Question

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)

Answer 1

D.(1,5,2)

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

Answer 2

D. (1, 5, 2)

Explanation:

It is convenient to let a graphing calculator do this. See the attachment.