Unit-1 Matrices 2024-2025 PDF

 Parul University Faculty of Engineering & Technology Department of Applied Sciences and Humanities 1st Year B. Tech Programme (All Branches)[2024-25...

 Parul University Faculty of Engineering & Technology Department of Applied Sciences and Humanities 1st Year B. Tech Programme (All Branches)[2024-25] Mathematics– 1(303191101) Unit – 1 MATRICES (Lecture Note) Matrix: A Matrix is a rectangular array of numbers (or functions) enclosed in brackets. These number or functions are called entries or elements of the matrix. For example: 4 −2 1 sin 𝑥 cos 𝑥 [ ],[ ] 0 3 5 − cos 𝑥 sin 𝑥 Trace of a matrix: If 𝐴 is a square matrix, the trace of 𝐴, denoted by 𝑡𝑟(𝐴)and is defined to be the sum of entries on the main diagonal of 𝐴. The trace of 𝐴 is undefined if𝐴 is not a square matrix. For example: 4 5 If 𝐴 = [ ], then 𝑡𝑟(𝐴) = 4 + 6 = 10. 10 6 Symmetric matrix: - For any square matrix A, if 𝐴 = 𝐴𝑇 , then it is known as symmetric matrix. For example: 1 3 4 1 3 4 If 𝐴 = [3 2 7] 𝑡ℎ𝑒𝑛 𝐴𝑇 = [3 2 7] 4 7 4 4 7 4 Here, we can see that so𝐴 = 𝐴𝑇 ; hence 𝐴is symmetric matrix. Skew-symmetric matrix: - For any square matrix A, if 𝐴 = −𝐴𝑇 then it is known as Skew symmetric matrix. For example:  0 −3 4 0 3 −4 0 −3 4 𝐴=[ 3 0 7] ⇒ 𝐴𝑇 = [−3 0 −7] = − [ 3 0 7] = −𝐴𝑇 −4 −7 0 4 7 0 −4 −7 0 Here, we can see that 𝐴 = −𝐴𝑇 ; so 𝐴is skew-symmetric matrix. Singular and non-singular matrix: - For any square matrix A, if|𝐴| ≠ 0, then it is known as non-singular matrix and if |𝐴| = 0 then it is known as singular matrix. 2 1 Example 1: - If 𝐴 = [ ] ⇒ |𝐴| = 8 − 8 = 0 ⇒ Singular Matrix 8 4 1 2 Example 2: - If 𝐴 = [ ] ⇒ |𝐴| = 4 − 6 = −2 ≠ 0 ⇒ Non − Singular Matrix 3 4 Orthogonal Matrix:The matrix is said to be an orthogonal matrix if the product of a matrix and its transpose gives an identity value. i.e.𝐴𝐴𝑇 = 𝐼 Example: Given𝐴 is an orthogonal matrix because −1 0 −1 0 −1 0 −1 0 1 0 𝐴=[ ] Then 𝐴𝑇 = [ ] and 𝐴𝐴𝑇 = [ ][ ]=[ ]=𝐼 0 1 0 1 0 1 0 1 0 1 System of linear equation Linear Equations: Any straight line in the 𝑥𝑦-plane can be represented algebraically by equation of the form 𝑎𝑥 + 𝑏𝑦 = 𝑐, where 𝑎&𝑏 are real numbers. A system of linear equation is a collection of one or more linear equations involving the same variables. A linear system of m linear equations in n variables: An arbitrary system of m linear equations in n variables 𝑥1 , 𝑥2 , 𝑥3 , ….. 𝑥𝑛 is a set of equations of the form 𝑛 ∑ 𝑎𝑖𝑗 𝑥𝑗 = 𝑏𝑖 (𝑖 = 1,2,3, … , 𝑚, 𝑗 = 1,2,3, … , 𝑛) 𝑗=1 A system of linear equations has either 1. No solutions, or 2. Exactly one solution, or 3. Infinitely many solutions  Geometrical representation. Exactly one solution No solution Infinitely many solutions Note (i) The system is said to be consistent if we get infinitely many solutions or unique solution. (ii) The system is said to be inconsistent if we get No solution.  Augmented matrix A system of 𝑚 equations in 𝑛 unknowns can be abbreviated by writing only the rectangular array of numbers. 𝑎11 𝑎12 …𝑎1𝑛 |𝑏1 𝑎21 𝑎22…𝑎2𝑛 |𝑏2 [ ] ⋮ ⋮ ⋮ ⋮ ⋮ 𝑎𝑚1 𝑎𝑚2 …𝑎𝑚𝑛 |𝑏𝑚 This is known as augmented matrix. For example: Find the augmented matrix for each of the following system of linear equations: 2𝑥1 + +2𝑥3 = 1 3𝑥1 − 𝑥2 + 4𝑥3 = 7 6𝑥1 + 𝑥2 − 𝑥3 = 0 2 0 2 |1 Then, augmented matrix is given by[3 −1 4 |7]. 6 1 −1|0 Condition of Consistency for non-homogeneous system: (1) If there is a zero row to left of the augmentation bar but the last entry of this row is non-zero then the system has no solution. 1 0 2|1 For example: [ 0 1 4|7] 𝟎 𝟎 𝟎|𝟒 (2) If at least one of the columns on the left of the augmentation bar has zero element pivot entry, then the system has infinitely many solutions. 1 0 2|1 For example: [ 0 1 4|7] 𝟎 𝟎 𝟎|𝟎 (3) If all the rows having the leading entry 1 then the system has unique solution. 1 0 2|1 For example: [ 0 1 4|7] 𝟎 𝟎 𝟏|𝟒  Row-Echelon (RE)form and Row-Reduced Echelon (RRE) formof a matrix Definition: A rectangular matrix is in row-echelon form (or echelon form) if it has the following three properties: 1. The first element in each row must be non-zero and equals to 1, that is called leading entry 1. 