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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  Definition: Matrix  A ractangular array of m rows and n columns, enclosed by brackets [ ] is called a matrix of order 𝑚 × 𝑛. A matrix of order 3 × 3 is expressed as, 𝐴 = [ 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 ]  An element 𝑎𝑖𝑗 denotes, ith row and jth column 𝑎23 denotes, 2nd row and 3rd column.  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 × 𝑛. For example, 𝐴 = [1 4]1×2 𝐴 = [2 4 −3]1×3 2. Column Matrix: A matrix having single column is called column matrix. Its order is 𝑛 × 1. For example, 𝐴 = [ 2 1 ] 2×1 𝐴 = [ 4 −2 6 ] 3×1 3. Square Matrix: A matrx in which the number of rows is equal to number columns is called a square matrix.
3. 3. ∴ 𝑚 = 𝑛. ∴ No. of rows = No. of columns For example, 𝐴 = [ 1 3 −4 2 ] 2×2 𝐴 = [ −3 2 1 2 3 1 3 1 −5 ] 3×3 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 −5 ]
4. 4. 𝐴 = [ 3 0 0 0 −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 = [ −2 0 0 −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.  Transposeof matrix A is denoted by 𝐴′ or 𝐴 𝑇 . For example, 𝐴 = [ 2 1 4 7 6 −3 4 1 0 ] ∴ 𝐴′ = 𝐴 𝑇 = [ 2 7 4 1 6 1 4 −3 0 ]  ( 𝐴′)′ = 𝐴 9. Symmetric Matrix: A square matrix 𝐴 = [𝑎𝑖𝑗] is said to be symmetric, if 𝑎𝑖𝑗 = 𝑎𝑗𝑖 of each pair (i, j) .  For Symmetric matrix 𝐴′ = 𝐴 For example, 𝐴 = [ 4 2 0 2 −3 5 0 5 6 ] ∴ 𝐴′ = [ 4 2 0 2 −3 5 0 5 6 ]
5. 5. 10. Skew Symmetric Matrix: A square matrix 𝐴 = [𝑎𝑖𝑗] is said to be skew symmetric if 𝑎𝑖𝑗 = −𝑎𝑗𝑖 for each pair (i, j). For Skew symmetric matrix 𝐴′ = −𝐴. For example, 𝐴 = [ 0 3 5 −3 0 4 −5 −4 0 ] 𝐴′ = [ 0 −3 −5 3 0 −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 × 3) − (9 × 2) = 18 − 18 = 0 𝐴 = | 1 2 3 4 1 5 3 6 9 | ∴ | 𝐴| = 1(9− 30) − 2(36− 15) + 3(24 − 3). | 𝐴| = −21 − 42+ 63 | 𝐴| = 0 If | 𝐴| = 0 is called singular matrix. If | 𝐴| ≠ 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 −4 5 ] 𝐵 = [ 2 3 −4 5 ] ∴ 𝐴 = 𝐵 13. Negative matrix:
6. 6. For two matrices, if their correspondingelements are equal but opposite, they are called Negative matrix. For example, 𝐴 = [ 1 −2 3 −5 ] ∴ (−𝐴) = [ −1 2 −3 5 ] 14. Orthogonalmatrix: For square matrix A if Productof matirx A & its transpose matrix A’ (i.e. AA’) is Identity matrix (I). 