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MAST10007 Linear Algebra
SCHOOL OF MATHEMATICS AND STATISTICS
These notes have been made in accordance with the provisions of Part VB of the copyright act for teaching purposes of the University. These notes are for the use of students of the University of Melbourne enrolled in MAST10007 Linear Algebra. 1 Topic 1: Linear equations [AR 1.1 and 1.2] 1.1. Systems of linear equations. Row operations. 1.2. Reduction of systems to row-echelon form. 1.3. Reduction of systems to reduced row-echelon form. 1.4. Consistent and inconsistent systems. 2 1.1 Systems of linear equations. Row operations Linear equations Definition (Linear equation and linear system) A linear equation in n variables, x1, x2, . . . , xn, is an equation of the form a1x1 + a2x2 + · · ·+ anxn = b, where a1, . . . , an and b are constants. A finite collection of linear equations in the variables x1, x2, . . . , xn is called a system of linear equations or a linear system. Example x1 + 5x2 + 6x3 = 100 x2 − x3 = −1 −x1 + x3 = 11 3 Definition (Homogeneous linear system) A system of linear equations is called homogeneous if all the constants on the right hand side are zero. Example x1 + 5x2 + 6x3 = 0 x2 − x3 = 0 −x1 + x3 = 0 Definition (Solution of a system of linear equations) A solution to a system of linear equations in the variables x1, . . . , xn is a set of values of these variables which satisfy every equation in the system. 4 Example 1. Data fitting using a polynomial Find the coefficients c , α, β, γ such that the cubic polynomial y = c + αx + βx2 + γx3 passes through the points specified below. x y −0.1 0.90483 0 1 0.1 1.10517 0.2 1.2214
Solution: Substituting the points (x , y) into the cubic polynomial gives: 0.90483 = c − 0.1α + 0.01β − 0.001γ 1 = c 1.10517 = c + 0.1α + 0.01β + 0.001γ 1.2214 = c + 0.2α + 0.04β + 0.008γ 5 So c = 1. We can rewrite the system as 3 equations in 3 unknowns. −0.1α+ 0.01β − 0.001γ = −0.09517 0.1α + 0.01β + 0.001γ = 0.10517 0.2α + 0.04β + 0.008γ = 0.2214 Note: To solve these equations by hand is tedious, so we can use a computer package such as MATLAB. 6 Example 2. (Revision) Solve the following linear system. 2x − y = 3 x + y = 0 Solution: Graphically Solution is the intersection of the lines: (x , y) = (1,−1). Note: ◮ Need accurate sketch. ◮ Not practical for three or more variables. 7 Elimination 2x − y = 3 (1) x + y = 0 ⇒ y = −x (2) Substituting (2) in (1) gives 2x + x = 3 ⇒ 3x = 3 ⇒ x = 1 Substituting into (2) gives y = −1. Note: Elimination of variables will always give a solution, but we need to do this systematically and not in an adhoc manner, for three or more variables. 8 Definition (Coefficient matrix and augmented matrix) The coefficient matrix for a linear system is the matrix formed from the coefficients in the equations. The augmented matrix for a linear system is the matrix formed from the coefficients in the equations and the constant terms. These are separated by a vertical line. Example 3. Write down an augmented matrix for the following linear system: 2x − y = 3 x + y = 0 Solution: [ 2 −1 3 1 1 0 ] 9 Row Operations To find the solutions of a linear system, we perform row operations to simplify the augmented matrix. An essential condition is that whichever row operations we perform, we can recover the solutions to the original system from the new matrix. Definition (Elementary row operations) The elementary row operations are: 1. Interchanging two rows. 2. Multiplying a row by a non-zero constant. 3. Adding a multiple of one row to another row. Note: The matrices produced after each row operation are not equal but are equivalent, meaning that the solution is the same for the system represented by each augmented matrix. We use the symbol ∼ to denote equivalence of matrices. 10 Example 4. Solve the following system using elementary row operations: 2x − y = 3 x + y = 0 Solution:[ 2 −1 3 1 1 0 ] R2 ↔ R1 ∼ [ 1 1 0 2 −1 3 ] R2 → R2 − 2R1 ∼ [ 1 1 0 0 −3 3 ] R2 → −13R2 ∼ [ 1 1 0 0 1 −1 ] R1 → R1 − R2 ∼ [ 1 0 1 0 1 −1 ] So x = 1, y = −1. 