Module 2 Linear Algebra PDF

Module 2 Linear Algebra Elementary transformations associated with a matrix Interchange any two rows(columns) Multiplication of any row(column) by a nonzero constant Addition to any row(column), a constant multiple of any other row(col...

Module 2 Linear Algebra Elementary transformations associated with a matrix Interchange any two rows(columns) Multiplication of any row(column) by a nonzero constant Addition to any row(column), a constant multiple of any other row(column) Equivalent matrices: Two matrices A & B of the same order are said to be equivalent if one matrix can be obtained from the other by a finite number of successive elementary row(column) transformations and is denoted by 𝑨~𝑩 Rank of a matrix: Let A be a non-zero matrix of order mn. A positive integer ‘r’ is said to be the rank of A, if the following conditions are satisfied A has at least one non-zero minor of order ‘r’ Every minor of A whose order is greater than ‘r’ is equal to zero i.e. Rank is the order of any highest order non-vanishing minor of the matrix. The rank of a matrix A in echelon form is equal to the number non-zero rows and is denoted by 𝜌(𝐴) Echelon form of a matrix: All zero rows are below non zero rows The first non-zero element in any non-zero row is to the right of first non-zero element in the previous row. 1. Find rank of the following matrices by using elementary row transformations 𝟏 𝟐 𝟑 i. 𝑨 = [𝟏 𝟒 𝟐] 𝟐 𝟔 𝟓 Solution: 1 2 3 𝐴 = [1 4 2] 2 6 5 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1 1 2 3 ~ [0 2 −1] 0 2 −1 𝑅3 → 𝑅3 − 𝑅2 1 2 3 ~ [0 2 −1] 0 0 0 𝝆(𝑨) = 𝟐  𝟏 𝟒 𝟓 ii. 𝑨 = [𝟐 𝟔 𝟖 ] 𝟐 𝟕 𝟐𝟐 𝑅2 → 𝑅2 − 2𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1 1 4 5 ~ [0 −2 −2] 0 −1 12 𝑅3 → 2𝑅3 − 𝑅2 1 4 5 ~ [0 −2 −2] 0 0 26 𝝆(𝑨) = 𝟑 𝟐 𝟏 𝟑 𝟓 iii. 𝑨=[𝟒 𝟐 𝟏 𝟑 ] 𝟖 𝟒 𝟕 𝟏𝟑 𝟏𝟔 𝟖 −𝟔 −𝟐 Solution: 2 1 3 5 𝐴=[4 2 1 3 ] 8 4 7 13 16 8 −6 −2 𝑅2 → 𝑅2 − 2𝑅1 , 𝑅3 → 𝑅3 − 4𝑅1 , 𝑅4 → 𝑅4 − 8𝑅1 2 1 3 5 ~ [0 0 −5 −7 ] 0 0 −5 −7 0 0 −30 −42 𝑅3 → 𝑅3 − 𝑅2 , 𝑅4 → 𝑅4 − 6𝑅2 2 1 3 5 ~ [0 0 −5 −7] 0 0 0 0 0 0 0 0 𝝆(𝑨) = 𝟐 𝟐 𝟑 −𝟏 −𝟏 iv. 