Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.

3x-4y-5z=-27

5x+2y-2z=11

5x-4y+4z=-7

A. (1,5,51)

B. (10,5,51)

C. (10,51,23)

D. (1,5,2)

D. (1,5,2)

Step-by-step explanation:

Lets write the augmented matrix by writing the coefficients of all the variables

3 - 4 - 5 - 27 Row 1

5 2 - 2 11 Row 2

5 - 4 4 - 7 Row 3

WE need to get

1 0 0

0 1 0

0 0 1 then last column will be the value of x, y, z

To get identity matrix we use row operations

(R1=R1/3)

1 - 4/3 - 5/3 - 9 Row 1

5 2 - 2 11 Row 2

5 - 4 4 - 7 Row 3

Subtract row 1 multiplied by 5 from row 2 (R2=R2 - (5) R1)

1 - 4/3 - 5/3 - 9 Row 1

0 26/3 19/3 56 Row 2

5 - 4 4 - 7 Row 3

Subtract row 1 multiplied by 5 from row 3 (R3=R3 - (5) R1)

1 - 4/3 - 5/3 - 9 Row 1

0 26/3 19/3 56 Row 2

0 8/3 37/3 38 Row 3

Multiply row 2 by 326 (R2 = (3/26) R2)

1 - 4/3 - 5/3 - 9 Row 1

0 1 19/26 84/13 Row 2

0 8/3 37/3 38 Row 3

Add row 2 multiplied by 4/3 to row 1 (R1=R1 + (4/3) R2)

1 0 - 9/13 - 5/13 Row 1

0 1 19/26 84/13 Row 2

0 8/3 37/3 38 Row 3

Subtract row 2 multiplied by 8/3 from row 3 (R3=R3 - (8/3) R2)

1 0 - 9/13 - 5/13 Row 1

0 1 19/26 84/13 Row 2

0 0 135/13 270/13 Row 3

Multiply row 3 by 13/135 (R3 = (13/135) R3)

1 0 - 9/13 - 5/13 Row 1

0 1 19/26 84/13 Row 2

0 0 1 2 Row 3

Add row 3 multiplied by 9/13 to row 1 (R1=R1 + (9/13) R3)

1 0 0 1 Row 1

0 1 19/26 84/13 Row 2

0 0 1 2 Row 3

Subtract row 3 multiplied by 19/26 from row 2 (R2=R2 - (19/26) R3)

1 0 0 1 Row 1

0 1 0 5 Row 2

0 0 1 2 Row 3

From the above matrix we can say that x=1, y=5 and z = 2