Linear algebra. Systems of linear equations

Transcription

1 Linear algebra

2 Outline 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 27

3 Outline 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 27

4 Outline 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 27

5 Outline 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 3 / 27

6 Basic notation Basic notation Definition The system of equations a 11x 1 + a 12x a 1nx n = b 1 a 21x 1 + a 22x a 2nx n = b 2. a m1x 1 + a m2x a mnx n = b m, (1) where numbers a ij, i = 1,..., m, j = 1,..., n are called coefficients, numbers b 1,..., b m are called absolute (or constant) terms and x 1,..., x n are unknowns, is called the system of linear algebraic equations. Lucie Doudová (UoD Brno) Linear algebra 4 / 27

7 Basic notation Basic notation Notation a 11 a a 1n a 21 a a 2n Matrix of the system: A =... a m1 a m2... a mn a 11 a a 1n b 1 a 21 a a 2n b 2 Augmented matrix of the system: R =.... a m1 a m2... a mn b m Vector of unknowns: x = (x 1,..., x n) Vector of absolute terms: b = (b 1,..., b m) Lucie Doudová (UoD Brno) Linear algebra 5 / 27

8 Basic notation Basic notation Remark We can rewrite system (1) in the shorter way: A x = b Definition If b 1 = = b n = 0 then system of linear equations is called a homogenous system. We mark it S 0(m, n). Otherwise, the system of linear equations is called non-homogenous. We mark it S(m, n). Lucie Doudová (UoD Brno) Linear algebra 6 / 27

9 Basic notation Basic notation Example x 1-2x 2 + x 3 = 0 2x 1-3x 2 - x 3 = 0 Homogenous system of 2 linear equations for 3 unknowns x 1, x 2, x 3. Example x 1 + x 2 - x 3 = 1 x 1-2x 2 - x 3 = 0 x 1-3x 2 + 2x 3 = -2 Non-homogenous system of 3 linear equations for 3 unknowns x 1, x 2, x 3. Lucie Doudová (UoD Brno) Linear algebra 7 / 27

10 Solvability of system Basic notation Definition Ordered n-tuple X = (r 1, r 2,..., r n) is called solution of equation (1) if after replacing r 1 with x 1,..., r n with x n in (1) we get m valid identity between numbers. System is solved if we known all its solutions. System which has at least 1 solution is called solvable, in other case it is called insolvable. Lucie Doudová (UoD Brno) Linear algebra 8 / 27

11 Solvability of system Basic notation Example x 1 + 2x 2 = 3 2x 1 x 2 = 1 solvable system: x = (1, 1) Example x 1 + x 2 = 4 x 1 + x 2 = 5 insolvable system Example 2x 1 + x 2 = 3 4x 1 + 2x 2 = 6 solvable system: x = (t, 3 2t), t R Lucie Doudová (UoD Brno) Linear algebra 9 / 27

12 Solvability of system Basic notation Forbenius theorem The system of n linear equations has a solution if and only if r(a) = r(r). If r(a) = r(r) = n then the solution is unique. If r(a) = r(r) < n then the system (1) has infinitely many solutions. Lucie Doudová (UoD Brno) Linear algebra 10 / 27

13 Equivalent modifications Basic notation Definition 1 Two systems S(m 1, n) and S(m 2, n) are called equivalent if they have identical sets of solutions. 2 Modification which transform system S(m, n) to a equivalent system is called a equivalent modification. Theorem (Equivalent modifications) Let S be given system of linear equations (1). Let S be a new system formed from system S by following equivalent modifications. 1 Interchanging of two equations of system S. 2 Multiplying of a equation of system S by number k 0. 3 Adding k-multiple of one equation of system S to another equation of system S. 4 Adding of arbitrary linear combination of equations of system S to another equation of system S. 5 Inserting of equation which is linear combination of other equations of system S. 6 Omitting of equation which is linear combination of other equations of system S. Then both systems S and S have the same set of solution. We say that they are equivalent. Lucie Doudová (UoD Brno) Linear algebra 11 / 27

14 Gaussian elimination Gaussian elimination Gaussian elimination We transform the augmented matrix R of system S(m, n) (by equivalent modifications) to a row echelon form. We write the system of equations which correspond to modified augmented matrix R. We solve this system of equations by back substitution. Lucie Doudová (UoD Brno) Linear algebra 12 / 27

15 Gaussian elimination Gaussian elimination Example Solve the system of equations. x 1 + 2x 2 + x 3 = 1 2x 1 + x 2 x 3 = 4 x 1 x 2 x 3 = x 1 + 2x 2 + x 3 = 1 x 2 + x 3 = 2 x 3 = 4 x 1 = 1 x 2 = 2 x 3 = 4 x = (1, 2, 4) Lucie Doudová (UoD Brno) Linear algebra 13 / 27

16 Gaussian elimination Gaussian elimination Example Different way how to compute this example. x 1 + 2x 2 + x 3 = 1 2x 1 + x 2 x 3 = 4 x 1 x 2 x 3 = 1 x = (1, 2, 4) Lucie Doudová (UoD Brno) Linear algebra 14 / 27

