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# BSC_COMPUTER _SCIENCE_UNIT-3_DISCRETE MATHEMATICS

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Class: B.Sc CS.
Subject: Discrete Mathematics
Unit-3 Basics Of Matrix

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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 οΆ Definition: Matrix ο A ractangular array of m rows and n columns, enclosed by brackets [ ] is called a matrix of order π Γ π. A matrix of order 3 Γ 3 is expressed as, π΄ = [ π11 π12 π13 π21 π22 π23 π31 π32 π33 ] ο An element πππ denotes, ith row and jth column π23 denotes, 2nd row and 3rd column. ο Matrices are denoted by capital letters A,B,C,β¦β¦..etc. οΆ Types of Matrices: 1. Row Matrix: A matrix having only single row is called row matrix. Its order is 1 Γ π. For example, π΄ = [1 4]1Γ2 π΄ = [2 4 β3]1Γ3 2. Column Matrix: A matrix having single column is called column matrix. Its order is π Γ 1. For example, π΄ = [ 2 1 ] 2Γ1 π΄ = [ 4 β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. β΄ π = π. β΄ No. of rows = No. of columns For example, π΄ = [ 1 3 β4 2 ] 2Γ2 π΄ = [ β3 2 1 2 3 1 3 1 β5 ] 3Γ3 4. Null Matrix: A matrix whose all elements are zero, is called null matrix. For example, π΄ = [ 0 0 0 0 ] π΄ = [ 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. πΌ = [ 1 0 0 1 ] πΌ = [ 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, π΄ = [ 1 0 0 β5 ]
4. 4. π΄ = [ 3 0 0 0 β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, π΄ = [ 5 0 0 0 5 0 0 0 5 ] π΄ = [ β2 0 0 β2 ] 8. Transpose Matrix: For a given matrix A, if rows and column are interchanged ,the new matrix obtained π΄β is called transposeof a matrix. ο Transposeof matrix A is denoted by π΄β² or π΄ π . For example, π΄ = [ 2 1 4 7 6 β3 4 1 0 ] β΄ π΄β² = π΄ π = [ 2 7 4 1 6 1 4 β3 0 ] ο ( π΄β²)β² = π΄ 9. Symmetric Matrix: A square matrix π΄ = [πππ] is said to be symmetric, if πππ = πππ of each pair (i, j) . ο For Symmetric matrix π΄β² = π΄ For example, π΄ = [ 4 2 0 2 β3 5 0 5 6 ] β΄ π΄β² = [ 4 2 0 2 β3 5 0 5 6 ]
5. 5. 10. Skew Symmetric Matrix: A square matrix π΄ = [πππ] is said to be skew symmetric if πππ = βπππ for each pair (i, j). For Skew symmetric matrix π΄β² = βπ΄. For example, π΄ = [ 0 3 5 β3 0 4 β5 β4 0 ] π΄β² = [ 0 β3 β5 3 0 β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, π΄ = | 6 2 9 3 | β΄ | π΄| = (6 Γ 3) β (9 Γ 2) = 18 β 18 = 0 π΄ = | 1 2 3 4 1 5 3 6 9 | β΄ | π΄| = 1(9β 30) β 2(36β 15) + 3(24 β 3). | π΄| = β21 β 42+ 63 | π΄| = 0 If | π΄| = 0 is called singular matrix. If | π΄| 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, π΄ = [ 2 3 β4 5 ] π΅ = [ 2 3 β4 5 ] β΄ π΄ = π΅ 13. Negative matrix:
6. 6. For two matrices, if their correspondingelements are equal but opposite, they are called Negative matrix. For example, π΄ = [ 1 β2 3 β5 ] β΄ (βπ΄) = [ β1 2 β3 5 ] 14. Orthogonalmatrix: For square matrix A if Productof matirx A & its transpose matrix Aβ (i.e. AAβ) is Identity matrix (I). π΄ = [ πππ π βπ πππ π πππ πππ π ] π΄β² = [ πππ π π πππ βπ πππ πππ π ] π΄π΄β² = [ 1 0 0 1 ] = πΌ 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, π΄ = [ 1 4 5 0 β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, π΄ = [ 1 0 0 4 2 0 3 β1 6 ] 17. Trace ofMatrix:
7. 7. For square matrix A sum of all diagonal elements are called trace of matrix A. For example, π΄ = [ β1 2 7 3 5 β8 1 2 7 ] π‘π( π΄) = β1 + 5 + 7 = 11 18. Idempotent Matrix: Matrix A is said to be Idempotent matrix if matrix π΄ satisfy the equation π΄2 = π΄. For example π΄ = [ 2 β2 β4 β1 3 4 1 β2 β3 ] 19. Involuntary Matrix: Matrix A is said to be Involuntary matrix if matrix A satisfy the equation π΄2 = πΌ. Since πΌ2 = πΌ always. Therefore Unit matrix is involuntary. 20. Conjugate Matrix: Let π΄ = [ 1 + π 2 + 3π 4 7 + 2π βπ 3 β 2π ] Conjugate of matrix A is π΄Μ π΄Μ = [ 1 β π 2 β 3π 4 7 β 2π π 3 + 2π ] Note: Transposeof the conjugate of a matrix A is denoted by π΄ π . 21.Unitary Matrix: A square matrix A is said to be unitary if π΄ π π΄ = πΌ For example,
8. 8. π΄ = [ 1+π 2 β1+π 2 β1βπ 2 1βπ 2 ] , π΄ π = [ 1βπ 2 1βπ 2 β1βπ 2 1+π 2 ], π΄. π΄ π = πΌ 22. Hermitian Matrix: A square matrix π΄ = (ππ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ππ = πππΜΜΜΜ For example, [ 1 2 + 3π 3 + π 2 β 3π 2 1 β 2π 3 β π 1 + 2π 5 ] 23. Skew Hermitian Matrix: A square matrix π΄ = (ππ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ππ = βπππΜΜΜΜ For example, [ π 2 β 3π 4 + 5π β(2 + 3π) 0 2π β(4 β 5π) 2π β3π ] 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, π΄ = | 1 2 1 3 4 5 β1 1 2 | Minor of element 2 = | 3 5 β1 2 |
9. 9. Minor of element 3 = | 2 1 1 2 | οΆ Operations On Matrices and itβs Properties: οΆ 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 π΄ = [ 4 2 5 1 3 β6 ] , π΅ = [ 1 0 2 3 1 4 ] Then π΄ + π΅ = [ 4 + 1 2 + 0 5 + 2 1 + 3 3 + 1 β6 + 4 ] π΄ + π΅ = [ 5 2 7 4 4 β2 ]. ο If π΄ = [πππ], π΅ = [πππ] then π΄ + π΅ = [πππ + πππ] οΆ 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 οΆ 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. π΄ β π΅ = [πππ β πππ] Thus [ 8 6 4 1 2 0 ] β [ 3 5 1 7 6 2 ] = [ 8 β 3 6 β 5 4 β 1 1 β 7 2 β 6 0 β 2 ] = [ 5 1 3 β6 β4 β2 ] οΆ 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. π΄ = [ 2 3 4 4 5 6 6 7 9 ] 3π΄ = 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 ] οΆ 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 π΄ = [πππ] be an π Γ π matrix and π΅ = [πππ] be an π Γ π matrix. Then the productAB of these matrices is an π Γ π matrix πΆ = [πππ] where, πππ = ππ1 π1π + ππ2 π2π + ππ3 π3π + β―+ πππ π ππ οΆ Properties of Matrix Multiplication: a) Multiplication of matrix is not commutative. π¨π© β  π©π¨ b) Matrix multiplication is associative , if conformability is assured. π¨( π©πͺ) = ( π¨π©) πͺ c) Matrix multiplication is distributive with respectto addition. π¨( π© + πͺ) = π¨π© + π¨πͺ d) Multiplcation of matrix A by unit matrix. π¨π° = π°π¨ = π¨ e) Multiplicative inverse of a matrix exists if | π΄| β  0. π¨. π¨βπ = π¨βπ . π¨ = π° f) If A is a square then π¨ Γ π¨ = π¨ π , π¨ Γ π¨ Γ π¨ = π¨ π g) π¨ π = π° h) π° π = π°, where π is positive integer.
