Systems of Linear Equations

Elementary Row Operations There are three kinds of Elementary Row Operations. 1) Interchanging two rows. Multiplying a row by a non-zero constant. Adding a multiple of one row to another row. Fact: When applied to systems of linear equations: Elementary Row Operations do not change the solution set. i.e. Elementary Row Operations produce an equivalent system of linear equations. We will solve systems of linear equations using the method of Gaussian Elimination. This involves starting with the augmented matrix of a system, performing elementary row operations to obtain an equivalent system of linear equations which is easier to solve. As the name suggests Gaussian Elimination involves eliminating more and more variables at each step.

Example 1 Row-Echelon Form 1) The first non zero element in each row is a 1. (Leading one) Any row, below the first row, has more leading zeros (initial zeros) than the row above it, except in the case of two complete rows of zeros. In terms of equations, this means that more and more variables are eliminated from the system as one goes down. These systems can be easily solved by back-substitution. Example 1 This represents the system w+4x+y =5 x+y -2z=7 y+3z=-1 1 4 5 -2 3 -1 w=5-4x-y=5-4(1+5z)-(-1-3z)=2-23z x=0-y+2z=-(-1-3z)+2z=1+5z y=-1-3z z is called a free variable.

Example 2 ~ ~ ~ x+3y+2z=6 2x+5y+4z=11 3x+8y+6z=17 1 3 2 6 5 4 11 8 17 -2R1 ~ -1 -3R1 -R2 -1 1 3 2 6 -1 1 3 2 6 ×-1 ~ ~

An equivalent system is x+3y+2z=6 x = 6-3y-2z = 6-3-2z = 3-2z Back Substitute All solutions are given by x = 3-2z y 1 z is called a free variable. It can take any value and we can obtain all possible solutions by assigning different values to z. Any symbol can be used instead of z. E.g. we can put z=t and write the solution as x y = z 3-2t 1 t This can also be written as

This system has an infinite number of solutions. x y = z 3 1 + -2t t = 3 1 + -2t t = 3 1 + t -2 1 This system has an infinite number of solutions.

Example 3 ~ ~ x+y+3z=5 2x+y+4z=7 3x-y+z=8 1 3 5 2 4 7 -1 8 1 3 5 -2R1 ~ -1 -2 -3 -3R1 -4R2 -4 -8 -7 1 3 5 -1 -2 -3 The last line implies 0x+0y+0z = 5 which is clearly impossible. This system is inconsistent. i.e. It has no solutions. ~ 5

Example 4 ~ ~ ~ ~ 3x+7y+3z=-1 x+2y+2z= 1 5x+11y+6z= 0 3 7 -1 1 2 5 11 1 2 3 7 -1 5 11 6 -3R1 ~ -5R1 1 2 1 2 ~ ~ 1 -3 -4 -R2 1 -3 -4 ×-1 1 -4 -5 -1 1 2 ~ 1 -3 -4 1

An equivalent system is x+2y+2z= 1 Back Substitute y=-4+3z=-4+3=-1 z=1 x y = z 1 -1 This system has an unique solution. This method, called Gaussian Elimination, can be used for systems of any size.

Example 5 ~ ~ ~ w+x+2y-z= 3 2w+x+4y = 7 3w+x+6y+z=11 Performing row operations starting with the augmented matrix. 1 2 -1 3 4 7 6 11 1 2 -1 3 -2R1 ×-1 ~ -1 2 1 -3R1 -2 4 2 1 2 -1 3 -2 1 2 -1 3 ~ 1 -2 -1 ~ +2R2 -2 4 2 w+x+2y-z = 3 x - 2z = -1 w = 3-x-2y+z=3-(-1+2z)-2y+z= 4-2y-z Back Substitute An equivalent system is x = -1+2z

Solutions are w x y z 4-2y-z -1+2z y z 4 -1 -2 1 -1 2 1 = = + y + z

Example 6 ~ ~ w+3x+7y = 11 4w+3x+y-9z = -1 3w - 6y-9z=-11 Performing row operations starting with the augmented matrix. 1 3 7 11 4 -9 -1 -6 -11 1 3 7 11 -4R1 ~ -9 -27 -45 -3R1 -R2 -9 -27 -44 1 2 -1 3 The last line implies 0w+0x+0y+0z = 1 which is clearly impossible. This system is inconsistent. i.e. It has no solutions. ~ -9 -27 -45 1

Example 7 ~ ~ ~ x - y-4z=-4 2x + y+ z= 4 4x + y - z= 4 3x+2y+3z= 8 1 -1 -4 2 4 3 8 -2R1 -4R1 -3R1 1 -1 -4 3 9 12 5 15 20 1 -1 -4 3 4 5 15 20 -5R2 × 1/3 ~ ~ 1 -1 -4 3 4 x-y-4z = -4 y+3z = 4 ~ An equivalent system is

Solutions are (putting z = t): x-y-4z = -4 y+3z = 4 y = 4-3z x = -4+y+4z = -4+(4-3z)+4z = z Back Substitute Solutions are (putting z = t): x y = z t 4-3t = 4 + t -3t = 4 + t 1 -3 .

Example 8 ~ ~ 2w+3x + y =6 2x + 3y + z= 6 2w - x - y = 0 2x - y - z= 0 1 6 -1 -R1 2 3 1 6 -4 -2 -6 -1 +2R2 -R2 2 3 1 6 4 -4 -2 -6 +R3 ~ ~

~ ~ 2w+3x +y = 6 2x+3y+z = 6 2y+z = 3 w = ½{ 6-3x-y}=9/8-(1/8)z 6 4 ×1/2 2 3 1 6 +R3 ~ ~ 2w+3x +y = 6 2x+3y+z = 6 2y+z = 3 An equivalent system is Back Substitute w = ½{ 6-3x-y}=9/8-(1/8)z x = ½{ 6-3y-z}=½{ 6-3×½(3-z)-z}=½{ 6-3×½(3-z)-z}=3/4+z/4 y = ½(3-z)