2. All the leading 1’s must be on the right-hand side of the matrix. 3. If any zero row is available, then it must be below to the all-leading 1. If the matrix satisfies the 4th property (i.e., In each column except leading 1 if all entries are zero) then row-echelon form (RE form) becomes reduced row echelon form (RRE form). Example 1:Which of the following matrices are in row-echelon or echelon form? 1 1 0 1 1 1 0 0 0 1 0 0 1 2 0 1 0 0 0 2 0 2 2    (a) 0 1 0 (b) 0 1 0 (c) 0 1 0 (d)   (e) 0 1 1 0       0 0 0 3 3  0 0 0 0  0 0 1 0 0 0 0 2 0     0 0 0 0 4  0 0 0 1  Example 2:Which of the following matrices are in reduced row-echelon or reduced echelon form? 1 0 0 1 0 0 0 1 0  1 0 0 1 0 0 (a) 0 1 0 (b) 0 1 0 (c)   0 0 1 (d)   0 0 1 (e)   0 0 0    0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0  0 0  (f) 1 0 0 (g)     0 0  0 0 0 Methods of solving system of linear equations Gauss-Elimination Gauss-Jordan Method Method Row-Echelon (RE) Reduced Row Form Echelon (REE) Form  Examples: Solve the following system using Gauss-Elimination Method Case-1: Unique solution Case-2: No solution Example 1: Solve the following system by Example 2: Solve the following system of gauss- Elimination method equationby Gauss elimination. 𝑥+𝑦+𝑧 = 6 𝑥 + 2𝑦 + 3𝑧 = 14 −2𝑏 + 3𝑐 = 1 2𝑥 + 4𝑦 + 7𝑧 = 30 3𝑎 + 6𝑏 − 3𝑐 = −2 Solution: The matrix form of the given system is 6𝑎 + 6𝑏 + 3𝑐 = 5 1 1 1 𝑥 6 [1 2 3] [𝑦] = Solution: 2 4 7 𝑧 30 The matrix form of the given system is The augmented matrix is 0 −2 3 𝑥 1 1 1 1| 6 [3 6 −3] [ 𝑦 ] = [ −2] [𝐴|𝐵] = [1 2 3|14] 6 6 3 𝑧 5 2 4 7|30 The augmented matrix is Now, to convert the given augmented matrix in row-echelon form we apply elementary 0 −2 3 | 1 operations as following. [3 6 −3|−2] 6 6 3|5 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1 𝑅1 ↔ 𝑅2 1 1 1| 6 [0 1 2| 8 ] 3 6 −3|−2 0 2 5|18 [0 −2 3 | 1 ] 6 6 3|5 𝑅3 → 𝑅3 − 2𝑅2 𝑅1 → (1⁄3)𝑅1 1 1 1|6 [0 1 2|8] 1 2 −1|−2/3 0 0 1|2 [0 −2 3 | 1 ] 6 6 3| 5 The corresponding system of equation is 𝑅3 → 𝑅3 − 6𝑅1 𝑥+𝑦+𝑧 = 6 𝑦 + 2𝑧 = 8 1 2 −1|−2/3 𝑧=2 [0 −2 3 | 1 ] By using back substitution of 𝑧 = 2 in 𝑦 + 2𝑧 = 0 −6 9 | 9 8, we get 𝑦 = 4 and 𝑧 = 2 & 𝑦 = 4 in 𝑥 + 𝑦 + 𝑅3 → 𝑅3 − 3𝑅2 𝑧 = 6 we get 𝑥 = 0. 𝑥 = 0, 𝑦 = 4, 𝑧 = 2 is unique solution of given 1 2 −1|−2/3 system. [0 −2 3 | 1 ] 0 0 0| 6 The system of linear equation is Case-3: Infinitely many solutions Example 3:Solve the following system by 𝑎 + 2𝑏 − 𝑐 = −2/3 Gauss elimination method. −2𝑏 + 3𝑐 = 1 4𝑥 − 2𝑦 + 6𝑧 = 8 𝑥 + 𝑦 − 3𝑧 = −1 0𝑎 + 0𝑏 + 0𝑐 = 6 is not possible. 15𝑥 − 3𝑦 + 9𝑧 = 21 This shows that the system has no solution. Solution: The matrix form of the given system is _______________________________________ 4 −2 6 𝑥 8 Example 4: [1 𝑦 1 −3] [ ] = [−1] Solve the following system by gauss elimination 15 −3 9 𝑧 21 method. The augmented matrix is −1 3 4 + + = 30 4 −2 6 | 8 𝑥 𝑦 𝑧 [𝐴|𝐵] = [ 1 1 −3|−1] 15 −3 9 | 21 3 2 1 + − =9 𝑥 𝑦 𝑧 𝑅1 ↔ 𝑅2 2 1 2 − + = 10 1 1 −3|−1 𝑥 𝑦 𝑧 [4 −2 6 | 8 ] 15 −3 9 | 21 Solution: 1 1 1 𝑅2 → 𝑅2 − 4𝑅1 , 𝑅3 → 𝑅3 − 15𝑅1 Let 𝑢 = 𝑥 , 𝑣 = 𝑦 , 𝑤 = 𝑧 1 1 −3|−1 Then the system of equations [0 −6 18 | 12 ] 0 −18 54 | 36 −𝑢 + 3𝑣 + 4𝑤 = 30 𝑅2 → (−1/6)𝑅1 , 𝑅3 → (−1/6)𝑅3 3𝑢 + 2𝑣 − 𝑤 = 9 1 1 −3|−1 2𝑢 − 𝑣 + 2𝑤 = 10 [0 1 −3|−2] 0 1 −3|−2 The matrix form of the system is 𝑅3 → 𝑅3 − 𝑅2 −1 3 4 𝑢 30 [3 𝑣 2 −1] [ ] = [ 9 ] 2 −1 2 𝑤 10  1 1 −3|−1 The augmented matrix is [0 1 −3|−2] 0 0 0|0 −1 3 4 |30 [3 2 −1| 9 ] The corresponding system of equations is 2 −1 2 |10 𝑅1 → (−1)𝑅1 𝑥 + 𝑦 − 3𝑧 = −1 1 −3 −4|−30 [3 2 −1| 9 ] 𝑦 − 3𝑧 = −2 2 −1 2 | 10 Assigning the free variable 𝑧 an arbitrary value 𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1 𝑡, 1 −3 −4|−30 𝑦 = 3𝑡 − 2, [0 11 11 | 99 ] 0 5 10 | 70 𝑥 = −1 − 3𝑡 + 2 + 3𝑡 = 1 1 1 Hence, 𝑥 = 1, 𝑦 = 3𝑡 − 2, 𝑧 = 𝑡 is solution of 𝑅2 → ( ) 𝑅2 , 𝑅3 → ( ) 𝑅3 11 5 the given system of equations. 1 −3 −4|−30 Since 𝑡 is arbitrary real number, The system has [0 1 1| 9 ] infinitely many solutions. 