𝐴 = [ 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 ] 𝐴′ = [ 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 −𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 ] 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 −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 −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 = [ −1 2 7 3 5 −8 1 2 7 ] 𝑡𝑟( 𝐴) = −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 −2 −4 −1 3 4 1 −2 −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 −𝑖 3 − 2𝑖 ] Conjugate of matrix A is 𝐴̅ 𝐴̅ = [ 1 − 𝑖 2 − 3𝑖 4 7 − 2𝑖 𝑖 3 + 2𝑖 ] Note: Transposeof the conjugate of a matrix A is denoted by 𝐴 𝜃 . 21.Unitary Matrix: A square matrix A is said to be unitary if 𝐴 𝜃 𝐴 = 𝐼 For example,
8. 8. 𝐴 = [ 1+𝑖 2 −1+𝑖 2 −1−𝑖 2 1−𝑖 2 ] , 𝐴 𝜃 = [ 1−𝑖 2 1−𝑖 2 −1−𝑖 2 1+𝑖 2 ], 𝐴. 𝐴 𝜃 = 𝐼 22. Hermitian Matrix: A square matrix 𝐴 = (𝑎𝑖𝑗) is called Hermitian matrix, if every i-jth element of A is equal to conjugate complex j-ith element of A. In other words, 𝑎𝑖𝑗 = 𝑎𝑗𝑖̅̅̅̅ For example, [ 1 2 + 3𝑖 3 + 𝑖 2 − 3𝑖 2 1 − 2𝑖 3 − 𝑖 1 + 2𝑖 5 ] 23. Skew Hermitian Matrix: A square matrix 𝐴 = (𝑎𝑖𝑗) 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 , 𝑎𝑖𝑗 = −𝑎𝑗𝑖̅̅̅̅ For example, [ 𝑖 2 − 3𝑖 4 + 5𝑖 −(2 + 3𝑖) 0 2𝑖 −(4 − 5𝑖) 2𝑖 −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 −1 1 2 | Minor of element 2 = | 3 5 −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 −6 ] , 𝐵 = [ 1 0 2 3 1 4 ] Then 𝐴 + 𝐵 = [ 4 + 1 2 + 0 5 + 2 1 + 3 3 + 1 −6 + 4 ] 𝐴 + 𝐵 = [ 5 2 7 4 4 −2 ].  If 𝐴 = [𝑎𝑖𝑗], 𝐵 = [𝑏𝑖𝑗] then 𝐴 + 𝐵 = [𝑎𝑖𝑗 + 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 − 𝐵 = [𝑎𝑖𝑗 − 𝑎𝑗𝑖] Thus [ 8 6 4 1 2 0 ] − [ 3 5 1 7 6 2 ] = [ 8 − 3 6 − 5 4 − 1 1 − 7 2 − 6 0 − 2 ] = [ 5 1 3 −6 −4 −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 × 2 3 × 3 3 × 4 3 × 4 3 × 5 3 × 6 3 × 6 3 × 7 3 × 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 = [𝑎𝑖𝑗] be an 𝑚 × 𝑛 matrix and 𝐵 = [𝑏𝑖𝑗] be an 𝑛 × 𝑝 matrix. Then the productAB of these matrices is an 𝑚 × 𝑝 matrix 𝐶 = [𝑐𝑖𝑗] where, 𝑐𝑖𝑗 = 𝑎𝑖1 𝑏1𝑗 + 𝑎𝑖2 𝑏2𝑗 + 𝑎𝑖3 𝑏3𝑗 + ⋯+ 𝑎𝑖𝑛 𝑏 𝑛𝑗  Properties of Matrix Multiplication: a) Multiplication of matrix is not commutative. 𝑨𝑩 ≠ 𝑩𝑨 b) Matrix multiplication is associative , if conformability is assured. 𝑨( 𝑩𝑪) = ( 𝑨𝑩) 𝑪 c) Matrix multiplication is distributive with respectto addition. 𝑨( 𝑩 + 𝑪) = 𝑨𝑩 + 𝑨𝑪 d) Multiplcation of matrix A by unit matrix. 𝑨𝑰 = 𝑰𝑨 = 𝑨 e) Multiplicative inverse of a matrix exists if | 𝐴| ≠ 0. 𝑨. 𝑨−𝟏 = 𝑨−𝟏 . 𝑨 = 𝑰 f) If A is a square then 𝑨 × 𝑨 = 𝑨 𝟐 , 𝑨 × 𝑨 × 𝑨 = 𝑨 𝟑 g) 𝑨 𝟎 = 𝑰 h) 𝑰 𝒏 = 𝑰, where 𝑛 is positive integer.