11. 11. i) ( π¨π©)β² = π©β² π¨β² j) ( π¨π©)βπ = π©βπ π¨βπ οΆ Example-1. If π¨ = [ π π π π π βπ π π π ], π© = [ π π π π π π π π βπ ], πͺ = [ π π π π βπ π βπ π βπ ] Find ππ¨ β ππ© + πͺ. Solution: 2π΄ β 3π΅ + πΆ = 2 [ 2 2 2 2 1 β3 1 0 4 ] β 3[ 3 3 3 3 0 5 9 9 β1 ] + [ 4 4 4 5 β1 5 β7 8 β1 ] 2π΄ β 3π΅ + πΆ = [ 4 4 4 4 2 β6 2 0 8 ] + [ β9 β9 β9 β9 0 β15 β27 β27 3 ] + [ 4 4 4 5 β1 5 β7 8 β1 ]. 2π΄ β 3π΅ + πΆ = [ 4 β 9 + 4 4 β 9 + 4 4 β 9 + 4 4 β 9 + 5 2 + 0 β 1 β6 β 15 + 5 2 β 27 β 7 0 β 27 + 8 8 + 3 β 1 ]. 2π΄ β 3π΅ + πΆ = [ β1 β1 β1 0 1 β16 β32 β19 10 ]. οΆ Example-2. If π¨ = [ π π π π π π ] , π© = [ π βπ π π π π ] find matrixπΏ from πΏ + π¨ + π© = π Solution: π + π΄ + π΅ = 0 ο β΄ π + [ 1 2 1 3 4 2 ] + [ 3 β2 4 1 5 0 ] = 0. ο β΄ π + [ 4 0 5 4 9 2 ] = [ 0 0 0 0 0 0 ] ο β΄ π = [ β4 0 β5 β4 β9 β2 ] οΆ 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. ο Then π΄ = 1 2 ( π΄ + π΄β²) + 1 2 (π΄ β π΄β² ) ο Now, ( π΄ + π΄β²)β² = π΄β² + ( π΄β²)β² = π΄β² + π΄ = π΄ + π΄β² ο β΄ π΄ + π΄β² is a symmetric matrix. ο Also, ( π΄ β π΄β²)β² = π΄β² β ( π΄β²)β² = π΄β² β π΄ = β(π΄ β π΄β² ) ο β΄ ( π΄ β π΄β²) ππ 1 2 (π΄ β π΄β² ) is an anti symmetric matrix. ο β΄ π¨ = π π ( π¨ + π¨β²) + π π (π¨ β π¨β² ) οΆ Example-4.Express π¨ = [ π βπ βπ π π π π π π ] as the sum of a lower triangular matrix and upper triangular matrix. Solution: Let π΄ = πΏ + π ο [ 1 β2 β3 3 0 5 5 6 1 ] = [ π 0 0 π π 0 π π π ] + [ 1 π π 0 1 π 0 0 1 ] ο [ 1 β2 β3 3 0 5 5 6 1 ] = [ π + 1 0 0 π + 0 π + 1 0 + π π + 0 π + 0 π + 1 ] ο Equating the correspondingelements on both the sides, we get π + 1 = 1 π = β2 π = β3 π = 3 π + 1 = 0 π = 5 π = 5 π = 6 π + 1 = 1 ο On Solving these equations, we get π = 0 π = β2 π = β3 π = 3 π = β1 π = 5 π = 5 π = 6 π = 0 ο Hence πΏ = [ 0 0 0 3 β1 0 5 6 0 ] & π = [ 1 β2 β3 0 1 5 0 0 1 ]