0 1 2 | 14 -_---------------------------------------------------- 𝑅3 → 𝑅3 − 2𝑅2 Example 5: Consider the following system 1 −3 −4|−30 [0 1 1| 9 ] 𝑥+𝑦+𝑧 = 6 0 0 1| 5 𝑥 + 2𝑦 + 3𝑧 = 10 The corresponding system of equations is 𝑥 + 2𝑦 + 𝜆𝑧 = 𝜇 𝑢 − 3𝑣 − 4𝑤 = −30 For what values of 𝜆 and𝜇 the system has (i) 𝑣+𝑤 =9 infinitely many solutions (ii) unique solution and 𝑤=5 (iii) no solution. By doing back substitution we get Solution: The Augmented matrix is 1 𝑣+5=9⇒𝑣 =4⇒𝑦 = 1 1 1| 6 4 [1 2 3|10] 1 1 2 𝜆| 𝜇 𝑢 − 12 − 20 = −30 ⇒ 𝑢 = 2 ⇒ 𝑥 = 2 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 𝑅1  1 1 1 1 1 1 | 6 Hence, 𝑥 = 2 , 𝑦 = 4 , 𝑧 = 5 is required unique [0 1 2 | 4 ] solution of the system. 0 1 𝜆 − 1|𝜇 − 6 _______________________________________ 𝑅3 → 𝑅3 − 𝑅2 Exercise: Solve the following system of equations 1 1 1 | 6 by using Gauss elimination method. [0 1 2 | 4 ] 0 0 𝜆 − 3|𝜇 − 10 (1) 𝑥 + 𝑦 + 2𝑧 = 9 Ans: (i) If 𝜆 − 3 = 0 and 𝜇 − 10 = 0, that is if 𝜆 = 3 2𝑥 + 4𝑦 − 3𝑧 = 1 𝑥 = 1, 𝑦 = 2, 𝑧 = 3 and 𝜇 = 10 then the system has infinitely many solutions. 3𝑥 + 6𝑦 − 5𝑧 = 0 (ii)If𝜆 − 3 = 0 then the system has a unique solution. That is 𝜆 ≠ 3 and 𝜇 can possess any (2)3𝑥 + 𝑦 − 3𝑧 = 13 Ans: No solution as the real value. augmented matrix in 2𝑥 − 3𝑦 + 7𝑧 = 5 row-echelon form is (iii)If 𝜆 − 3 = 0 and𝜇 − 10 ≠ 0, that is if 𝜆 = 3 and 𝜇 ≠ 10 then the system does not have any 2𝑥 + 19𝑦 − 47𝑧 = 32 1 1/3 −1 |13/3 solution. [0 1 −27/11| 1 ] 0 0 0 | 5 (3) 2𝑥 + 2𝑦 + 2𝑧 = 0 Ans: Infinitely many solutions.The solution −2𝑥 + 5𝑦 + 2𝑧 = 1 −3𝑘−1 1−4𝑘 set is {( , , 𝑘)/ 7 7 8𝑥 + 𝑦 + 4𝑧 = −1 k R}.  Examples: Solve the following system using Gauss-Jordan Method Case-1: Unique Solution Case-2: Infinitely many Solutions (1) 𝑥 + 𝑦 + 2𝑧 = 8 (2) 𝑥 + 2𝑦 − 3𝑧 = −2 −𝑥 − 2𝑦 + 3𝑧 = 1 3𝑥 − 𝑦 − 2𝑧 = 1 3𝑥 − 7𝑦 + 4𝑧 = 10 2𝑥 + 3𝑦 − 5𝑧 = −3 Solution: The matrix form of the system is Solution:The matrix form of the system is 1 1 2 𝑥 8 1 2 −3 𝑥 −2 [−1 −2 3] [𝑦] = [ 1 ] [3 −1 −2] [𝑦] = [ 1 ] 3 −7 4 𝑧 10 2 3 −5 𝑧 −3 The augmented matrix is The augmented matrix is 1 1 2| 8 1 2 −3|−2 [𝐴|𝐵] = [−1 −2 3| 1 ] [𝐴|𝐵] = [3 −1 −2| 1 ] 3 −7 4|10 2 3 −5|−3 𝑅2 → 𝑅2 + 𝑅1 , 𝑅3 → 𝑅3 − 3𝑅1 𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1 1 1 2| 8 1 2 −3|−2 [0 −1 5| 9 ] [0 −7 7 | 7 ] 0 −10 −2|−14 0 −1 1 | 1 𝑅2 → (−1)𝑅2 𝑅2 → (−1/7)𝑅2 1 1 2| 8 1 2 −3|−2 [0 1 −5| −9 ] [0 1 −1|−1] 0 −10 −2|−14 2 −1 1 | 1 𝑅3 → 𝑅3 + 10𝑅2 𝑅3 → 𝑅3 + 𝑅2 1 1 2 | 8 [0 1 −5 | −9 ] 0 0 −52|−104 1 2 −3|−2 [0 1 −1|−1] 𝑅3 → (−1/52)𝑅3 0 0 0|0 1 1 2|8 𝑅1 → 𝑅1 − 2𝑅2 [0 1 −5|−9] 0 0 1|2  1 0 −1| 0 [0 1 −1|−1] 𝑅2 → 𝑅2 + 5𝑅3 , 𝑅1 → 𝑅1 − 2𝑅3 0 0 0|0 1 1 0|4 [0 1 0|1] 0 0 1|2 The corresponding system of equations gives 𝑅1 → 𝑅1 − 𝑅2 𝑥−𝑧 =0 1 0 0|3 𝑦 − 𝑧 = −1 [0 1 0|1] 0 0 1|2 Assigning the free variable 𝑧 an arbitrary value 𝑡, The corresponding system of equation is 𝑧=𝑡 𝑥 = 3, 𝑦 = 1, 𝑧 = 2 which is a unique solution of 𝑥=𝑧=𝑡 the given system of equations. 𝑦 =𝑧−1=𝑡−1 ___________________________________ Hence, 𝑥 = 𝑡, 𝑦 = 𝑡 − 1, 𝑧 = 𝑡 is solution of the Case-3: No Solution given system of equations and since 𝑡 is arbitrary real number, The system has infinitely many (3)𝑥 + 𝑦 + 𝑧 = 1 solutions. 3𝑥 − 𝑦 − 𝑧 = 4 ____________________________ 𝑥 + 5𝑦 + 5𝑧 = −1 Exercise: Solve the following system of equations by using Gauss- Jordan method. Solution:The matrix form of the system is 1 1 1 𝑥 1 [3 −1 −1] [𝑦] = [ 4 ] (1) 2𝑦 + 3𝑧 = 7 Ans: Unique solution 1 5 5 𝑧 −1 3𝑥 + 6𝑦 − 12𝑧 = −3 𝑥 = −1, 𝑦 = 2, 𝑧 = 1 The augmented matrix is 5𝑥 − 2𝑦 + 2𝑧 = −7 1 1 1|1 [𝐴|𝐵] = [3 −1 −1| 4 ] 1 5 5 |−1 (2) −2𝑦 + 3𝑧 = 1 Ans: No Solution 𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 𝑅1 3𝑥 + 6𝑦 − 3𝑧 = −2 1 1 1|1 [0 −4 −4| 1 ] 6𝑥 + 6𝑦 + 3𝑧 = 5 0 4 4 |−2  (3) 𝑥 − 𝑦 + 𝑧 = 1 Ans: Infinitely many solutions.The solution 𝑅3 → 𝑅3 + 𝑅2 2𝑥 + 𝑦 − 𝑧 = 2 set is {(1, 𝑘, 𝑘)/ k 1 1 1|1 5𝑥 − 2𝑦 + 2𝑧 = 5 R}. [0 −4 −4| 1 ] 0 0 0 |−1 Observe the 3rd row in last matrix gives 0𝑥 + 0𝑦 + 0𝑧 = −1 which is not possible. This shows that the system has no solution. Practice Problem: 1. Solve the system of linear equations using Gauss-Jordan method (Winter 2017) 𝑥 + 2𝑦 + 𝑧 = 5; −𝑥 − 𝑦 + 𝑧 = 2; 𝑦 + 3𝑧 = 1 Solution: Matrix form of the system is 1 2 1 𝑥 5 [−1 −1 1] [𝑦] = 0 1 3 𝑧 1 1 2 1 𝑥 5 Let 𝐴 = [−1 −1 1] 𝑋 = [𝑦] 𝐵 = 0 1 3 𝑧 1 1 2 15 [𝐴|𝐵] = [−1 −1 12] 0 1 31 𝑅2 → 𝑅2 + 𝑅1 1 2 15 ~ [0 1 27] 0 1 31 𝑅1 → 𝑅1 − 2𝑅2 𝑅3 → 𝑅3 − 𝑅2 1 0 −3−9 ~ [0 1 2 7 ] 0 0 1 −6  𝑅1 → 𝑅1 + 3𝑅3 𝑅2 → 𝑅2 − 2𝑅3 1 0 0−27 ~ [0 1 0 19 ] 0 0 1 −6 ∴ The solution of the system is 𝑥 = −27, 𝑦 = 19, 𝑧 = −6 2) Solve the system of linear equations 𝒙 − 𝟐𝒚 + 𝒛 = 𝟏; −𝒙 + 𝒚 − 𝒛 = 𝟎; 𝟐𝒙 − 𝒚 + 𝒛 = −𝟏 using Gauss Elimination method. (Winter -2018) The Matrix form of the system of equation 1 −2 1 𝑥 1 𝑦 [−1 1 −1] [ ] = [ 0 ] 2 −1 1 𝑧 −1 1 −2 1 𝑥 1 Let A= [−1 1 −1] 𝑋 + [𝑦] 𝐵 = [ 0 ] 2 −1 1 𝑧 −1 1 −2 1 1 [𝐴|𝐵] = [−1 1 −1 0 ] 2 −1 1 −1 𝑅2 → 𝑅1 + 𝑅2 𝑅3 → 𝑅3 − 2𝑅1 1 −2 1 1 ~ [0 −1 0 1 ] 0 3 −1−3 𝑅2 → −𝑅2 1 −2 1 1 ~ [0 1 0 −1] 0 3 −1−3 𝑅3 → 𝑅3 − 3𝑅2 1 −2 1 1 ~ [0 1 0 −1] 0 0 −1 0 −𝑧 = 0 => 𝑧 = 0 𝑦 = −1 𝑥 − 2𝑦 + 𝑧 = 1  𝑥 = −1 Hence, the required solution is 𝑥 = −1, 𝑦 = −1 𝑎𝑛𝑑 𝑧 = 0 3) Solve the system of linear equation using Gauss Elimination method 𝑥 + 𝑦 + 𝑧 = 6; 𝑥 + 2𝑦 + 3𝑧 = 14; 2𝑥 + 4𝑦 + 7𝑧 = 30(Winter- 2019) 1 1 16 [𝐴|𝐵] = [1 2 314] 2 4 730 𝑅2 → 𝑅2 − 𝑅1 𝑅3 → 𝑅3 − 2𝑅1 1 1 16 ~ [0 1 2 8 ] 0 2 518 𝑅3 → 𝑅3 − 2𝑅2 1 1 16 ~ [0 1 28] 0 0 12 𝑧=2 𝑦 + 2𝑧 = 8 ∴𝑦=4 𝑥=0 Therefore, required solution is 𝑥 = 0; 𝑦 = 4; 𝑧 = 8 Exercise: (i) Solve the system of linear equations using Gauss Elimination method 𝑥 + 𝑦 + 2𝑧 = 9; 2𝑥 + 4𝑦 − 3𝑧 = 1; 3𝑥 + 6𝑦 − 5𝑧 = 0 (Summer 2022) Answer: 𝑥 = 1; 𝑦 = 2; 𝑧 = 3 (ii) Solve the system of linear equation using Gauss Jordan method: −𝑥 + 3𝑦 + 4𝑧 = 30; 3𝑥 + 2𝑦 − 𝑧 = 9; 2𝑥 − 𝑦 + 2𝑧 = 10 (Summer 2019) Answer: 𝑥 = 2; 𝑦 = 4; 𝑧 = 5 (iii) Solve the following system of equation using Gauss Elimination Method.  𝑥 + 2𝑦 + 𝑧 = 5 ; 3𝑥 – 𝑦 + 𝑧 = 6 ; 𝑥 + 𝑦 + 4𝑧 = 7. ( Summer 2023) Answer: X=2,y=1,z=1 (iv) Solve the following system of equations by using Gauss elimination method 𝑥 + 𝑦 + 𝑧 = 6; 𝑥 + 2𝑦 + 3𝑧 = 14; 2𝑥 + 4𝑦 + 7𝑧 = 30. ( Summer 2023 ) Answer: X=0,y=4,z=2 HOMOGENEOUS EQUATIONS A system of linear equations in terms of 𝑥1, 𝑥2, 𝑥3, … 𝑥𝑛 having the matrix form AX=O, where A is 𝑚 × 𝑛 coefficient matrix, X is 𝑛 × 1 column matrix, O is a 𝑚 × 1 zero column matrix is called a system of homogeneous equations. For example:(i) 𝑥 + 𝑦 + 𝑧 = 0 ; 𝑥 + 2𝑦 − 𝑧 = 0 ; 𝑥 + 3𝑦 + 2𝑧 = 0 (ii) 𝑥 + 𝑦 = 0 ; 𝑥 + 2𝑦 = 0 Homogeneous equations are never inconsistent. They always have the solution “all variables = 0”. The solution (0, 0, …, 0) is often called the trivial solution. Any other solution is called nontrivial solution. Example-1: Solve the following system: Example-2: Solve the following system 4𝑥 + 3𝑦 − 𝑧 = 0 3𝑥 + 4𝑦 + 𝑧 = 0 5𝑥 + 𝑦 − 4𝑧 = 0 Solution:The matrix form of the system is Solution: 4 3 −1 𝑥 0 [3 4 1 ] [𝑦] = 5 1 −4 𝑧 0 The augmented matrix is [𝐴|𝐵] = 4 3 −1|0 [3 4 1 |0] 5 1 −4|0  = = = = = = The last equation does not give any information about the equations. Let. = ___________________________ Exercise:Solve the following system of equations. (1) Ans: Infinitely many solutions. 𝑥+𝑦−𝑧+𝑤 =0 The solution set is The required solution is𝑥 = 0, 𝑦 = 0, 𝑧 = 0 which is 𝑥 − 𝑦 + 2𝑧 − 𝑤 = 0 {(t/4, −7𝑡/4, 𝑡)/ t trivial solution. 3𝑥 + 𝑦 +𝑤 =0 R}. (2) 2𝑥 + 𝑦 + 3𝑧 = 0 Ans: Trivial solution 𝑥 + 2𝑦 =0 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 𝑦+𝑧 =0 Rank of a Matrix The positive integer 𝑟 is said to be a rank of a matrix 𝐴 if it possesses the following properties: (1) There is at least one minor of order 𝑟 which is non-zero. (2) Every minor of order greater than 𝑟 is zero. Notes: 1. Rank of matrix 𝐴 is denoted by 𝜌(𝐴) 2. The rank of matrix remains unchanged by elementary transformation 3. 