11. 11. i) ( 𝑨𝑩)′ = 𝑩′ 𝑨′ j) ( 𝑨𝑩)−𝟏 = 𝑩−𝟏 𝑨−𝟏  Example-1. If 𝑨 = [ 𝟐 𝟐 𝟐 𝟐 𝟏 −𝟑 𝟏 𝟎 𝟒 ], 𝑩 = [ 𝟑 𝟑 𝟑 𝟑 𝟎 𝟓 𝟗 𝟗 −𝟏 ], 𝑪 = [ 𝟒 𝟒 𝟒 𝟓 −𝟏 𝟓 −𝟕 𝟖 −𝟏 ] Find 𝟐𝑨 − 𝟑𝑩 + 𝑪. Solution: 2𝐴 − 3𝐵 + 𝐶 = 2 [ 2 2 2 2 1 −3 1 0 4 ] − 3[ 3 3 3 3 0 5 9 9 −1 ] + [ 4 4 4 5 −1 5 −7 8 −1 ] 2𝐴 − 3𝐵 + 𝐶 = [ 4 4 4 4 2 −6 2 0 8 ] + [ −9 −9 −9 −9 0 −15 −27 −27 3 ] + [ 4 4 4 5 −1 5 −7 8 −1 ]. 2𝐴 − 3𝐵 + 𝐶 = [ 4 − 9 + 4 4 − 9 + 4 4 − 9 + 4 4 − 9 + 5 2 + 0 − 1 −6 − 15 + 5 2 − 27 − 7 0 − 27 + 8 8 + 3 − 1 ]. 2𝐴 − 3𝐵 + 𝐶 = [ −1 −1 −1 0 1 −16 −32 −19 10 ].  Example-2. If 𝑨 = [ 𝟏 𝟐 𝟏 𝟑 𝟒 𝟐 ] , 𝑩 = [ 𝟑 −𝟐 𝟒 𝟏 𝟓 𝟎 ] find matrix𝑿 from 𝑿 + 𝑨 + 𝑩 = 𝟎 Solution: 𝑋 + 𝐴 + 𝐵 = 0  ∴ 𝑋 + [ 1 2 1 3 4 2 ] + [ 3 −2 4 1 5 0 ] = 0.  ∴ 𝑋 + [ 4 0 5 4 9 2 ] = [ 0 0 0 0 0 0 ]  ∴ 𝑋 = [ −4 0 −5 −4 −9 −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.  Then 𝐴 = 1 2 ( 𝐴 + 𝐴′) + 1 2 (𝐴 − 𝐴′ )  Now, ( 𝐴 + 𝐴′)′ = 𝐴′ + ( 𝐴′)′ = 𝐴′ + 𝐴 = 𝐴 + 𝐴′  ∴ 𝐴 + 𝐴′ is a symmetric matrix.  Also, ( 𝐴 − 𝐴′)′ = 𝐴′ − ( 𝐴′)′ = 𝐴′ − 𝐴 = −(𝐴 − 𝐴′ )  ∴ ( 𝐴 − 𝐴′) 𝑜𝑟 1 2 (𝐴 − 𝐴′ ) is an anti symmetric matrix.  ∴ 𝑨 = 𝟏 𝟐 ( 𝑨 + 𝑨′) + 𝟏 𝟐 (𝑨 − 𝑨′ )  Example-4.Express 𝑨 = [ 𝟏 −𝟐 −𝟑 𝟑 𝟎 𝟓 𝟓 𝟔 𝟏 ] as the sum of a lower triangular matrix and upper triangular matrix. Solution: Let 𝐴 = 𝐿 + 𝑈  [ 1 −2 −3 3 0 5 5 6 1 ] = [ 𝑎 0 0 𝑏 𝑐 0 𝑑 𝑒 𝑓 ] + [ 1 𝑝 𝑞 0 1 𝑟 0 0 1 ]  [ 1 −2 −3 3 0 5 5 6 1 ] = [ 𝑎 + 1 0 0 𝑏 + 0 𝑐 + 1 0 + 𝑟 𝑑 + 0 𝑒 + 0 𝑓 + 1 ]  Equating the correspondingelements on both the sides, we get 𝑎 + 1 = 1 𝑝 = −2 𝑞 = −3 𝑏 = 3 𝑐 + 1 = 0 𝑟 = 5 𝑑 = 5 𝑒 = 6 𝑓 + 1 = 1  On Solving these equations, we get 𝑎 = 0 𝑝 = −2 𝑞 = −3 𝑏 = 3 𝑐 = −1 𝑟 = 5 𝑑 = 5 𝑒 = 6 𝑓 = 0  Hence 𝐿 = [ 0 0 0 3 −1 0 5 6 0 ] & 𝑈 = [ 1 −2 −3 0 1 5 0 0 1 ]
