Systems of Linear Equations

Example 1 ~ ~ ~ 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 2 ~ ~ x+y+3z=5 2x+y+4z=7 3x-y+z=8 1 3 5 2 4 7 -1 8 1 3 2 6 -2R1 ~ -1 -2 -3 -3R1 -4R2 -4 -8 -7 1 3 2 6 -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 3 ~ ~ ~ 3x+7y+3z=-1 x+2y+2z= 1 5x+11y+6z= 0 3 7 -1 1 2 5 11 6 1 2 3 7 -1 5 11 6 -3R1 ~ -5R1 1 2 1 2 ~ ~ 1 -3 -4 1 -3 -4 ×-1 -1 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 4 ~ ~ ~ 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 5 ~ ~ 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 6 ~ ~ ~ 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 .