13. 13. οΆ Example-5 If π¨ = [ π π π π π π ] and π© = [ π π π π π π ] find π¨π© and π©π¨. Solution: π΄π΅ = [ 1 2 3 4 5 6 ] [ 1 2 2 1 1 2 ] ο π΄π΅ = [ (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) ] ο π΄π΅ = [ (1 + 4 + 3) (2 + 2 + 6) (4 + 10 + 6) (8 + 5 + 12) ] ο π΄π΅ = [ 8 10 20 25 ] Now π΅π΄ = [ 1 2 2 1 1 2 ][ 1 2 3 4 5 6 ] ο π΄ = [ (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) ] ο π΄ = [ (1 + 8) (2 + 10) (3 + 12) (2 + 4) (4 + 5) (6 + 6) (1 + 8) (2 + 10) (3 + 12) ] ο π΄ = [ 9 12 15 6 9 12 9 12 15 ] οΆ Example-6. If π¨ = [ π π βπ π ] , π© = [ π π π π ] and πͺ = [ βπ π π π ]. Verify that ( π¨π©) πͺ = π¨(π©πͺ) and π¨( π© + πͺ) = π¨π© + π¨πͺ. Solution: π΄π΅ = [ 1 2 β2 3 ] Γ [ 2 1 2 3 ] ο π΄π΅ = [ (1)(2) + (2)(2) (1)(1) + (2)(3) (β2)(2) + (3)(2) (β2)(1) + (3)(3) ] ο π΄π΅ = [ 6 7 2 7 ]
14. 14. ο π΅πΆ = [ 2 1 2 3 ] Γ [ β3 1 2 0 ] ο π΅πΆ = [ β6 + 2 2 + 0 β6 + 6 2 + 0 ] ο π΅πΆ = [ β4 2 0 2 ] ο π΄πΆ = [ 1 2 β2 3 ] Γ [ β3 1 2 0 ] ο π΄πΆ = [ β3 + 4 1 + 0 6 + 6 β2 + 0 ] ο π΄πΆ = [ 1 1 12 β2 ] ο π΅ + πΆ = [ 2 1 2 3 ] + [ β3 1 2 0 ] ο π΅ + πΆ = [ 2 + (β3) 1 + 1 2 + 2 3 + 0 ] ο π΅ + πΆ = [ β1 2 4 3 ] i. ( π΄π΅) πΆ = [ 6 7 2 7 ] Γ [ β3 1 2 0 ] ο ( π΄π΅) πΆ = [ β18 + 14 6 + 0 β6 + 14 2 + 0 ] ο ( π΄π΅) πΆ = [ β4 6 8 2 ] .β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..(1) ο & π΄( π΅πΆ) = [ 1 2 β2 3 ] Γ [ β4 2 0 2 ] ο π΄( π΅πΆ) = [ β4 + 0 2 + 4 8 + 0 β4 + 6 ] ο π΄( π΅πΆ) = [ β4 6 8 2 ] β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(2) Thus from (1) and (2), we get ο ( π΄π΅) πΆ = π΄(π΅πΆ) ii. π΄( π΅ + πΆ) = [ 1 2 β2 3 ][ β1 2 4 3 ]
15. 15. ο π΄( π΅ + πΆ) = [ β1 + 8 2 + 6 2 + 12 β4 + 9 ] ο π΄( π΅ + πΆ) = [ 7 8 14 5 ] β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(3) ο π΄π΅ + π΄πΆ = [ 6 + 1 7 + 1 2 + 12 7 β 2 ] ο π΄π΅ + π΄πΆ = [ 7 8 24 5 ] β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..