17 Gaussian elimination Gaussian elimination Example x 1 + x 2 + 2x 3 = 1 2x 1 5x 3 = 2 4x 1 + 2x 2 x 3 = from the last row we have equation: = 2 insolvable system Lucie Doudová (UoD Brno) Linear algebra 15 / 27

18 Gaussian elimination Gaussian elimination Example x 1 + x 2 + 2x 3 x 4 = 2 x 2 + x 4 = 3 2x 1 + x 2 + 4x 3 3x 4 = 1 3x 1 + 2x 2 + 6x 3 4x 4 = This system has infinitely many solutions. ( ) Lucie Doudová (UoD Brno) Linear algebra 16 / 27

19 Gaussian elimination Gaussian elimination Example ( ) x 1 + x 2 + 2x 3 x 4 = 2 x 2 + x 4 = 3 x 1 + x 2 = 2 2x 3 + x 4 x 2 = 3 x 4 x 3 = a, x 4 = b; a and b are arbitrary real numbers. x 1 + x 2 = 2 2a + b x 2 = 3 b x 1 = 1 2a + 2b x 2 = 3 b x = ( 1 2a + 2b, 3 b, a, b) Lucie Doudová (UoD Brno) Linear algebra 17 / 27

20 Jordan method Gaussian elimination Example Equivalent method of computation. ( ) ( ( x 3 = a, x 4 = b; a and b are arbitrary real numbers. ) ) x = ( 1 2a + 2b, 3 b, a, b) This method is called the Jordan method. Lucie Doudová (UoD Brno) Linear algebra 18 / 27

21 Cramer s rule Cramer s rule Theorem (Cramer s rule) Let determinant of matrix of system (1) be nonzero. Then system (1) has just one solution. Let D be determinant of matrix of system and D i determinants which we get from D by replacing of i-th column by column of absolute terms. Then x i = D i D, i = 1,..., n Lucie Doudová (UoD Brno) Linear algebra 19 / 27

22 Cramer s rule Cramer s rule Example x y = 4 x + y = 6 ( ) 1 1 A =, D = A = 2 ( ) 4 1 A 1 =, D = A = 10 ( ) 1 4 A 2 =, D = A 2 = 2 x = D1 D = 5, y = D2 D = 1 x = (5, 1) Lucie Doudová (UoD Brno) Linear algebra 20 / 27

23 Cramer s rule Cramer s rule Example x 1 2x 2 + x 3 = 0 3x 1 5x 2 2x 3 = 3 7x 1 3x 2 + x 3 = 16 D = D 2 = = 49, D1 = = 98, D3 = = 147 = 49 x = (3, 2, 1) x 1 = = 3, x2 = = 2, x3 = = 1 Lucie Doudová (UoD Brno) Linear algebra 21 / 27

24 Outline Examples 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 22 / 27

25 Examples Examples Solve systems of linear equations x + 3y + 3z = 1 5x + 3y + 2z = 1 x + 4y + 3z = 2 2x y 2z = 1 x y + z = 2 x 2y + 5z = 4 x 1 + 3x 2 + x 3 x 4 = 2 2x 1 2x 2 + x 4 = 3 2x 1 + 3x 2 + x 3 3x 4 = 6 3x 1 + 4x 2 x 3 + 2x 4 = 0 Lucie Doudová (UoD Brno) Linear algebra 23 / 27

26 Examples Examples Solve systems of linear equations x 1 2x 2 + x 3 + x 4 = 1 x 1 2x 2 + x 3 x 4 = 1 x 1 2x 2 + x 3 + 5x 4 = 5 x + y z = 0 3x + y + z = 0 x y 2z = 0 x + y z = 0 y + 2z 2u = 0 x + z u = 0 Lucie Doudová (UoD Brno) Linear algebra 24 / 27

27 Outline List of tasks for students 1 Basic notation Gaussian elimination Cramer s rule 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 25 / 27

28 List of tasks for students List of tasks for students Example 1: Example 2: Example 3: 2x 4y + 3y = 1 x 2y + 4z = 3 3x y + 5z = 2 3x 1 + 2x 2 x 3 + x 4 = 0 2x 1 x 2 + x 3 + 2x 4 = 2 x 1 2x 2 + 2x 3 x 4 = 16 4x 1 2x 2 2x 3 + 3x 4 = 5 x 1 + 2x 2 + 3x 3 + x 4 2x 5 = 1 3x 1 + x 3 x 4 + 4x 5 = 0 x 2 2x 3 + 3x 4 + x 5 = 2 4x 1 + 3x 2 + 2x 3 + 3x 4 + 3x 5 = 7 x 1 + 2x 2 + 5x 3 + 4x 4 + x 5 = 3 Lucie Doudová (UoD Brno) Linear algebra 26 / 27

29 List of tasks for students List of tasks for students Example 4: Example 5: Example 6: x 1 x 2 + x 3 + x 4 2x 5 = 0 2x 1 + x 2 x 3 x 4 + 2x 5 = 1 3x 1 + 3x 3 3x 3 3x 4 + 6x 5 = 2 4x 1 + 5x 2 5x 3 5x x 5 = 3 2x + y 2z = 0 x + 2y + 2z = 0 3x + y 4z = 0 2x y z = 0 x + y 2z = 0 x 2y + 2z = 0 Lucie Doudová (UoD Brno) Linear algebra 27 / 27