𝑨=[𝟏 −𝟏 −𝟐 −𝟒] 𝟑 𝟏 𝟑 −𝟐 𝟔 𝟑 𝟎 −𝟕 2 3 −1 −1 Solution:𝐴 = [1 −1 −2 −4] 3 1 3 −2 6 3 0 −7 𝑅1 ↔ 𝑅2 1 −1 −2 −4 ~ [2 3 −1 −1] 3 1 3 −2 6 3 0 −7 𝑅2 → 𝑅2 − 2𝑅1 , 𝑅3 → 𝑅3 − 3𝑅1 , 𝑅4 → 𝑅4 − 6𝑅1 1 −1 −2 −4 ~ [0 5 3 7] 0 4 9 10 0 9 12 −7 4 9 𝑅3 → 𝑅3 − 𝑅2 , 𝑅4 → 𝑅4 − 𝑅2 5 5 1 −1 −2 −4 0 5 3 7 33 22 ~ 0 0 5 5 0 0 33 22 [ 5 5] 𝑅4 → 𝑅4 − 𝑅3 𝝆(𝑨) = 𝟑 v. Find the values of k such that the matrix A may have rank equal to a) 3 b)2 1 1 1 1 𝐴 = [1 2 4 𝑘] 1 4 10 𝑘 2 Practice problems: Find rank of the following matrices 4 0 2 1 i. 𝐴 = [2 1 3 4 ] (Ans:4) 2 3 4 7 2 3 1 4 0 1 2 −2 ii. 𝐴 = [4 0 2 6] (Ans:2) 2 1 3 1 2 1 −1 −1 iii. 𝐴 = [1 −1 −2 −4] (Ans: 3) 3 1 3 −2 6 3 0 −7 Solution of system of linear equations-Consistency: Let 𝑎11 𝑥1 + 𝑎12 𝑥2 + 𝑎13 𝑥3 + ⋯ + 𝑎1𝑛 𝑥𝑛 = 𝑏1 𝑎21 𝑥1 + 𝑎22 𝑥2 + 𝑎23 𝑥3 + ⋯ + 𝑎2𝑛 𝑥𝑛 = 𝑏2 ----------------------------------------------------- ----------------------------------------------------- 𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + 𝑎𝑚3 𝑥3 + ⋯ + 𝑎𝑚𝑛 𝑥𝑛 = 𝑏𝑚 be the system of linear equations. The above system of equations can be written in the matrix form AX=B The system is said to be Homogeneous if 𝑏1 = 𝑏2 = 𝑏3 = ⋯ = 𝑏𝑛 = 0 otherwise said to be Non homogeneous system of linear equations Solution: It is the set of values of 𝑥1 , 𝑥2 , 𝑥3 , … 𝑥𝑛 that satisfy all the equations of the system If 𝑥1 = 𝑥2 = 𝑥3 = … = 𝑥𝑛 = 0 is called trivial solution of the homogeneous system of equations otherwise it is said to be non trivial solution. Consistent: The system of linear equations has solution Inconsistent: The system of linear equations has no solution. Here (n-r) unknowns can be taken arbitrarily. To determine whether the system is consistent or inconsistent, consider augmented matrix [𝐴: 𝑩] and find its rank. Unique solution: if 𝜌([𝐴: 𝐵] ) = 𝜌(𝐴) = 𝑟 = 𝑛 Infinite number of solutions: 𝜌([𝐴: 𝐵] ) = 𝜌(𝐴) = 𝑟 < 𝑛 No solution: 𝜌([𝐴: 𝐵] ) ≠ 𝜌(𝐴) Test the following system of equations for consistency i. 𝑥 + 𝑦 + 𝑧 = −3 3𝑥 + 𝑦 − 2𝑧 = −2 3𝑥 + 4𝑦 + 7𝑧 = 7 Solution: 1 1 1 : −3 [𝐴: 𝑩] = [3 1 −2 : −2] 3 4 7 : 7 𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 3𝑅1 1 1 1 : −3 [𝐴: 𝑩] = [0 −2 −5 : 7] 0 1 4 : 16 𝑅3 → 𝑅3 + 𝑅2 1 1 1 : −3 [𝐴: 𝑩] = [0 −2 −5 : 7] 0 0 3 : 39 𝜌([𝐴: 𝐵] ) = 𝜌(𝐴) = 3 Hence the given system of equations is consistent and has unique solution. ii. 