(4) Thus from (3) and (4), we get π΄( π΅ + πΆ) = π΄π΅ + π΄πΆ οΆ Example-7. If π¨ = [ π π π π π π π π π ] show that π¨ π β ππ¨ β ππ° = π where I, 0 are the unit matrix and the null matrix of order 3 respectively. Use this result to find π¨βπ . Solution: Here, we have π΄ = [ 1 2 2 2 1 2 2 2 1 ] ο π΄2 = [ 1 2 2 2 1 2 2 2 1 ][ 1 2 2 2 1 2 2 2 1 ] ο π΄2 = [ 9 8 8 8 9 8 8 8 9 ] ο π΄2 β 4π΄ β 5πΌ = [ 9 8 8 8 9 8 8 8 9 ] β 4[ 1 2 2 2 1 2 2 2 1 ] β 5[ 1 0 0 0 1 0 0 0 1 ] ο π΄2 β 4π΄ β 5πΌ = [ 9 β 4 β 5 8 β 8 + 0 8 β 8 + 0 8 β 8 + 0 9 β 4 β 5 8 β 8 + 0 8 β 8 + 0 8 β 8 + 0 9 β 4 β 5 ] ο π΄2 β 4π΄ β 5πΌ = [ 0 0 0 0 0 0 0 0 0 ]
16. 16. ο π΄2 β 4π΄ β 5πΌ = 0 βΉ 5πΌ = π΄2 β 4π΄ ο Both the side multiplying by π΄β1 , we get ο 5π΄β1 = π΄ β 4πΌ ο 5π΄β1 = [ 1 2 2 2 1 2 2 2 1 ] β 4[ 1 0 0 0 1 0 0 0 1 ] ο 5π΄β1 = [ β3 2 2 2 β3 2 2 2 β3 ] ο π΄β1 = 1 5 [ β3 2 2 2 β3 2 2 2 β3 ] οΆ Adjoint of a square Matrix: Let the determinant of the square matrix π΄ be | π΄| = [ π1 π2 π3 π1 π2 π3 π1 π2 π3 ] ο The matrix formed by the co-factors ofthe elements in | π΄| ππ  [ π΄1 π΄2 π΄3 π΅1 π΅2 π΅3 πΆ1 πΆ2 πΆ3 ] ο Where π΄1 = | π2 π3 π2 π3 |= π2 π3 β π3 π2, ο π΄2 = β| π1 π3 π1 π3 | = βπ1 π3 β π3 π1, ο π΄3 = | π1 π2 π1 π2 | = π1 π2 β π2 π1, ο π΅1 = β | π2 π3 π2 π3 | = βπ2 π3 + π3 π2, ο π΅2 = | π1 π3 π1 π3 | = π1 π3 β π3 π1, ο π΅3 = β| π1 π2 π1 π2 | = βπ1 π2 + π2 π1, ο πΆ1 = | π2 π3 π2 π3 | = π2 π3 β π3 π2,
17. 17. ο πΆ2 = β| π1 π3 π1 π3 | = βπ1 π3 + π3 π1, ο πΆ3 = | π1 π2 π1 π2 | = π1 π2 β π2 π1, ο Then the transposeof the matrix of co-factors [ π΄1 π΅1 πΆ1 π΄2 π΅2 πΆ2 π΄3 π΅3 πΆ3 ] Is called the adjoint of the matrix π΄ and is written as πππ π΄. Note:For π Γ π order matrix Adj A is defined as: If π΄ = [ π1 π2 π1 π2 ] then Adj A = [ π2 βπ2 π1 π1 ] i.e. Change location of elelment of principal diagonal and change sign of elements of subsidary diagonal. οΆ Property of Adjoint Matrix: The productof a matrix π΄ and its adjoint is equal to unit matrix multiplied by the determinant π΄. οΆ Example-1: If π¨ = [ π π π π ] then find π¨βπ . Solution: Since here, given matrix π΄ is of the Order 2 Γ 2. ο β΄ πππ π΄ = [ 3 β2 β7 5 ] οΆ Example-2: For matrix π¨ = [ π π π π π βπ π π π ] find co-factormatrix πππ π¨. Solution: Let π΄ = [πππ] ο First we will find co-factors ofeach element. ο π΄11 = | 3 β1 5 0 | = 0 β (β5) = 5 ο π΄12 = β| 0 β1 2 0 | = β[0 β (β2)] = β2
