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1 System of Linear Equations and Matrices 1.1 Linear Equation y= mx .1 is an equation, in which variable y is expressed in terms of x and the constant m , is called Linear Equation. In Linear Equation exponents of the variable is always ‘ one’. Equation 1 is also called equation of line. 1.2Linear Equation in n variables: a1x1 a2 x2 a3 x3 .... an xn b .2 where x1 , x2 , x3 ,..., xn are variables and a1 , a2 , a13 ,..., an and b are constants. 1.3 Linear System: A Linear System of m linear equations and n unknowns is: a11x1 a12 x2 a13 x3 .... a1n xn b1 a21x1 a22 x2 a23 x3 .... a2 n xn b2 a31x1 a32 x2 a33 x3 .... a3n xn b3 ....................................................... .3 ....................................................... ........................................................ am1 x1 am 2 x2 am 3 x3 .... amn xn bm where x1 , x2 , x3 ,..., xn are variables or unknowns and a’s and b’s are constants. 2 1.4 Solution: Solution of the linear system (3) is a sequence of n numbers s1, s2 , s3 ,..., sn , which satisfies system (3) when we substitute x1 s1 , x2 s2 , x3 s3 ,..., xn sn . Example.1. Solve the system of equations Solution: x - 3y 3 1 2x y 8 2 -2E1 + E2 2x 6 y 6 2x y 8 ______________ +7y = 14 y=2 From equation 1 x = -3 +3y x = -3 + 6 = 3 Solution is x = 3 and y = 2 Check Substitute the solution in Equations 1 and 2 Equation 1 3 – 3(2) = 3 – 6 = -3 Equation 2 2(3) +2 = 6 + 2 = 8 . Example.2. Solve the system of equations Solution: x - 3y 7 1 2x - 6 y 7 2 2E1 - E2 2x - 6 y - 7 2 x 6 y - 14 _____________________ 0 + 0 = -21 This makes no sense as 0 -21, hence there is no solution. 3 NOTE: Inconsistent , the system of equations is inconsistent, if the system has no solution. Consistent, the system of equations is consistent if the system has at least one solution. Example: Inconsistent and consistent system of equations For the system of linear equations which is represented by straight lines: a1 x - b1 y c1 l1 a2 x - b 2 y c2 l2 There are three possibilities: No solution solutions [inconsistent] one solution [consistent] infinite many [consistent] Note:1. A system will have unique solution (only one solution)when number of unknowns is equal to number of equations Note:2. A system is over determined , if there are more equations then unknowns and it will be mostly inconsistent. Note:3. A system is under determined if there are less equations then unknowns and it may turn inconsistent. 4 1.5 Augmented Matrix System of linear equations: a11x1 a12 x2 a13 x31 b1 a21x1 a22 x2 a23 x31 b2 a31x1 a32 x2 a33 x31 b3 can be written in the form of matrices product a11 a12 a 21 a22 a31 a32 a13 x1 b1 a23 x2 b2 a33 x3 b3 or we may write it in the form AX=b, a11 a12 where A= a21 a22 a31 a32 a13 a23 , X = a33 x1 x 2 x3 a11 a12 Augmented matrix is A : b a21 a22 a31 a32 b1 , b = b2 b3 a13 a23 a33 b1 b2 b3 Example: 4. Write the matrix and augmented form of the system of linear equations 3x – y + 6z = 6 x+y+ z=2 2x + y +4z = 3 Solution: Matrix form of the system is 3 1 2 1 1 1 6 x 6 1 y 2 4 z 3 5 3 1 6 6 Augmented form is A : b 1 1 1 2 . 2 1 4 3 1.5 Elementary Row operations: Elementary row operations are steps for solving the linear system of equations: I. II. III. Interchange two rows Multiply a row with non zero real number Add a multiple of one row to another row Note: Elementary row operations produce same results when operated either on a system or on its augmented matrix form. 1.6 Methods for solving System of Linear equations 1. 2. Gaussian Elimination Method Gauss – Jorden Elimination Method 1.7 Gaussian Elimination Method STEP 1. by using elementary row operations a11 a12 a 21 a22 a31 a32 a13 a23 a33 b1 1 b2 0 b3 0 A12 1 0 A13 B1 A23 B2 1 B3 STEP 2. Find solution by back – substitutions. 6 Example:3. Solve the system of linear equations by Gaussion- elimination method x1 + x2 + 2x3 = 8 - x1 - 2x2 + 3x3 = 1 3x1 - 7 x2 + 4x3 = 10 Solution: Augmented matrix is 1 2 8 1 1 2 3 1 3 7 4 10 STEP 1. 2 8 1 1 0 1 5 9 0 10 2 - 14 2 1 1 0 1 5 0 0 52 R1+R2, -3R1+R3 8 - 9 -R2, 10R2+R3 - 104 8 1 1 2 0 1 5 9 -R3/52 0 0 1 2 Equivalent system of equations form is: x1 + x2 + 2x3 = 8 x2 - 5x3 = -9 x3 = 2 STEP 2. Back Substitution Solution is x1 = 3, x3 = 2 x2 = 5x3 -9 =10 – 9 =1 x1 = - x2 - 2x3 + 8 = -1 – 4 +8 = 3 x2 = 1, x3 = 2. 