𝜌(𝐴𝑇 ) = 𝜌(𝐴) 4. The rank of the product of two matrices always less than or equal to the rank of either matrix(i.e., 𝜌(𝐴𝐵) ≤ 𝜌(𝐴) or 𝜌(𝐴𝐵) ≤ 𝜌(𝐵)). Methods for finding Rank of a Matrix ❖ Method-1: Rank of a Matrix by Determinant Matrix Consider a square matrix 𝐴 of order 𝑟. Step-1: Find the determinant of 𝐴. If 𝑑𝑒𝑡(𝐴) ≠ 0 then 𝜌(𝐴) = 𝑟. Otherwise 𝜌(𝐴) < 𝑟. Step-2: Find the all-possible minors of order 𝑟 − 1. If any one of them is non-zero then order is 𝑟 − 1, otherwise 𝜌(𝐴) < 𝑟 − 1. Step-3: By continuing this process upto the non-zero determinant. Example 1: Find the rank the following matrices by determinant method: 𝟐 𝟑 𝟒 (1) 𝑨 = [𝟒 𝟑 𝟏] 𝟏 𝟐 𝟒 2 3 4 Solution: Given, 𝐴 = [4 3 1] then 𝑑𝑒𝑡(𝐴) ≠ 0. Hence, the 𝜌(𝐴) = 3 1 2 4  𝟏 𝟐 𝟑 (2) 𝑨 = [𝟐 𝟑 𝟒] 𝟑 𝟓 𝟕 1 2 3 Solution: Given, 𝐴 = [2 3 4] then 𝑑𝑒𝑡(𝐴) = 0. Hence, the rank of 𝐴 is less than 3. 3 5 7 3 4 Now, minor of 1 = | | = 21 − 20 = 1 ≠ 0. Hence, 𝜌(𝐴) = 2. 5 7 𝟒 𝟐 𝟑 (3) 𝑨 = [ 𝟖 𝟒 𝟔] 𝟑 −𝟐 −𝟏 − 𝟐 4 2 3 Solution: Given, 𝐴 = [ 8 4 6 ] then 𝑑𝑒𝑡(𝐴) = 0. Hence 𝜌(𝐴) < 3. 3 −2 −1 − 2 Consider all the minors of order 2, i.e., 2 3 4 3 4 2 2 3 4 3 4 2 3 3| = 0 | | = 0, | | = 0, | | = 0, | | = 0, | | = 0, | 8 4 4 6 8 6 −2 −1 −1 − −2 − 2 2 Here, all the minors of order 2 are zero. There rank is less than 2. Hence, 𝜌(𝐴) = 1.  1 2 −1 −4  (4) A =  2 4 3 5   −1 −2 6 −7  Solution: Here, the order of matrix 𝐴 is 3 × 4. Hence the rank of 𝐴 is maximum 3 as we can find the square matrix of order 3. Therefore, consider all the minors of order 3, i.e., 1 2 −1 2 −1 −4 1 2 −4 1 −1 −4 |2 4 3 | = 0, | 4 3 5 | = 0, | 2 4 5 | = 0, | 2 3 5 | = −120 −1 −2 6 −2 6 −7 −1 −2 −7 −1 6 −7 Here, one minor of rank 3 is not equal to zero. Hence, 𝜌(𝐴) = 3. ❖ Method-2: Rank of a Matrix by Row Echelon Form The Rank of a Matrix in Row Echelon Form is equal to the number of non-zero rows of the matrix. 1 3 −1 For example: = |0 1 4 | ; the matrix 𝐴 is in Row Echelon form with two non-zero 0 0 0 rows. Hence, rank of matrix 𝐴 is 2. Example 1: Find the rank the following matrices by reducing to Row Echelon Form: 5 3 14 4   1 2 3 −1  −2 −1 −3 −1 (1) A = 0 1 2 1  (2) A =   1 −1 2 0  1 0 1 1    0 1 1 −1 Solution: Given Solution: Given 5 3 14 4  A = 0 1 2 1   1 2 3 −1  −2 −1 −3 −1 1 −1 2 0  A=  1 0 1 1   By applying row-operations  0 1 1 −1 1 −1 2 0  By applying row-operations R13 : 0 1 2 1  5 3 14 4  1 2 3 −1 0 3 3 −3 1 −1 2 0  R2 + 2 R1 , R3 − R1 :   0 −2 −2 2  R3 − 5R1 : 0 1 2 1    0 8 4 4  0 1 1 −1 1 2 3 −1 1 −1 2 0  0 1 1 −1 R3 − 8R2 : 0 1 2 1  R24 :   0 −2 −2 2  0 0 −12 −4    0 3 3 −3   1 −1 2 0  1 2 3 −1  1   0 1 1 −1  −  R3 : 0 1 2 1  R3 + 2 R2 , R4 − 3R2 :    12  0 0 1 1  0 0 0 0       3 0 0 0 0   The equivalent matrix is in Row-Echelon The equivalent matrix is in Row-Echelon Form. Form. Number of non-zero rows = 3.Hence, Number of non-zero rows = 2.Hence, 𝜌(𝐴) = 3 𝜌(𝐴) = 2 Exercise: (1) Find the ranks of 𝐴, 𝐵, 𝐴𝐵 and verify 𝜌(𝐴𝐵) ≤ 𝜌(𝐴) or 𝜌(𝐴𝐵) ≤ 𝜌(𝐵)) where 2 4 1 1 2 3 𝐴 = |3 6 2|, 𝐵 = |3 1 2| 4 8 3 4 3 5 (2) Find the rank the following matrices by reducing to Row Echelon Form: 3 −2 0 −1 −7  1 1 −1 1  0 2 2 1 −5  (I) A = 1 −1 2 −1 , (II) A =  1 −2 −3 −2 1 3 1 0 1    0 1 2 1 6 Practice Problem: Find The Rank of the following Matrices. 1 2 3 1)[4 5 6] (Winter 2017)(Summer 2023) Answer: 2 7 8 9 1 1 −1 1 2) [1 −1 2 −1] (Winter 2021) Answer: 2 3 1 0 1 1 2 3 3) [2 3 4] (Winter 2021) Answer: 2 3 4 5  Important Results (1) If 𝜌(𝐴) ≠ 𝜌(𝐴|𝐵) then the system is inconsistent. (2) If 𝜌(𝐴) = 𝜌(𝐴|𝐵) then the system is consistent. (3) If 𝜌(𝐴) < 𝑛 then there are infinitely many solutions (𝑛 is the number of unknowns) (4) If 𝜌(𝐴) = 𝑛 then there is a unique solution. Example: Find the number of parameters in the general solution of 𝐴𝑋 = 𝑂 if 𝐴 is a 5 × 7 matrix of rank 3. Solution: Here, 𝜌(𝐴) = 3 and 𝑛 = 7. Hence, number of parameters = 𝑛 − 𝜌(𝐴) = 7 − 3 = 4. Eigen values and Eigen vectors Let 𝐴 be 𝑛 × 𝑛 matrix, then there exists a real number 𝜆 and a nonzero vector 𝑋 such that 𝐴𝑋 = 𝜆𝑋 then, 𝜆 is called as the eigen value or characteristic value or proper roots of the matrix 𝐴, and 𝑋 is called as eigen vector or characteristic vector or real vector corresponding to eigen value 𝜆 of the matrix 𝐴. Notes 1. An eigen vector is never the zero vector. 