11 1.2 Reduction of systems to row-echelon form Definition (Leading entry) The leftmost non-zero entry in each row of the matrix is called the leading entry. Definition (Row-echelon form) A matrix is in row-echelon form if: 1. For any row with a leading entry, all elements below that entry and in the same column as it, are zero. 2. For any non-zero two rows, the leading entry of the lower row is further to the right than the leading entry in the higher row. 3. Any row that consists solely of zeros is lower than any row with non-zero entries. 12 Examples[ 1 −2 3 4 5 ] row-echelon form 1 0 0 30 4 1 2 0 0 0 3 row-echelon form 0 0 0 2 4 0 0 3 1 6 0 0 0 0 0 2 −3 6 −4 9 not row-echelon form 13 Gaussian elimination Gaussian elimination is a systematic way to reduce a matrix to row-echelon form using row operations. Note: The row-echelon form obtained is not unique. Gaussian elimination 1. Interchange rows, if necessary, to bring a non-zero number to the top of the first column with a non-zero entry. 2. Add suitable multiples of the top row to lower rows so that all entries below the leading entry are zero. (Multiplying a row by a constant is also allowed and is often useful.) 3. Start again at Step 1 applied to the matrix without the first row. To solve a linear system we read off the equations from the row-echelon matrix and then solve the equations to find the unknowns, starting with the last equation. This final step is called back substitution. 14 Example 5. Solve the following system of linear equations using Gaussian elimination: x − 3y + 2z = 11 2x − 3y − 2z = 13 4x − 2y + 5z = 31 Solution: Step 1. Write the system in augmented matrix form. 1 −3 2 112 −3 −2 13 4 −2 5 31 15 Step 2. Use elementary row operations to reduce the augmented matrix to row-echelon form. 1 −3 2 112 −3 −2 13 4 −2 5 31 R2 → R2 − 2R1 R3 → R3 − 4R1 ∼ 1 −3 2 110 3 −6 −9 0 10 −3 −13 R2 → 13R2 ∼ 1 −3 2 110 1 −2 −3 0 10 −3 −13 R3 → R3 − 10R2 ∼ 1 −3 2 110 1 −2 −3 0 0 17 17 16 Step 3. Read off the equations from the augmented matrix and use back substitution to solve for the unknowns, starting with the last equation. This gives us three equations: 17z = 17 ⇒ z = 1 y − 2z = −3 ⇒ y = 2− 3 = −1 x − 3y + 2z = 11 ⇒ x = −3− 2 + 11 = 6 17 1.3 Reduction of systems to reduced row-echelon form Definition (Reduced row-echelon form) A matrix is in reduced row-echelon form if the following three conditions are satisfied: 1. It is in row-echelon form. 2. Each leading entry is equal to 1 (called a leading 1). 3. In each column containing a leading 1, all other entries are zero. 18 Examples[ 1 −2 3 −4 5 ] reduced row-echelon form [ 1 2 0 0 0 1 ] reduced row-echelon form 1 0 0 30 1 1 2 0 0 0 3 is not in reduced row-echelon form 1 0 0 2 4 0 1 0 1 6 0 0 0 0 0 0 0 1 −4 9 is not in reduced row-echelon form 19 Gauss-Jordan elimination Gauss-Jordan elimination is a systematic way to reduce a matrix to reduced row-echelon form using row operations. Note: The reduced row-echelon form obtained is unique. Gauss-Jordan elimination 1. Use Gaussian elimination to reduce matrix to row-echelon form. 2. Multiply each non-zero row by an appropriate number to create a leading 1 (type 2 row operations). 3. Use row operations (of type 3) to create zeros above the leading entries. 20 Example 6. Solve the following system of linear equations using Gauss-Jordan elimination: x − 3y + 2z = 11 2x − 3y − 2z = 13 4x − 2y + 5z = 31 Solution: Step 1. Use Gaussian elimination to reduce the augmented matrix to row-echelon form. From example 5, we have: 1 −3 2 112 −3 −2 13 4 −2 5 31 ∼ 1 −3 2 110 1 −2 −3 0 0 17 17 21 Step 2. Divide rows by their leading entry. 1 −3 2 110 1 −2 −3 0 0 17 17 R3 → 117R3 ∼ 1 −3 2 110 1 −2 −3 0 0 1 1 Step 3. Add multiples of the last row to the rows above to make the entries above its leading 1 equal to zero. 1 −3 2 110 1 −2 −3 0 0 1 1 R1 → R1 − 2R3R2 → R2 + 2R3 ∼ 1 −3 0 90 1 0 −1 0 0 1 1