13. 13.  Example-5 If 𝑨 = [ 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 ] and 𝑩 = [ 𝟏 𝟐 𝟐 𝟏 𝟏 𝟐 ] find 𝑨𝑩 and 𝑩𝑨. Solution: 𝐴𝐵 = [ 1 2 3 4 5 6 ] [ 1 2 2 1 1 2 ]  𝐴𝐵 = [ (1 × 1) + (2 × 2) + (3 × 1) (1 × 2) + (2 × 1) + (3 × 2) (4 × 1) + (5 × 2) + (6 × 1) (4 × 2) + (5 × 1) + (6 × 2) ]  𝐴𝐵 = [ (1 + 4 + 3) (2 + 2 + 6) (4 + 10 + 6) (8 + 5 + 12) ]  𝐴𝐵 = [ 8 10 20 25 ] Now 𝐵𝐴 = [ 1 2 2 1 1 2 ][ 1 2 3 4 5 6 ]  𝐴 = [ (1 × 1) + (2 × 4) (1 × 2) + (2 × 5) (1 × 3) + (2 × 6) (2 × 1) + (1 × 4) (2 × 2) + (1 × 5) (2 × 3) + (1 × 6) (1 × 1) + (2 × 4) (1 × 2) + (2 × 5) (1 × 3) + (2 × 6) ]  𝐴 = [ (1 + 8) (2 + 10) (3 + 12) (2 + 4) (4 + 5) (6 + 6) (1 + 8) (2 + 10) (3 + 12) ]  𝐴 = [ 9 12 15 6 9 12 9 12 15 ]  Example-6. If 𝑨 = [ 𝟏 𝟐 −𝟐 𝟑 ] , 𝑩 = [ 𝟐 𝟏 𝟐 𝟑 ] and 𝑪 = [ −𝟑 𝟏 𝟐 𝟎 ]. Verify that ( 𝑨𝑩) 𝑪 = 𝑨(𝑩𝑪) and 𝑨( 𝑩 + 𝑪) = 𝑨𝑩 + 𝑨𝑪. Solution: 𝐴𝐵 = [ 1 2 −2 3 ] × [ 2 1 2 3 ]  𝐴𝐵 = [ (1)(2) + (2)(2) (1)(1) + (2)(3) (−2)(2) + (3)(2) (−2)(1) + (3)(3) ]  𝐴𝐵 = [ 6 7 2 7 ]
14. 14.  𝐵𝐶 = [ 2 1 2 3 ] × [ −3 1 2 0 ]  𝐵𝐶 = [ −6 + 2 2 + 0 −6 + 6 2 + 0 ]  𝐵𝐶 = [ −4 2 0 2 ]  𝐴𝐶 = [ 1 2 −2 3 ] × [ −3 1 2 0 ]  𝐴𝐶 = [ −3 + 4 1 + 0 6 + 6 −2 + 0 ]  𝐴𝐶 = [ 1 1 12 −2 ]  𝐵 + 𝐶 = [ 2 1 2 3 ] + [ −3 1 2 0 ]  𝐵 + 𝐶 = [ 2 + (−3) 1 + 1 2 + 2 3 + 0 ]  𝐵 + 𝐶 = [ −1 2 4 3 ] i. ( 𝐴𝐵) 𝐶 = [ 6 7 2 7 ] × [ −3 1 2 0 ]  ( 𝐴𝐵) 𝐶 = [ −18 + 14 6 + 0 −6 + 14 2 + 0 ]  ( 𝐴𝐵) 𝐶 = [ −4 6 8 2 ] .……………………………………………..(1)  & 𝐴( 𝐵𝐶) = [ 1 2 −2 3 ] × [ −4 2 0 2 ]  𝐴( 𝐵𝐶) = [ −4 + 0 2 + 4 8 + 0 −4 + 6 ]  𝐴( 𝐵𝐶) = [ −4 6 8 2 ] ……………………………………………….(2) Thus from (1) and (2), we get  ( 𝐴𝐵) 𝐶 = 𝐴(𝐵𝐶) ii. 𝐴( 𝐵 + 𝐶) = [ 1 2 −2 3 ][ −1 2 4 3 ]
15. 15.  𝐴( 𝐵 + 𝐶) = [ −1 + 8 2 + 6 2 + 12 −4 + 9 ]  𝐴( 𝐵 + 𝐶) = [ 7 8 14 5 ] ………………………………………….(3)  𝐴𝐵 + 𝐴𝐶 = [ 6 + 1 7 + 1 2 + 12 7 − 2 ]  𝐴𝐵 + 𝐴𝐶 = [ 7 8 24 5 ] …………………………………………..