𝑥+𝑦+𝑧 =6 𝑥 − 𝑦 + 2𝑧 = 5 3𝑥 + 𝑦 + 𝑧 = 8 Solution: 1 1 1 : 6 [𝐴: 𝑩] = [1 −1 2 : 5] 3 1 1 : 8 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 3𝑅1  1 1 1 : 6 ~ [0 −2 1 : −1 ] 0 −2 −2 : −10 𝑅3 → 𝑅3 − 𝑅2 1 1 1 : 6 ~ [0 −2 1 : −1 ] 0 0 −3 : −9 𝜌([𝐴: 𝐵] ) = 𝜌(𝐴) = 3 = 𝑛 Hence the given system of equations is consistent and has unique solution. 𝑥+𝑦+𝑧 =6 −2𝑦 + 𝑧 = −1 −3𝑧 = −9 Hence the unique solution is 𝑥 = 1, 𝑦 = 2, 𝑧 = 3 iii. 5𝑥 + 3𝑦 + 7𝑧 = 4 3𝑥 + 26𝑦 + 2𝑧 = 9 7𝑥 + 2𝑦 + 10𝑧 = 5 Solution: 5 3 7 : 4 [𝐴: 𝑩] = [3 26 2 : 9 ] 7 2 10 : 5 𝑅2 → 5𝑅2 − 3𝑅1 , 𝑅3 → 5𝑅3 − 7𝑅1 5 3 7 : 4 ~ [0 121 −11 : 33] 0 −11 1 : −3 𝑅3 → 11𝑅3 + 𝑅2 5 3 7 : 4 ~ [0 11 −1 : 3] 0 −11 1 : −3 𝑅3 → 𝑅3 + 𝑅2 5 3 7 : 4 ~ [0 11 −1 : 3] 0 0 0 : 0 𝜌([𝐴: 𝐵] ) = 𝜌(𝐴) = 2 < 𝑛 Hence the given system of equations is consistent and has infinite number of solutions. 5𝑥 + 3𝑦 + 7𝑧 = 4 11𝑦 − 𝑧 = 3 𝑘+3 7−16𝑘 Let 𝑧 = 𝑘  𝑦 = 11  𝑥= 11 iv. Find the values of  and  so that that the equations 𝑥+𝑦+𝑧 =6 𝑥 + 2𝑦 + 3𝑧 = 10 𝑥 + 2𝑦 + 𝜆 𝑧 = 𝜇 may have a) unique solution b)Infinite solution c) No solution Solution: 1 1 1 : 6 [𝐴: 𝑩] = [1 2 3 : 10] 1 2 𝜆 : 𝜇 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 𝑅1 1 1 1 : 6 ~ [0 1 2 : 4 ] 0 1 𝜆−1 : 𝜇−6 𝑅3 → 𝑅3 − 𝑅2 1 1 1 : 6 ~ [0 1 2 : 4 ] 0 0 𝜆 − 3 : 𝜇 − 10 Unique solution if 𝜆 ≠ 3 Infinite solution if 𝜆 = 3, 𝜇 = 10 No solution if 𝜆 = 3, 𝜇 ≠ 10 Practice problems: i. 𝑥 + 2𝑦 + 3𝑧 = 14 4𝑥 + 5𝑦 + 7𝑧 = 35 3𝑥 + 3𝑦 + 4𝑧 = 21 (Ans: x=k/3, y=7-5k/3, z=k) ii. 𝑥 − 4𝑦 + 7𝑧 = 14 3𝑥 + 8𝑦 − 2𝑧 = 13 7𝑥 − 8𝑦 + 26𝑧 = 5 (Ans: Inconsistent) iii. Find for what value of k the system of equation possesses a solution and solve in each case 𝑥+𝑦+𝑧 =1 𝑥 + 2𝑦 + 4𝑧 = 𝑘 𝑥 + 4𝑦 + 10𝑧 = 𝑘 2 Solution of system of equations by Gauss elimination method: In this method, the augmented matrix [𝐴: 𝑩] is reduced to an upper triangular matrix and the solution is obtained by back substitution. 𝑎11 𝑎12 𝑎13 : 𝑏1 [𝐴: 𝑩] = ~ [ 0 𝑎22 ′ 𝑎23 ′ : 𝑏2 ′ ] is upper triangular matrix. 0 0 𝑎33 ′ : 𝑏3 ′ Solve the following system of equations by Gauss elimination method iv. 