝐴22 = | 1 1 1 1 | = (1 βˆ’ 1) = 0 οƒ˜ 𝐴23 = βˆ’| 1 βˆ’1 1 0 | = βˆ’[0 βˆ’ (βˆ’1)] = βˆ’1 οƒ˜ 𝐴31 = | βˆ’1 1 βˆ’1 0 | = [0 βˆ’ (βˆ’1)] = 1 οƒ˜ 𝐴32 = βˆ’| 1 1 2 0 | = [βˆ’(0 βˆ’ 2)] = 2 οƒ˜ 𝐴33 = | 1 βˆ’1 2 βˆ’1 | = [βˆ’1 βˆ’ (βˆ’2)] = 1 ∴ π‘Žπ‘‘π‘— 𝐴 = [ 𝐴11 𝐴21 𝐴31 𝐴12 𝐴22 𝐴32 𝐴13 𝐴23 𝐴33 ] = [ βˆ’1 1 1 βˆ’2 0 2 1 βˆ’1 1 ] οƒ˜ π΄βˆ’1 = 𝐴𝑑𝑗 𝐴 | 𝐴| = 1 2 Γ— [ βˆ’1 1 1 βˆ’2 0 2 1 βˆ’1 1 ] οƒ˜ π΄βˆ’1 = [ βˆ’ 1 2 1 2 1 2 βˆ’1 0 1 1 2 βˆ’ 1 2 1 2 ]  Example-3. If a matrix 𝑨 satisfies a relation 𝑨 𝟐 + 𝑨 βˆ’ 𝑰 = 𝟎 prove that π‘¨βˆ’πŸ exists and that π‘¨βˆ’πŸ = 𝑰 + 𝑨, 𝑰 being an identity matrix. Solution: Here, 𝐴2 + 𝐴 βˆ’ 𝐼 = 0 οƒ˜ ∴ 𝐴2 + 𝐴 = 𝐼 οƒ˜ ∴ 𝐴( 𝐴 + 𝐼) = 𝐼
  21. 21. οƒ˜ ∴ | 𝐴|| 𝐴 + 𝐼| = | 𝐼| οƒ˜ ∴ | 𝐴| β‰  0 and so π΄βˆ’1 exists. οƒ˜ Again 𝐴2 + 𝐴 βˆ’ 𝐼 = 0 ⟹ 𝐴2 + 𝐴 = 𝐼 οƒ˜ Multiplying (1) by π΄βˆ’1 , we get π΄βˆ’1( 𝐴2 + 𝐴) = π΄βˆ’1 𝐼 ⟹ 𝐴 + 𝐼 = π΄βˆ’1 ∴ π΄βˆ’1 = 𝐼 + 𝐴  Reference BookandWebsite Name: 1. Polytechnic Mathematics -1 by Nirav Prakashan. 2. Introduction to Engineering Mathematics-1 by H.K. Dass and Dr.Rama Verma. (S.Chand) 3. Engineering Mathematics ( Pearson Fourth Edition) 4. A Textbookof Engineering mathematics by N.P.Bali and Dr.Manish goyal 5. http://aleph0.clarku.edu/~djoyce/ma130/elementary.pdf
  22. 22. EXERCISE-3 Q-1.Evaluate the following Questions: 1. If 𝐴 = [ 0 2 0 1 0 3 1 1 2 ], 𝐡 = [ 1 2 1 2 1 0 0 0 3 ] find ( 𝑖)2𝐴 + 3𝐡 ( 𝑖𝑖)3𝐴 βˆ’ 4𝐡 2. Express [ 1 2 0 3 7 1 5 9 3 ] as a sum of symmetric matrix and skew-symmetric matrix. 3. If 𝐴 = [ 2 3 1 0 ], 𝐡 = [ 4 1 2 βˆ’3 ] prove that ( 𝐴 + 𝐡) 𝑇 = 𝐴 𝑇 + 𝐡 𝑇 . 4. If 𝐴 = [ 2 βˆ’1 0 3 2 βˆ’4 5 1 9 ] and 𝐡 = [ 17 βˆ’1 3 βˆ’24 βˆ’1 βˆ’16 βˆ’7 1 1 ] and 4𝐴 + 3𝐢 = 𝐡, then find matrix 𝐢. Q-2. Evaluate the following Questions: 1. If 𝐴 = [ 0 1 2 1 2 3 2 3 4 ] and 𝐡 = [ 1 βˆ’2 βˆ’1 0 2 βˆ’1 ] Obtain the product 𝐴𝐡 and explain why 𝐡𝐴 is not defined. 2. If 𝐴 = [ 1 2 βˆ’1 3 0 2 4 5 0 ] and 𝐡 = [ 1 0 0 2 1 0 0 1 3 ] verify that ( 𝐴𝐡)β€² = 𝐡′ 𝐴′ . 3. Compute 𝐴𝐡 if 𝐴 = [ 1 2 3 4 5 6 ] and 𝐡 = [ 2 5 3 3 6 4 4 7 5 ]. 4. Verify that 𝐴 = 1 3 [ 1 2 2 2 1 βˆ’2 βˆ’2 2 βˆ’1 ] is Orthogonal. Q-3. Evaluate the following Questions:
  23. 23. 1. If 𝐴 = [ 2 5 3 3 1 2 1 2 1 ] then find 𝐴𝑑𝑗 𝐴. 2. If 𝐴 = [ 1 1 2 1 9 3 1 4 2 ] then find π΄βˆ’1 . 3. If 𝐴 = [ 1 1 1 1 2 3 1 4 9 ], 𝐡 = [ 2 5 3 3 1 2 1 2 1 ], then show that ( 𝐴𝐡)βˆ’1 = π΅βˆ’1 π΄βˆ’1 . 4. If 𝐴 = [ βˆ’4 βˆ’3 βˆ’3 1 0 1 4 4 3 ] Prove that 𝐴𝑑𝑗 𝐴 = 𝐴.

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