2. The matrix [A –𝜆In] is known as the characteristic matrix of A. 3. The determinant of (A –𝜆In) after expansion gives the polynomial in 𝜆, it is known as the characteristic polynomial of the matrix A of order 𝑛 × 𝑛 and is of degree n. 4. |𝐴 − 𝜆𝐼𝑛 |=0 is called the characteristic equation of matrix A. 5. The root of the characteristic equation is known as characteristic value or eigenvalue of the matrix. 6. The set of all characteristic roots (eigen values) of the matrix A is called the spectrum of A. 7. Let A be 𝑛 × 𝑛 matrix and 𝜆 be an eigen value for A. Then the set𝐸𝜆 = {𝑋/𝐴𝑋 = 𝜆𝑋} is called the eigen space of 𝜆.  Results 1. The eigen values of a diagonal matrix are its diagonal elements. 2. The sum of eigen values of an 𝑛 × 𝑛 matrix is its trace and their product is |𝐴|. 3. For the upper triangular (lower triangular) 𝑛 × 𝑛matrix 𝐴, the eigen values are its diagonal elements. 1 4 Example 1:If𝐴 = | |, find the eigen values for the given matrices: 3 2 (i) 𝐴, (ii) 𝐴𝑇 , (iii) 𝐴−1 , (iv) 4𝐴−1, (v) 𝐴2 , (vi) 𝐴2 − 2𝐴 + 𝐼, (vii) 𝐴3 + 2𝐼 1 4 Solution: Given, 𝐴 = | | 3 2 The characteristic equation of matrix 𝐴 is A −  I2 = 0 1−  4  = 0  (1 −  )( 2 −  ) − 12 = 0   2 − 3 − 10 = 0  (  − 5 )(  + 2 ) = 0 3 2−   = 5 or  = −2 Eigenvalues of 𝐴 = 𝜆 5, −2 Eigenvalues of 𝐴𝑇 = 𝜆𝑇 5, −2 Eigenvalues of 𝐴−1 = 𝜆−1 1 1 ,− 5 2 Eigenvalues of 4𝐴−1 = 4𝜆−1 4 , −2 5 Eigenvalues of 𝐴2 = 𝜆2 25, 4 Eigenvalues of 𝐴2 − 2𝐴 + 𝐼 = 𝜆2 − 2𝜆 + 1 16, 9 Eigenvalues of 𝐴3 + 2𝐼 = 𝜆3 + 2 127, −6 3 2 Example 2:Find the eigen values of 𝐴 = | | 3 8 3 2 Solution: Given 𝐴 = | |, then the characteristic equation of matrix 𝐴 is 3 8  A −  I2 = 0 3− 2  = 0  ( 3 −  )( 8 −  ) − 6 = 0   2 − 11 + 18 = 0  (  − 9 )(  − 2 ) = 0 3 8−  1 = 9 or 2 = 2 Types of Eigen Values Eigenvalues are non- Eigenvalues are Eigenvalues are repeated, whether repeated and the repeated and the matrix is symmetric matrix is non- matrix is symmetric or non-symmetric symmetric −2 −8 −12 Example 3: Find the eigen values and eigen vector of the matrix A= [ 1 4 4 ] 0 0 1 Solution: The characteristic equation is |𝐴 − 𝜆𝐼𝑛 |=0 −2 −  −8 −13 1 4− 4 =0 0 0 1−    3 − S1 2 + S2 − A = 0 S1 = tr ( A) = −2 + 4 + 1 = 3 S2 = Sum of minors of diagonal entries 4 4 −2 −12 1 4 = + + = 4−2+0 = 2 0 1 0 1 0 0 A = −2 ( 4 ) + 8 (1) − 12 ( 0 ) = −8 + 8 = 0 ∴ characteristic equation is  3 − 3 2 + 2 = 0 We suppose z = k , y = 0, x + 4 z = 0   (  2 − 3 + 2 ) = 0   = 0 or  2 − 3 + 2 = 0  z = k , y = 0, x = −4 z = −4k   = 0 or (  − 2 )(  − 1) = 0   = 0 or  = 1or  = 2 Here, one can observe that all eigenvalues are non- repeated and matrix is non-symmetric. When 1 = 0 Therefore, eigen vector space for 1 = 0 ( Continue on next page …) When 3 = 2 Therefore, we suppose x + 4 y + 6 z = 0, − 2 z = 0, y = k  z = 0, y = k , x = −4k Therefore, eigen vector space is Therefore, eigen vector space for 1 = 0 is When 2 = 1 We suppose z = 0, y = k , x + 2 y = 0   z = 0, y = k , x = −2 z = −2k Therefore, eigen vector space for 3 = 2 ( Continue on previous page …) Algebraic multiplicity and Geometric multiplicity Let 𝐴 be 𝑛 × 𝑛 matrix and 𝜆 be an eigen value for 𝐴. If 𝜆 occurs ( k  1) times then 𝑘 is called the Algebraic multiplicity of 𝜆, and the number of basis vectors is called Geometric multiplicity. −2 2 −3 Example: Find eigen values and eigen vectors of the matrix. 𝐴 = [ 2 1 −6]. Also determine −1 −2 0 algebraic and geometric multiplicity. Solution: The characteristic equation is |𝐴 − 𝜆𝐼𝑛 | = 0. Algebraic Multiplicity of 𝜆 = −3 is 2 and of 𝜆 = 5 is 1. We solve the following homogeneous system: Case I: When𝜆1 = 5 Case II: When 𝜆2 = −3, 𝜆3 = −3 = = = = = which is in Row-Echelon from. We suppose x2 = k1 , x3 = k2 , x1 + 2 x2 − 3x3 = 0  x1 = −2k1 + 3k2 = Therefore, eigen space is for𝜆2 = −3, 𝜆3 = −3 is k ( −2,1, 0 ) + k ( 3, 0,1) / k , k 1 2 1 2  R Hence, Geometric multiplicity of 𝜆2 = −3 is 2 and of𝜆 = 5 is 1. =  = which is in Row-Echelon from. x3 = k , x2 + 2 x3 = 0  x2 = −2k , We suppose x1 + x3 = 0  x1 = −k Therefore, eigen space is for 𝜆1 = 5 is k ( −1, −2,1) / k  R 0 1 1 Example: Find eigen values and eigen vectors of the matrix. 