(4) Thus from (3) and (4), we get 𝐴( 𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶  Example-7. If 𝑨 = [ 𝟏 𝟐 𝟐 𝟐 𝟏 𝟐 𝟐 𝟐 𝟏 ] show that 𝑨 𝟐 − 𝟒𝑨 − 𝟓𝑰 = 𝟎 where I, 0 are the unit matrix and the null matrix of order 3 respectively. Use this result to find 𝑨−𝟏 . Solution: Here, we have 𝐴 = [ 1 2 2 2 1 2 2 2 1 ]  𝐴2 = [ 1 2 2 2 1 2 2 2 1 ][ 1 2 2 2 1 2 2 2 1 ]  𝐴2 = [ 9 8 8 8 9 8 8 8 9 ]  𝐴2 − 4𝐴 − 5𝐼 = [ 9 8 8 8 9 8 8 8 9 ] − 4[ 1 2 2 2 1 2 2 2 1 ] − 5[ 1 0 0 0 1 0 0 0 1 ]  𝐴2 − 4𝐴 − 5𝐼 = [ 9 − 4 − 5 8 − 8 + 0 8 − 8 + 0 8 − 8 + 0 9 − 4 − 5 8 − 8 + 0 8 − 8 + 0 8 − 8 + 0 9 − 4 − 5 ]  𝐴2 − 4𝐴 − 5𝐼 = [ 0 0 0 0 0 0 0 0 0 ]
16. 16.  𝐴2 − 4𝐴 − 5𝐼 = 0 ⟹ 5𝐼 = 𝐴2 − 4𝐴  Both the side multiplying by 𝐴−1 , we get  5𝐴−1 = 𝐴 − 4𝐼  5𝐴−1 = [ 1 2 2 2 1 2 2 2 1 ] − 4[ 1 0 0 0 1 0 0 0 1 ]  5𝐴−1 = [ −3 2 2 2 −3 2 2 2 −3 ]  𝐴−1 = 1 5 [ −3 2 2 2 −3 2 2 2 −3 ]  Adjoint of a square Matrix: Let the determinant of the square matrix 𝐴 be | 𝐴| = [ 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 𝑐1 𝑐2 𝑐3 ]  The matrix formed by the co-factors ofthe elements in | 𝐴| 𝑖𝑠 [ 𝐴1 𝐴2 𝐴3 𝐵1 𝐵2 𝐵3 𝐶1 𝐶2 𝐶3 ]  Where 𝐴1 = | 𝑏2 𝑏3 𝑐2 𝑐3 |= 𝑏2 𝑐3 − 𝑏3 𝑐2,  𝐴2 = −| 𝑏1 𝑏3 𝑐1 𝑐3 | = −𝑏1 𝑐3 − 𝑏3 𝑐1,  𝐴3 = | 𝑏1 𝑏2 𝑐1 𝑐2 | = 𝑏1 𝑐2 − 𝑏2 𝑐1,  𝐵1 = − | 𝑎2 𝑎3 𝑐2 𝑐3 | = −𝑎2 𝑐3 + 𝑎3 𝑐2,  𝐵2 = | 𝑎1 𝑎3 𝑐1 𝑐3 | = 𝑎1 𝑐3 − 𝑎3 𝑐1,  𝐵3 = −| 𝑎1 𝑎2 𝑐1 𝑐2 | = −𝑎1 𝑐2 + 𝑎2 𝑐1,  𝐶1 = | 𝑎2 𝑎3 𝑏2 𝑏3 | = 𝑎2 𝑏3 − 𝑎3 𝑏2,
17. 17.  𝐶2 = −| 𝑎1 𝑎3 𝑏1 𝑏3 | = −𝑎1 𝑏3 + 𝑎3 𝑏1,  𝐶3 = | 𝑎1 𝑎2 𝑏1 𝑏2 | = 𝑎1 𝑏2 − 𝑎2 𝑏1,  Then the transposeof the matrix of co-factors [ 𝐴1 𝐵1 𝐶1 𝐴2 𝐵2 𝐶2 𝐴3 𝐵3 𝐶3 ] Is called the adjoint of the matrix 𝐴 and is written as 𝑎𝑑𝑗 𝐴. Note:For 𝟐 × 𝟐 order matrix Adj A is defined as: If 𝐴 = [ 𝑎1 𝑎2 𝑏1 𝑏2 ] then Adj A = [ 𝑏2 −𝑎2 𝑏1 𝑎1 ] 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 = [ 𝟓 𝟐 𝟕 𝟑 ] then find 𝑨−𝟏 . Solution: Since here, given matrix 𝐴 is of the Order 2 × 2.  ∴ 𝑎𝑑𝑗 𝐴 = [ 3 −2 −7 5 ]  Example-2: For matrix 𝑨 = [ 𝟐 𝟏 𝟓 𝟎 𝟑 −𝟏 𝟐 𝟓 𝟎 ] find co-factormatrix 𝒂𝒅𝒋 𝑨. Solution: Let 𝐴 = [𝑎𝑖𝑗]  First we will find co-factors ofeach element.  𝐴11 = | 3 −1 5 0 | = 0 − (−5) = 5  𝐴12 = −| 0 −1 2 0 | = −[0 − (−2)] = −2