𝑥+𝑦+𝑧 =9 𝑥 − 2𝑦 + 3𝑧 = 8 2𝑥 + 𝑦 − 𝑧 = 3 Solution: 1 1 1 : 9 [𝐴: 𝑩] = [1 −2 3 : 8 ] 2 1 −1 : 3 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1  1 1 1 : 9 ~ [0 −3 2 : −1 ] 0 −1 −3 : −15 𝑅3 → −3𝑅3 + 𝑅2 1 1 1 : 9 ~ [0 −3 2 : −1 ] 0 0 11 : 44 𝑥+𝑦+𝑧 =9 −3𝑦 + 2𝑧 = −1 11𝑧 = 44 Hence the solution is 𝑥 = 2, 𝑦 = 3, 𝑧=4 v. 𝑥1 + 𝑥2 + 𝑥3 + 4𝑥4 = −6 𝑥1 + 7𝑥2 + 𝑥3 + 𝑥4 = 12 𝑥1 + 𝑥2 + 6𝑥3 + 𝑥4 = −5 5𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 =4 Solution: 1 1 1 4 : −6 [𝐴 ∶ 𝐵] = [1 7 1 1 : 12 ] 1 1 6 1 : −5 5 1 1 1 : 4 1 1 1 4 : −6 0 6 0 −3 : 18 ~[ ] 0 0 5 −3 : 1 0 −4 −4 −19 : 34 1 1 1 4 : −6 0 2 0 −1 : 6 ~[ ] 0 0 5 −3 : 1 0 −4 −4 −19 : 34 1 1 1 4 : −6 0 2 0 −1 : 6 ~[ ] 0 0 5 −3 : 1 0 0 −4 −21 : 46 1 1 1 4 : −6 0 2 0 −1 : 6 ] ~[ 0 0 5 −3 : 1 0 0 0 −117 : 234 𝑥1 + 𝑥2 + 𝑥3 + 4𝑥4 = −6 2𝑥2 − 𝑥4 = 6 5𝑥3 − 3𝑥4 = 1 −117𝑥4 = 234 Hence the solution is 𝑥1 = 1, 𝑥2 = 2, 𝑥3 = −1, 𝑥4 = −2  vi. 2𝑥1 − 𝑥2 + 3𝑥3 = 1 −3𝑥1 + 4𝑥2 − 5𝑥3 = 0 9 23 13 𝑥1 + 3𝑥2 − 6𝑥3 = 0 (Ans: 𝑥1 = , 𝑥2 = , 𝑥3 = ) 34 34 34 vii. 3𝑥 + 3𝑦 + 2𝑧 = 1 𝑥 + 2𝑦 = 4 10𝑦 + 3𝑧 = −2 2𝑥 − 3𝑦 − 𝑧 = 5 (Ans: x=2, y=1, z=-4) LU decomposition method(Crout’s method) In this method the coefficient matrix A is decomposed into the product of two matrices L and U which are lower and upper triangular matrices respectively. Note: The diagonal elements of U are taken 1 for convenience. Let 𝐴𝑋 = 𝐵 be the system of equations. where 𝑎11 𝑎12 𝑎13 𝑎 𝐴 = [ 21 𝑎22 𝑎23 ] 𝑎31 𝑎32 𝑎33 𝐴 = 𝐿𝑈 𝑎11 𝑎12 𝑎13 𝑙11 0 0 1 𝑢12 𝑢13 𝑎 ∴ [ 21 𝑎22 𝑎23 ] = [𝑙21 𝑙22 0 ] [0 1 𝑢23 ] 𝑎31 𝑎32 𝑎33 𝑙31 𝑙32 𝑙33 0 0 1 𝑎11 𝑎12 𝑎13 𝑙11 𝑙11 𝑢12 𝑙11 𝑢13 [𝑎21 𝑎22 𝑎23 ] = [𝑙21 𝑙21 𝑢12 + 𝑙22 𝑙21 𝑢13 + 𝑙22 𝑢23 ] 𝑎31 𝑎32 𝑎33 𝑙31 𝑙31 𝑢12 + 𝑙32 𝑙31 𝑢13 + 𝑙32 𝑢23 + 𝑙33 Equating the corresponding elements on both the sides, we get L & U Now, 𝐴𝑋 = 𝐵 (𝐿𝑈)𝑋 = 𝐵 𝐿(𝑈𝑋) = 𝐵 Let 𝑈𝑋 = 𝑌 𝐿𝑌 = 𝐵 Hence find Y Now by considering 𝑈𝑋 = 𝑌, the solution X can be obtained. Use LU decomposition method (Crout’s method) to solve the following system of equations i. 2𝑥 + 𝑦 + 4𝑧 = 12 4𝑥 + 11𝑦 − 𝑧 = 33 8𝑥 − 3𝑦 + 2𝑧 = 20 Solution: 2 1 4 𝑥 12 𝐴 = [4 11 −1] 𝑋 = [𝑦] 𝐵 = 8 −3 2 𝑧 20 𝐿𝑈 = 𝐴 𝑙11 0 0 1 𝑢12 𝑢13 2 1 4 [𝑙21 𝑙22 0 ] [0 1 𝑢23 ] = [4 11 −1] 𝑙31 𝑙32 𝑙33 0 0 1 8 −3 