Question

Solve the system using Gaussian elimination. 6x−y+3z=−9 5x + 5y−5z = 20 3x−y+4z = −5 A (2,12,−3)open paren 2 comma 12 comma negative 3 close paren B (2,2,0)open paren 2 comma 2 comma 0 close paren C (−1,6,1)open paren negative 1 comma 6 comma 1 close paren D (−2,6,3)

Answer 1

C. (-1,6,1)

Explanation:

Given the system of equation

6x−y+3z=−9 ...... 1

5x + 5y−5z = 20 ...... 2

3x−y+4z = −5 ...... 3

Reduce the equations to two variables with 2unknown

Multiply equation 3 by 2 and add to 1.

Equation 3×2; 6x−2y+8z = −10 ...... 4

Subtract 4 from 1

-y-(-2y)+3z-8z = -9-(-10)

-y+2y-5z = -9+10

y-5z = 1 ...... 5

Also multiply eqn 2 by 3 and 3 by 5

Equation 2×3; 15x+15y-15z= 60

Equation 3×5; 15x-5y+20z = -25

Subtract

15y+5y-15z-20z = 60+25

20y-35z = 85

Divide through by 5:

4y-7z = 17 ...... 6

...................................

Solve 5 and 6

y-5z = 1 ...... 5 × 4

4y-7z = 17 ...... 6 × 1

....................................

4y-20z = 4

4y-7z = 17

Subtract

-20z+7z = 4-17

-13z = -13

z = -13/-13

z = 1

Substitute z = 1 into equation 5 to get y:

From 5: y-5z = 1

y - 5(1) = 1

y-5 = 1

y = 1+5

y = 6

Substitute y = 6 and z = 1 into equation 3 to get x;

From 3: 3x−y+4z = −5

3x-6+4(1) = -5

3x -6+4 = -5

3x-2 = -5

3x = -5+2

3x = -3

x = -3/3

x = -1

Hence x = -1, y = 6 and z = 1

The correct answer is (-1, 6, 1)