18. 18.  𝐴13 = | 0 3 2 5 | = (0 − 6) = −6  𝐴21 = −| 1 5 5 0 | = −(0 − 25) = 25  𝐴22 = | 2 5 2 0 | = (0 − 10) = −10  𝐴23 = −| 2 1 2 5 | = −(10 − 2) = −8  𝐴31 = | 1 5 3 −1 | = (−1 − 15) = −16  𝐴32 = −| 2 5 0 −1 | = −(−2 − 0) = 2  𝐴33 = | 2 1 0 3 | = (6 − 0) = 6 ∴ 𝑎𝑑𝑗 𝐴 = [ 𝐴11 𝐴21 𝐴31 𝐴12 𝐴22 𝐴32 𝐴13 𝐴23 𝐴33 ] = [ 5 25 −16 −2 −10 2 −6 −8 6 ]  Inverse of a Matrix: If A and B are two square matrices of the same order, such that 𝐴𝐵 = 𝐵𝐴 = 𝐼 Then 𝐵 is called the inverse of 𝐴 i.e. 𝐵 = 𝐴−1 and 𝐴 is the invese of B.  Condition for a square matrix 𝐴 to possess aninverse is that matrix 𝐴 is non-singular. i.e. | 𝑨| ≠ 𝟎  If 𝐴 is square matrix and 𝐵 its inverse, then 𝐴𝐵 = 𝐼. Taking determinant of both sides, we get | 𝐴𝐵| = | 𝐼| 𝑜𝑟 | 𝐴|| 𝐵| = 𝐼 From this relation it is clear that | 𝐴| ≠ 0 i.e. the matrix 𝐴 is non singular. To find the inverse matrix with the help of adjoint matrix:
19. 19.  We know that 𝐴. ( 𝐴𝑑𝑗𝐴) = | 𝐴| 𝐼  𝐴. 1 | 𝐴| ( 𝐴𝑑𝑗 𝐴) = 𝐼 (Provided | 𝐴| ≠ 0) ………………………..(1)  Since 𝐴. 𝐴−1 = 𝐼 …………………………………………….(2)  From (1) and (2), we have ∴ 𝑨−𝟏 = 𝟏 | 𝑨| ( 𝑨𝒅𝒋 𝑨) = 𝑰  Example-1: If 𝑨 = [ 𝟏 𝟐 −𝟏 𝟑 ] find 𝑨−𝟏 . Solution: | 𝐴| = | 1 2 −1 3 | = 3 − (−2) = 5 ≠ 0  ∴ 𝐴−1 is possible.  𝐴𝑑𝑗 𝐴 = [ 3 −2 1 1 ]  ∴ 𝐴−1 = 𝑎𝑑𝑗 𝐴 | 𝐴|  𝐴−1 = 1 5 [ 3 −2 1 1 ]  𝐴−1 = [ 3 5 − 2 5 1 5 1 5 ]  Example-2: If 𝑨 = [ 𝟏 −𝟏 𝟏 𝟐 −𝟏 𝟎 𝟏 𝟎 𝟏 ] then find 𝑨−𝟏 . Solution: Let given matrix is 𝐴.  ∴ | 𝐴| = | 1 −1 1 2 −1 0 1 0 1 |  ∴ | 𝐴| = 1(−1 − 0) + 1(2 − 0) + 1[0− (−1)]  ∴ | 𝐴| = −1 + 2 + 1 = 2 ≠ 0  ∴ 𝐴−1 is possible.  First we will calculate, co-factors ofeach element.