𝐴 = [1 0 1]. Also determine algebraic 1 1 0 and geometric multiplicity. Solution: The characteristic equation is|𝐴 − 𝜆𝐼𝑛 | = 0. Algebraic Multiplicity of 𝜆 = −1 is 2 and of 𝜆 = 2 is 1. Case-1:𝜆1 = 2 Case-2: 𝜆2 = −1, 𝜆3 = −1 Let𝑥3 = 𝑘1, 𝑥2 = 𝑘2, ⇒ 𝑥1 + 𝑥2 + 𝑥3 = 0 ⇒ 𝑥1 = −𝑘1 −𝑘2,. Therefore, eigen space is for𝜆2 = −1, 𝜆3 = −1 is k1 ( −1, 0,1) + k2 ( −1,1, 0 ) / k1 , k2  R Hence, Geometric Multiplicity of𝜆2 = −1 is 2 and 𝜆1 = 2 of is 1. Let. Therefore, eigen space is for 𝜆1 = 2 is k (1,1,1) / k  R  1 2 2 Example: Determine algebraic and geometric multiplicity of matrix 𝐴 = [ 0 2 1]. −1 2 2 Answer: 𝜆 = 1,2,2 therefore algebraic multiplicity of 𝜆 = 2 is 2 and geometric multiplicity is 1. For 𝜆 = 1 , A.M. is 1 and G.M. is 1. Practice Problem: −𝟐 −𝟖 −𝟏𝟐 1) Find Eigen value and Eigen vector for following matrix[ 𝟏 𝟒 𝟒 ] (Oct- 𝟎 𝟎 𝟏 2021, May -2023) Solution: |A-λI|=0 −2 − 𝜆 −8 −12 | 1 4−𝜆 4 |=0 0 0 1−𝜆 ∴ (−2 − 𝜆)((4 − 𝜆) × (1 − 𝜆) − 4 × 0) − (−8)(1 × (1 − 𝜆) − 4 × 0) + (−12)(1 × 0 − (4 − 𝜆) × 0) = 0 ∴ (−2 − 𝜆)((4 − 5𝜆 + 𝜆2 ) − 0) + 8((1 − 𝜆) − 0) − 12(0 − 0) = 0 ∴ (−2 − 𝜆)(4 − 5𝜆 + 𝜆2 ) + 8(1 − 𝜆) − 12(0) = 0 ∴ (−8 + 6𝜆 + 3𝜆2 − 𝜆3 ) + (8 − 8𝜆) − 0 = 0 ∴ (−𝜆3 + 3𝜆2 − 2𝜆) = 0 ∴ −𝜆(𝜆 − 1)(𝜆 − 2) = 0 ∴ 𝜆 = 0𝑜𝑟(𝜆 − 1) = 0𝑜𝑟(𝜆 − 2) = 0 ∴ 𝑇ℎ𝑒 𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝐴 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝜆 = 0,1,2 1. Eigenvectors for λ=0 2. Eigenvectors for λ=1 3. Eigenvector for λ=2 −4 −4 −2 𝑣1 = [ 1 ] 𝑣2 = [ 0 ] 𝑣3 = [ 1 ] 0 1 0 2) Find the Eigen value and Eigenvector of the matrix 𝟐 𝟏 𝟎 [−𝟏 𝟎 𝟏] (Nov-2022 &May-2022) 𝟎 𝟎 𝟏 𝟏 𝟐 𝟐 3) Find the Eigen values and Eigenvectors of the matrix𝐴=⌊ 𝟎 𝟐 𝟏⌋ (May - 2023) −𝟏 𝟐 𝟐 4) ( Summer 2023 ) 5) ( summer 2022) Note Theorem: Every square matrix can be decomposed as a sum of symmetric and skew-symmetric matrices. Proof: Let 𝐴 be 𝑚 × 𝑛 matrix. 1 1 Let 𝐵 = 2 (𝐴 + 𝐴𝑇 ) 𝑎𝑛𝑑 𝐶 = = 2 (𝐴 − 𝐴𝑇 ) be two matrices. Obviously, 𝐴 = 𝐵 + 𝐶 1 𝑇 1 1 1 Now,𝐵 𝑇 = [2 (𝐴 + 𝐴𝑇 )] = 2 [(𝐴 + 𝐴𝑇 )]𝑇 = 2 [𝐴𝑇 + (𝐴𝑇 )𝑇 ] = 2 (𝐴𝑇 + 𝐴) = 𝐵 As 𝐵 𝑇 = 𝐵 , 𝐵 is symmetric. 𝑇 1 1 1 1 𝐶 = [ (𝐴 − 𝐴 )] = [(𝐴 − 𝐴𝑇 )]𝑇 = [𝐴𝑇 − (𝐴𝑇 )𝑇 ] = (𝐴𝑇 − 𝐴) = −𝐶 𝑇 𝑇 2 2 2 2 Therefore,𝐶 𝑇 = −𝐶 , 𝐶 is skew- symmetric. Therefore, 𝐴 is a sum of symmetric and skew-symmetric matrices. Caley –Hamilton Theorem Every square matrix satisfies its own characteristic equation i.e. The theorem states that, for a square matrix 𝐴 of order 𝑛, if |𝐴 − 𝜆𝐼𝑛 | = 0.  1 4 Example (i): Verify Caley-Hamilton theorem and hence find the inverse of 𝐴 = [ ] and 𝐴4. 2 3 Solution: The characteristic equation for given matrix is |𝐴 − 𝜆𝐼2 | = 0. Now, by putting 𝜆 = 𝐴, we have 1 4 1 4 1 4 1 0 9 16   4 16 5 0 0 0 A2 − 4 A − 5I =     −4  −5 = − − =  =O 2 3 2 3 2 3 0 1  8 17  8 12  0 5  0 0  Hence, Cayley-Hamilton theorem verified. Now, by using Cayley-Hamilton theorem, we have A2 − 4 A − 5I = 0 , by applying 𝐴−1 on both the sides A−1 ( A2 − 4 A − 5I ) = A−1 ( 0 )  A − 4 I − 5 A−1 = 0  5 A−1 = A − 4 I 1 1  1 4  1 0   1  −3 4   A−1 = ( A − 4I ) =    −4  =   5 5  2 3 0 1   5  2 −1 And for 𝐴4 , applying 𝐴2 both the sides A2 ( A2 − 4 A − 5 I ) = A 2 ( 0 )  A4 − 4 A3 − 5 A2 = 0  A4 = 4 A3 + 5 A2  41 84  9 16   209 416   A4 = 4   +5   A4 =    42 83 8 17   208 417   2 1 1 Example (ii): Find the characteristics equationof the matrix 𝐴 = [0 1 0]and hence prove that 1 1 2. Solution: The characteristics equation is By Caley-Hamilton Theorem..............(1) Now,  1 2 −2 Exercise: (1) Verify Caley-Hamilton theorem and hence find the inverse of 𝐴 = [−1 3 0] 0 −2 1 and 𝐴4. 1 2 3 2 2 3 (2) Compute 𝐴9 − 6𝐴8 + 10𝐴7 − 3𝐴6 + 𝐴 + 𝐼, where 𝐴 = [−1 3 1] (Answer: [−1 4 1] ) 1 0 2 1 0 3 Practice Problem: 1) Using Cayley-Hamilton theorem for the following matrix find 𝑨−𝟏 𝟐 −𝟏 𝟏 ⌊−𝟏 𝟐 −𝟏⌋ (OCT-2021) 𝟏 −𝟏 𝟐 Solution: To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial p(t) of the matrix A. |𝐴 − 𝜆𝐼| = 0 2−𝜆 −1 −1 = | −1 2−𝜆 −1 | 1 −1 2−𝜆 = (2 − 𝜆)((2 − 𝜆) × (2 − 𝜆) − (−1) × (−1)) − (−1)((−1) × (2 − 𝜆) − (−1) × 1) + (−1)((−1) × (−1) − (2 − 𝜆) × 1) = −𝜆3 + 6𝜆2 − 11𝜆 + 6 𝑝(𝑡) = −𝜆3 + 6𝜆2 − 11𝜆 + 6 The Cayley-Hamilton theorem yields that −𝐴3 + 6𝐴2 − 11𝐴 + 6𝐼 = 0 ∴6𝐼 = 𝐴3 − 6𝐴2 + 11𝐴 ∴ 6𝐼 = 𝐴(𝐴2 − 6𝐴 + 11𝐼) 1 ∴ 𝐴−1 = (𝐴2 − 6𝐴 + 11𝐼) 