7 1.8 Gauss – Jorden Elimination Method a11 a12 a 21 a22 a31 a32 a13 a23 a33 b1 1 0 0 B1 b2 0 1 0 B2 b3 0 0 1 B3 Example.4. Solve the system of linear equations by Gauss - Jorden elimination method x1 + x2 + 2x3 = 8 - x1 - 2x2 + 3x3 = 1 3x1 - 7 x2 + 4x3 = 10 Solution: Augmented matrix is 1 2 8 1 1 2 3 1 3 7 4 10 2 8 1 1 0 1 5 9 0 10 2 - 14 2 8 1 1 0 1 5 - 9 0 0 52 - 104 8 1 1 2 0 1 5 9 0 0 1 2 1 1 0 4 0 1 0 1 0 0 1 2 1 0 0 3 0 1 0 1 0 0 1 2 R1+R2, -3R1+R3 -R2, 10R2+R3 -R3/52 -2R3+R1, 5R3+R2 -R2+R1 8 Equivalent system of equations form is: x1 = 3 x2 = 1 x3 = 2 is the solution of the system. 1.9 Row Echelon Form A form of a matrix, which satisfies following conditions, is row echelon form i. ii. iii. iv. ‘1’ (leading entry) must be in the beginning of each row, ‘1’ must be on the right of the above leading entry, Below the leading entry all values must be zero, A row containing all zero values must be in the bottom. Examples: (i) 1 2 3 4 0 1 2 3 0 0 1 2 1 0 (ii) 0 0 2 3 4 1 2 3 0 1 2 0 0 0 0 1 2 3 4 (iii) 0 0 0 1 2 0 0 0 0 1 1.10 Reduced Row Echelon Form A form of a matrix, which satisfies following conditions, is row echelon form i. ii. iii. iv. ‘1’ (leading entry) must be in the beginning of each row, ‘1’ must be on the right of the above leading entry, All entries in the column containing leading entry must be zero, A row containing all zero values must be in the bottom. Examples 1 0 0 3 (i) 0 1 0 2 , 0 0 1 1 1 0 0 (ii) 0 1 0 , (iii) 0 0 1 0 0 0 0 1 2 0 1 0 0 1 3 0 0 0 0 0 0 0 0 9 Example:5. Use Gauss – Jorden method to solve the system of linear system x y 2 z w 1 2 x y 2 z 2 w 2 x 2 y 4z w 1 3x - 3w 3 Solution: Gauss-Jorden method is same as to reduce the augmented matrix to reduced row echelon from. Augmented matrix is 1 1 2 1 1 2 1 2 2 2 1 2 4 1 1 0 3 3 3 0 There is a leading entry ‘1’ in the first row, making all other entries in the first column zero 1 1 2 1 1 0 1 2 0 0 (-2R1+R2)/3 , R1+R3 , -3R1+R4 0 1 2 0 0 0 0 3 6 0 1 0 0 1 1 0 1 2 0 0 R2+R1 , -R2+R3 , -3R2+R4 0 0 0 0 0 0 0 0 0 0 is reduced row echelon form Equivalent matrix form is x w 1 y 2z 0 there are four variables x, y, w and z in the example, variables appearing as leading entries are called LEADING VARIABLES, and other variables are FREE VARIABLE x and y are leading variables and w and z are free variables. Let z = s and w = t , where s and t are real numbers , x = -1 +w = -1 + t y = 2z = 2s z=s w = t, There are infinite many solutions of the given system. 10 SYSTEM WITH NO SOLUTION Example: 6 . Solve the system of linear equations x 2y z 4u 1 x 3 y 7 z 2u 2 x 12 y 11z 16u 5 Solution: Augmented matrix is: 1 4 1 1 2 1 3 7 2 2 1 12 11 16 5 Reducing it to row echelon form (using Gaussian - elimination method) 1 4 1 1 2 0 5 6 6 1 0 10 12 12 4 1 2 1 4 1 0 5 6 6 1 0 0 0 0 3 R2- R1, R3-R1 -R3+2R2 Last equation is 0 x 0 y 0 z 0u 3 but 0 -3 hence there is no solution for the given system of linear equations. 11 Conditions on Solutions Example:7. For which values of ‘a’ will be following system x 2 y 3z 4 3x y 5 z 2 4 x y (a 14) z a 2 infinitely many solutions? No solution? Exactly one solution? 2 (i) (ii) (iii) Solution: Augmented matrix is 3 4 1 2 3 1 5 2 4 1 a 2 14 a 2 Reducing it to reduced row echelon form 3 4 1 2 0 7 14 10 0 7 a 2 2 a 14 R2-3R1, R3-4R1 3 4 1 2 10 17 R2, R3-R2 0 1 2 7 0 0 a 2 16 a 4 writing in the equation form, x 2 y 3z y 2z 4 1 107 2 (a 2 16) z a 4 or equation 3 can be written as (a 4) (a 4) z a 4 3 12 CASE I . a4 0z 0 x 2 y 3z 4 y 2 z 107 as number of equations are less than number of unknowns, hence the system has infinite many solutions, let z=t y 107 2t x 4 3t 4t 207 t 87 where ‘t’ is any real number. CASE II a 4 0z -8 , but 0 -8 , hence, there is no solution. CASE III a 4, a 4, let a 1 Equatins .3. (1 4)(1 4) z 1 4 - 15z -3 z 1 5 y 107 52 64 35 x 4 53 2( 64 ) 47 35 35 the system will have unique solution when a 4 and a -4 and for a=1 the solution is x 47 35 ,y 64 35 and z 15 . NOTE: (i) a=-4, no solution, (ii) a=4, infinite many solutions and (iii)a 4, a -4, exactly one solution . 13 Example:8. What conditions must a, b, and c satisfy in order for the system of equations x y 2z a x z b 2 x y 3z c to be consistent. Solution: The augmented matrix is 1 1 2 a 1 0 1 b reducing it to reduced row echelon form 2 1 3 c 2 a 1 1 0 1 1 b a R2-R1, R3-2R1 0 1 1 c 2a a 1 1 2 0 1 1 b a R3-R1 0 0 0 c a b The system will be consistent if only if c – a - b = 0 or c = a + b Thus the required condition for system to be consistent is c = a + b.