20. 20.  𝐴11 = | −1 0 0 1 | = (−1 − 0) = −1  𝐴12 = −| 2 0 1 1 | = [−(2− 0)] = −2  𝐴13 = | 2 −1 1 0 | = [0 − (−1)] = 1  𝐴21 = −| −1 1 0 1 | = [−(−1 − 0)] = 1  𝐴22 = | 1 1 1 1 | = (1 − 1) = 0  𝐴23 = −| 1 −1 1 0 | = −[0 − (−1)] = −1  𝐴31 = | −1 1 −1 0 | = [0 − (−1)] = 1  𝐴32 = −| 1 1 2 0 | = [−(0 − 2)] = 2  𝐴33 = | 1 −1 2 −1 | = [−1 − (−2)] = 1 ∴ 𝑎𝑑𝑗 𝐴 = [ 𝐴11 𝐴21 𝐴31 𝐴12 𝐴22 𝐴32 𝐴13 𝐴23 𝐴33 ] = [ −1 1 1 −2 0 2 1 −1 1 ]  𝐴−1 = 𝐴𝑑𝑗 𝐴 | 𝐴| = 1 2 × [ −1 1 1 −2 0 2 1 −1 1 ]  𝐴−1 = [ − 1 2 1 2 1 2 −1 0 1 1 2 − 1 2 1 2 ]  Example-3. If a matrix 𝑨 satisfies a relation 𝑨 𝟐 + 𝑨 − 𝑰 = 𝟎 prove that 𝑨−𝟏 exists and that 𝑨−𝟏 = 𝑰 + 𝑨, 𝑰 being an identity matrix. Solution: Here, 𝐴2 + 𝐴 − 𝐼 = 0  ∴ 𝐴2 + 𝐴 = 𝐼  ∴ 𝐴( 𝐴 + 𝐼) = 𝐼
21. 21.  ∴ | 𝐴|| 𝐴 + 𝐼| = | 𝐼|  ∴ | 𝐴| ≠ 0 and so 𝐴−1 exists.  Again 𝐴2 + 𝐴 − 𝐼 = 0 ⟹ 𝐴2 + 𝐴 = 𝐼  Multiplying (1) by 𝐴−1 , we get 𝐴−1( 𝐴2 + 𝐴) = 𝐴−1 𝐼 ⟹ 𝐴 + 𝐼 = 𝐴−1 ∴ 𝐴−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 ], 𝐵 = [ 1 2 1 2 1 0 0 0 3 ] find ( 𝑖)2𝐴 + 3𝐵 ( 𝑖𝑖)3𝐴 − 4𝐵 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 ], 𝐵 = [ 4 1 2 −3 ] prove that ( 𝐴 + 𝐵) 𝑇 = 𝐴 𝑇 + 𝐵 𝑇 . 4. If 𝐴 = [ 2 −1 0 3 2 −4 5 1 9 ] and 𝐵 = [ 17 −1 3 −24 −1 −16 −7 1 1 ] and 4𝐴 + 3𝐶 = 𝐵, then find matrix 𝐶. Q-2. Evaluate the following Questions: 1. If 𝐴 = [ 0 1 2 1 2 3 2 3 4 ] and 𝐵 = [ 1 −2 −1 0 2 −1 ] Obtain the product 𝐴𝐵 and explain why 𝐵𝐴 is not defined. 2. If 𝐴 = [ 1 2 −1 3 0 2 4 5 0 ] and 𝐵 = [ 1 0 0 2 1 0 0 1 3 ] verify that ( 𝐴𝐵)′ = 𝐵′ 𝐴′ . 3. Compute 𝐴𝐵 if 𝐴 = [ 1 2 3 4 5 6 ] and 𝐵 = [ 2 5 3 3 6 4 4 7 5 ]. 4. Verify that 𝐴 = 1 3 [ 1 2 2 2 1 −2 −2 2 −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 𝐴−1 . 3. If 𝐴 = [ 1 1 1 1 2 3 1 4 9 ], 𝐵 = [ 2 5 3 3 1 2 1 2 1 ], then show that ( 𝐴𝐵)−1 = 𝐵−1 𝐴−1 . 4. If 𝐴 = [ −4 −3 −3 1 0 1 4 4 3 ] Prove that 𝐴𝑑𝑗 𝐴 = 𝐴.