2 2 0 0 1 1⁄2 2 𝐿 = [4 9 0 ] 𝑈 = [0 1 −1] 8 −7 −21 0 0 1 𝐿𝑌 = 𝐵 2 0 0 𝑦1 12 [4 9 0 ] [𝑦 2 ] = [ 33] 𝑦1 = 6, 𝑦2 = 1, 𝑦3 = 1 8 −7 −21 3 𝑦 20 𝑈𝑋 = 𝑌 1 1⁄ 𝑥 6 2 2 [0 1 −1 ] [𝑦 ] =  𝑥 = 3, 𝑦 = 2, 𝑧=1 0 0 1 𝑧 1 ii. 4𝑥1 + 𝑥2 + 𝑥3 = 4 𝑥1 + 4𝑥2 − 2𝑥3 = 4 3𝑥1 + 2𝑥2 − 4𝑥3 = 6 Solution: 4 1 1 𝑥1 4 𝐴 = [1 4 −2] 𝑥 𝑋 = [ 2] 𝐵 = [ 4] 3 2 −4 𝑥3 6 𝐿𝑈 = 𝐴 𝑙11 0 0 1 𝑢12 𝑢13 4 1 1 [𝑙21 𝑙22 0 ] [0 1 𝑢23 ] = [ 1 4 −2] 𝑙31 𝑙32 𝑙33 0 0 1 3 2 −4 4 0 0 1 1⁄ 1⁄ 15 4 4 𝐿 = [1 ⁄4 0 ] 𝑈 = [0 1 − ⁄5] 3 3 5⁄4 −4 0 0 1 𝐿𝑌 = 𝐵 4 0 0 𝑦 15 1 4 4 [ 1 ⁄4 0 ] [ 2 ] 𝑦1 = 1, 𝑦2 = , 𝑦3 = −1/2 𝑦 = 5 3 5⁄4 −4 𝑦3 6 𝑈𝑋 = 𝑌 1 1⁄4 1⁄4 𝑥1 1 1 [0 1 − 3⁄ ] [𝑥2 ] = [ 4/5 ]  𝑥1 = 1, 𝑥2 = , 𝑥3 = −1/2 5 𝑥3 −1/2 2 0 0 1 iii. 𝑥1 + 5𝑥2 + 𝑥3 = 14 2𝑥1 + 𝑥2 + 3𝑥3 = 13  3𝑥1 + 𝑥2 + 4𝑥3 = 17 (Ans: 𝑥1 = 1, 𝑥2 = 2, 𝑥3 = 3 ) Practice problems: Use LU decomposition method (Crout’s method) to solve the following system of equations i. 2𝑥 + 5𝑦 + 7𝑧 = 52 2𝑥 + 𝑦 − 𝑧 = 0 𝑥+𝑦+𝑧 =9 (Ans: x=1, y=3, z=5) ii. 2𝑥 + 3𝑦 − 𝑧 = 5 4𝑥 + 4𝑦 − 3𝑧 = 3 2𝑥 − 3𝑦 + 2𝑧 = 2 (Ans:x=1, y=2, z=3) Eigen values and eigen vectors: Given a square matrix A, if there exists a scalar  and a non zero column matrix X such that AX=X then  is called eigen value and X is called the eigen vector corresponding of  Here |𝐴 − 𝜆𝐼| = 0 is called characteristic equation of A. The roots of this equation are the eigen values. For each eigen value there exist a corresponding eigen vector 𝑋 ≠ 0. Rayleigh power method to find the dominant(largest) eigen value and the corresponding eigen vector: This is an iterative method to find the largest eigen value and the corresponding eigen vector. In this method, the resultant matrix is normalized(take out numerically largest element)in each step. The process is continued till two consecutive values are approximately same. 1 1 0 0 Note: The eigen vector can be initialized with 𝑜𝑟 𝑜𝑟 𝑜𝑟 1 0 0 1 1. Find the largest eigen value and the corresponding eigen vector of the matrix A, by using Rayleigh power method, by taking the initial vector as [1, 1, 1]𝑇 where 2 −1 0 𝐴 = [−1 2 −1] 0 −1 2 Solution: 𝐴= 2 −1 0 1 1 (0) (1) (1) 𝐴𝑋 = [−1 2 −1] = 1 = 𝝀1 𝑋1 0 −1 2 1 1 2 −1 0 1 1 (2) (2) 𝐴𝑋 (1) = [−1 2 ] −1 0[ ] = 2 [ −1] = 𝝀1 𝑋1 0 −1 2 1 1 2 −1 0 1 0.75 (3) (3) 𝐴𝑋 (2) = [−1 2 −1] [−1] = 4 [ −1 ] = 𝝀1 𝑋1 0 −1 2 1 0.75 2 −1 0 0.75 0.71 (4) (4) 𝐴𝑋 (3) = [−1 2 −1] [ −1 ] = 3.5 [ −1 ] = 𝝀1 𝑋1 0 −1 2 0.75 0.71  2 −1 0 0.71 0.708 (5) (5) 𝐴𝑋 (4) = [−1 ] [ 2 −1 −1 ] = 3.42 [ −1 ] = 𝝀1 𝑋1 0 −1 2 0.71 0.708 2 −1 0 0.708 0.7073 (6) (6) 𝐴𝑋 (5) = [−1 2 −1] [ −1 ] = 3.416 [ −1 ] = 𝝀1 𝑋1 0 −1 2 0.708 0.7073 2 −1 0 0.7073 0.7071 (7) (7) 𝐴𝑋 (6) = [−1 2 −1] [ −1 ] = 3.4146 [ −1 ] = 𝝀1 𝑋1 0 −1 2 0.7073 0.7071 𝟎. 𝟕𝟎𝟕𝟏 Hence the largest eigen value is 𝟑. 𝟒𝟏𝟒𝟔 and the corresponding eigen vector is [ −𝟏 ] 𝟎. 𝟕𝟎𝟕𝟏 Practice problems: Find the largest eigen value and the corresponding eigen vector of the following matrices, by using Rayleigh power method 1 2 0.46 i. 𝐴=[ ] (Ans: 5.38 & [ ]) 3 4 1 4 1 −1 𝟏 ii. 𝐴=[ 2 3 −1] (Ans: 6 & [ 𝟏 ]) −2 1 5 −𝟏 Diagonalisation:If A is a square matrix of order n having n linearly independent eigen vectors then there exists an nth order square matrix P such that 𝑃−1 𝐴𝑃 is a diagonal matrix. If A is asquare matrix of order 2, then it will have 2 eigen values 𝜆1 & 𝜆2 and the corresponding eigen vectors are 𝑋1 & 𝑋2. Then the modal matrix 𝑃 = [𝑋1 𝑋2 ] 𝜆1 𝑛 0 Note: 𝐴𝑛 = 𝑃𝐷 𝑛 𝑃−1 where 𝐷 𝑛 = [ ] 0 𝜆2 𝑛 −1 3 1. Reduce the matrix 𝐴 = [ ] to the diagonal form −2 4 Solution: |−𝟏 − 𝝀 𝟑 |=𝟎 −𝟐 𝟒−𝝀 𝜆2 − 3𝜆 + 2 = 0 𝜆 = 1, 2 are the eigen values i. For 𝝀 = 𝟏 [𝐴 − 𝜆𝐼][𝑋] = 𝑥 𝑦 2𝑥 − 3𝑦 = 0 = 3 2 3 𝑋=[ ] 2 ii. For 𝜆 = 2 𝑥 𝑦 3𝑥 − 3𝑦 = 0 = 1 1  1 𝑋=[ ] 1 3 1 ∴ 𝑃=[ ] 2 1 1 −1 𝑃−1 = [ ] −2 1 1 0 ∴ 𝑃−1 𝐴𝑃 = 𝐷 = [ ] 0 2 −1 2 2. Reduce the matrix 𝐴 = [ ] to the diagonal form hence find 𝑨𝟔 2 −1 Solution: |−1 − 𝜆 2 |=0 2 −1 − 𝜆 𝜆2 + 2𝜆 − 3 = 0 𝜆 = 1, −3 are the eigen values i. For 𝜆 = 1 [𝐴 − 𝜆𝐼][𝑋] = 𝑥 𝑦 𝑥 − 𝑦 = 0 = 1 1 1 𝑋=[ ] 1 ii. For 𝜆 = −3 𝑥 𝑦 𝑥 + 𝑦 = 0 = 1 −1 1 𝑋=[ ] −1 1 1 ∴ 𝑃=[ ] 1 −1 1 1 1 𝑃−1 = 2 [ ] 1 −1 1 0 ∴ 𝑃−1 𝐴𝑃 = 𝐷 = [ ] 0 −3 365 −364 ∴ 𝐴6 = [ ] −364 365 Practice problems: −1 3 i. Reduce the matrix 𝐴 = [ ] to the diagonal form hence find 𝐴4 −2 4 −19 7 ii. Reduce the matrix 𝐴 = [ ] to the diagonal form −42 16 11 −4 −7 iii. Reduce the matrix 𝐴 = [ 7 −2 −5] to the diagonal form hence find 𝐴5 10 −4 −6