solving system of equation by matrix method

Question

5) 4r − 4s + 4t = −4 6) x − 6y + 4z = −12

4r + s − 2t = 5 x + y − 4z = 12

−3r − 3s − 4t = −16 2x + 2y + 5z = −15

3) −x − 5y + z = 17 4) 4x + 4y + z = 24

−5x − 5y + 5z = 5 2x − 4y + z = 0

2x + 5y − 3z = −10 5x − 4y − 5z = 12

Answer 1

1)

4 -4 4 -4

4 1 -2 5

-3 -3 -4 -16

-row1+row2; then row1/4

==========================

1 -1 1 -1

0 5 -6 9

-3 -3 -4 -16

3 * row1 + row2

======================

1 -1 1 -1

0 5 -6 9

0 -6 -1 -19

row2*6; row3*5

================================

1 -1 1 -1

0 30 -36 54

0 -30 -5 -95

row2+row3

==============

1 -1 1 -1

0 30 -36 54

0 0 -41 -41

t=1 per row 3

Per row2, 30s - 36t = 54

30s - 36 = 54

30s = 90

s = 3

Per row1, r-s+t = -1

r-3+1=-1

r-2 = -1

r=1

the solution is {r=1,s=3,t=1}

(2)

1 -6 4 -12

1 1 -4 12

2 2 5 -15

-row1+row2; -2*row1+row2

============================

1 -6 4 -12

0 7 -8 24

0 14 -3 9

-2*row2+row3

============

1 -6 4 -12

0 7 -8 24

0 0 13 -39

row3 says z=-3

row2 says 7y-8z=24

7y -8(-3)=24

7y + 24=24

7y = 0

y=0

row1 says x - 6y + 4z = -12

x - 6(0) + 4(-3) = -12

x -12 = -12

x = 0

the solution set is {x=0,y=0,z=-3}

(3)

row2/-5

====================

-1 -5 1 17

1 1 -1 -1

2 5 -3 -10

row1+row2; 2*row1+row3

==========================

-1 -5 1 17

0 -4 0 16

0 -5 -1 24

per row2, y=-4

per row3, -5y -z = 24

(-5)(-4) - z = 24

20 -z = 24

-z = 4

z = -4

per row1: -x - 5y + z = 17

-x - 5(-4) + -4 = 17

-x + 20 + -4 = 17

-x + 16 = 17

-x = 1

x = -1

the solution set is {x=-1,y=-4,z=-4}

(4)

4 4 1 24

2 -4 1 0

5 -4 -5 12

-row1+row3; -2*row2+row1;

=============================

0 12 -1 24

2 -4 1 0

1 -8 -6 -12

-2*row3+row2

=============================

0 12 -1 24

0 12 13 24

1 -8 -6 -12

-row1+row2

================

0 12 -1 24

0 0 14 0

1 -8 -6 -12

z=0

12y - z = 24

12y =24

y=2

x - 8y - z = -12

x -8(2) - 0 = -12

x - 16 = -12

x = 4

solution {x=4, y=2, z=0}