6  𝑁𝑜𝑤, 𝑓𝑖𝑟𝑠𝑡 𝑤𝑒 𝑓𝑖𝑛𝑑 𝐴2 − 6𝐴 + 11𝐼 4 −3 −3 𝐴2 = [−5 6 −3] 5 −5 4 12 −6 −6 6 × 𝐴 = [−6 12 −6] 6 −6 12 11 0 0 11 × 𝐼 = [ 0 11 0 ] 0 0 11 1 Now, 𝐴−1 = 6 (𝐴2 − 6𝐴 + 11𝐼) −1 1 3 3 3 𝐴 = [1 5 3] 6 −1 1 3 −𝟐 𝟏 2) Using Cayley- Hamilton theorem, find 𝑨𝟐 and 𝑨−𝟏 from [ ] (May – 2022) 𝟐 𝟒 𝟒 𝟎 𝟏 3) State Cayley-Hamilton theorem and verify if for the matrix 𝐴=⌊−𝟐 𝟏 𝟎⌋ −𝟐 𝟎 𝟏 (May - 2023) 𝟏 −𝟏 4) Using Cayley-Hamilton theorem find 𝑨−𝟏 for A=[ ] (May - 2023) 𝟐 𝟑 Diagonalization of a matrix: An 𝑛 × 𝑛 matrix 𝐴 is diagonalizable if and only if 𝐴 has 𝑛 linearly independent eigenvectors. OR If 𝑛 × 𝑛 matrix 𝐴 has a basis of eigenvectors, then D = P −1 AP is diagonal, with the eigenvalues of 𝐴 as the entries on the main diagonal. Here,𝑃 is the matrix with these eigenvectors as column vectors. Also, D n = P −1 An P and An = PDn P −1 Example (i): Find a matrix 𝑃 that diagonalizes matrix 𝐴 and determine 𝑃−1 𝐴𝑃 Solution (i): = The characteristic equation is|𝐴 − 𝜆𝐼𝑛 | = 0 For  = 1 For  = 3 = = = = = = = = = = For  = 2 = = = Example (ii): Find a matrix P that diagonalizes A and determine P-1APwhere Also find A10 and find eigenvalues of A2. Solution (ii): The characteristic equation is = |𝐴 − 𝜆𝐼𝑛 | = 0 𝑦 = 𝑘, 6𝑥 = 0 => 𝑥 = 0 ∴ (𝑥, 𝑦) = {𝑘(0,1)/𝑘 ∈ 𝑅} Now, For  = 1 = Suppose 𝑥 = 𝑘, 6𝑥 − 2𝑦 = 0  𝑥 = 𝑘, 𝑦 = 3𝑘 ∴ (𝑥, 𝑦) = {𝑘(1,3)/𝑘 ∈ 𝑅} For  = −1 ( Continue on previous page …) Eigenvalues of A2 are: 12 = 1 and(−1)2 = 1. Quadratic Forms A homogeneous polynomial of second degree in real variables𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 is called Quadratic form. For example: (i) 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 is a quadratic form in the variables x and y (ii) 2𝑥1 𝑥2 + 2𝑥2 𝑥3 + 2𝑥3 𝑥1 + 𝑥3 2 is a quadratic form in the variables 𝑥1 , 𝑥2 , 𝑥3. A quadratic on 𝑅 𝑛 is a function 𝑄 define on 𝑅 𝑛 whose value at a vector 𝑥in 𝑅 𝑛 can be computed in n variables 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 by an expression of the form. Here, 𝐴 is known as the coefficient matrix. Where 𝐴 is an 𝑛 × 𝑛 symmetric matrix and is called matrix of the quadratic form. Matrix Representation of Quadratic Forms A quadratic form can be represented as a matrix product. For example: (I) (II) Example: (i) Find a real symmetric matrix C of the quadratic form. Solution: The coefficient matrix of 𝑄 is So, C = symmetric matrix = (ii) Express the following quadratic forms in matrix notation Solution: Transformation (Reduction) of Quadratic form to canonical form OR Diagonalizing Quadratic Forms: Procedure to Reduce Quadratic form to canonical form: 1. Identity the real symmetric matrix associated with the quadratic form. 2. Determine the eigenvalues of 𝐴. 3. The required canonical form is given by ………….. (1) where𝜆1 , 𝜆2 , ….. 𝜆𝑛 are the eigenvalues of 𝐴 and𝐷 = 𝑃𝑇 𝐴𝑃. The matrix 𝑃 is said to orthogonally diagonalize the quadratic form. and equation (1) is known as canonical form.  4. Form modal matrix 𝑃 (where 𝑥 = 𝑃𝑦) containing the 𝑛 eigenvectors of 𝐴 as 𝑛 column vectors. Example: Reduce the quadratic form into canonical form Solution: 4 Eigenvalues for𝐴 are3, − 3 , −1. The canonical form of the given quadratic form is Nature of quadratic form 𝑸 a. Positive definite if Q(x) > 0 for all x 0, b. Negative definite if Q(x) < 0 for all x 0, c. Indefinite if Q(x) assumes both positive and negative values. d. Positive semidefinite if Q(x) 0 for all x. e. Negative semidefinite if Q(x) 0 for all x. OR a. Positive definite if and only if the eigenvalues of A are positive, b. negative definite if and only if the eigenvalues of A are positive, c. Indefinite if and only if A has both positive and negative eigenvalues. d. Positive semi-definite if and only if A has only non-negative eigenvalues. e. Indefinite if and only if A has only non-positive eigenvalues. Example: Describe the nature of quadratic forms. 1. 2. (Summer 2023 ) Obtain the real symmetric matrix for Quadratic form 𝑄 = x2 + 6𝑥𝑦 + 2𝑦2 (Summer 2023 ) Express the quadratic form 𝑄(𝑥, 𝑦) = 2𝑥2 + 4𝑦2 + 2𝑥𝑦 in the matrix form.