18. 18. ο π΄13 = | 0 3 2 5 | = (0 β 6) = β6 ο π΄21 = β| 1 5 5 0 | = β(0 β 25) = 25 ο π΄22 = | 2 5 2 0 | = (0 β 10) = β10 ο π΄23 = β| 2 1 2 5 | = β(10 β 2) = β8 ο π΄31 = | 1 5 3 β1 | = (β1 β 15) = β16 ο π΄32 = β| 2 5 0 β1 | = β(β2 β 0) = 2 ο π΄33 = | 2 1 0 3 | = (6 β 0) = 6 β΄ πππ π΄ = [ π΄11 π΄21 π΄31 π΄12 π΄22 π΄32 π΄13 π΄23 π΄33 ] = [ 5 25 β16 β2 β10 2 β6 β8 6 ] οΆ Inverse of a Matrix: If A and B are two square matrices of the same order, such that π΄π΅ = π΅π΄ = πΌ Then π΅ is called the inverse of π΄ i.e. π΅ = π΄β1 and π΄ is the invese of B. ο Condition for a square matrix π΄ to possess aninverse is that matrix π΄ is non-singular. i.e. | π¨| β  π ο If π΄ is square matrix and π΅ its inverse, then π΄π΅ = πΌ. Taking determinant of both sides, we get | π΄π΅| = | πΌ| ππ | π΄|| π΅| = πΌ From this relation it is clear that | π΄| β  0 i.e. the matrix π΄ is non singular. To find the inverse matrix with the help of adjoint matrix:
19. 19. ο We know that π΄. ( π΄πππ΄) = | π΄| πΌ ο π΄. 1 | π΄| ( π΄ππ π΄) = πΌ (Provided | π΄| β  0) β¦β¦β¦β¦β¦β¦β¦β¦β¦..(1) ο Since π΄. π΄β1 = πΌ β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(2) ο From (1) and (2), we have β΄ π¨βπ = π | π¨| ( π¨ππ π¨) = π° οΆ Example-1: If π¨ = [ π π βπ π ] find π¨βπ . Solution: | π΄| = | 1 2 β1 3 | = 3 β (β2) = 5 β  0 ο β΄ π΄β1 is possible. ο π΄ππ π΄ = [ 3 β2 1 1 ] ο β΄ π΄β1 = πππ π΄ | π΄| ο π΄β1 = 1 5 [ 3 β2 1 1 ] ο π΄β1 = [ 3 5 β 2 5 1 5 1 5 ] οΆ Example-2: If π¨ = [ π βπ π π βπ π π π π ] then find π¨βπ . Solution: Let given matrix is π΄. ο β΄ | π΄| = | 1 β1 1 2 β1 0 1 0 1 | ο β΄ | π΄| = 1(β1 β 0) + 1(2 β 0) + 1[0β (β1)] ο β΄ | π΄| = β1 + 2 + 1 = 2 β  0 ο β΄ π΄β1 is possible. ο First we will calculate, co-factors ofeach element.
20. 20. ο π΄11 = | β1 0 0 1 | = (β1 β 0) = β1 ο π΄12 = β| 2 0 1 1 | = [β(2β 0)] = β2 ο π΄13 = | 2 β1 1 0 | = [0 β (β1)] = 1 ο π΄21 = β| β1 1 0 1 | = [β(β1 β 0)] = 1 ο π΄22 = | 1 1 1 1 | = (1 β 1) = 0 ο π΄23 = β| 1 β1 1 0 | = β[0 β (β1)] = β1 ο π΄31 = | β1 1 β1 0 | = [0 β (β1)] = 1 ο π΄32 = β| 1 1 2 0 | = [β(0 β 2)] = 2 ο π΄33 = | 1 β1 2 β1 | = [β1 β (β2)] = 1 β΄ πππ π΄ = [ π΄11 π΄21 π΄31 π΄12 π΄22 π΄32 π΄13 π΄23 π΄33 ] = [ β1 1 1 β2 0 2 1 β1 1 ] ο π΄β1 = π΄ππ π΄ | π΄| = 1 2 Γ [ β1 1 1 β2 0 2 1 β1 1 ] ο π΄β1 = [ β 1 2 1 2 1 2 β1 0 1 1 2 β 1 2 1 2 ] οΆ Example-3. If a matrix π¨ satisfies a relation π¨ π + π¨ β π° = π prove that π¨βπ exists and that π¨βπ = π° + π¨, π° being an identity matrix. Solution: Here, π΄2 + π΄ β πΌ = 0 ο β΄ π΄2 + π΄ = πΌ ο β΄ π΄( π΄ + πΌ) = πΌ
21. 21. ο β΄ | π΄|| π΄ + πΌ| = | πΌ| ο β΄ | π΄| β  0 and so π΄β1 exists. ο Again π΄2 + π΄ β πΌ = 0 βΉ π΄2 + π΄ = πΌ ο Multiplying (1) by π΄β1 , we get π΄β1( π΄2 + π΄) = π΄β1 πΌ βΉ π΄ + πΌ = π΄β1 β΄ π΄β1 = πΌ + π΄ οΆ 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 π΄ = [ 0 2 0 1 0 3 1 1 2 ], π΅ = [ 1 2 1 2 1 0 0 0 3 ] find ( π)2π΄ + 3π΅ ( ππ)3π΄ β 4π΅ 2. Express [ 1 2 0 3 7 1 5 9 3 ] as a sum of symmetric matrix and skew-symmetric matrix. 3. If π΄ = [ 2 3 1 0 ], π΅ = [ 4 1 2 β3 ] prove that ( π΄ + π΅) π = π΄ π + π΅ π . 4. If π΄ = [ 2 β1 0 3 2 β4 5 1 9 ] and π΅ = [ 17 β1 3 β24 β1 β16 β7 1 1 ] and 4π΄ + 3πΆ = π΅, then find matrix πΆ. Q-2. Evaluate the following Questions: 1. If π΄ = [ 0 1 2 1 2 3 2 3 4 ] and π΅ = [ 1 β2 β1 0 2 β1 ] Obtain the product π΄π΅ and explain why π΅π΄ is not defined. 2. If π΄ = [ 1 2 β1 3 0 2 4 5 0 ] and π΅ = [ 1 0 0 2 1 0 0 1 3 ] verify that ( π΄π΅)β² = π΅β² π΄β² . 3. Compute π΄π΅ if π΄ = [ 1 2 3 4 5 6 ] and π΅ = [ 2 5 3 3 6 4 4 7 5 ]. 4. Verify that π΄ = 1 3 [ 1 2 2 2 1 β2 β2 2 β1 ] is Orthogonal. Q-3. Evaluate the following Questions:
23. 23. 1. If π΄ = [ 2 5 3 3 1 2 1 2 1 ] then find π΄ππ π΄. 2. If π΄ = [ 1 1 2 1 9 3 1 4 2 ] then find π΄β1 . 3. If π΄ = [ 1 1 1 1 2 3 1 4 9 ], π΅ = [ 2 5 3 3 1 2 1 2 1 ], then show that ( π΄π΅)β1 = π΅β1 π΄β1 . 4. If π΄ = [ β4 β3 β3 1 0 1 4 4 3 ] Prove that π΄ππ π΄ = π΄.