Given the following equations find w x y and z 4w10x2y8 2wx4

Given the following equations, find w, x, y, and z. 4w+10x-2y=8 -2w+x-4y+z=-10 10w-2y+5z=12 3(w+2x)+2y+z-20=0

Solution

The given equations are 4w+10x-2y=8 or, 10x -2y +4w = 8…(1), -2w+x-4y+z=-10 or, x -4y +z-2w = -10…(2), 10w-2y+5z=12 or, -2y +5z+10w = 12…(3) and 3(w+2x)+2y+z-20=0 or, 3w+6x +2y+z = 20 or, 6x+2y+z+3w=20…(4). The given system of linear equations can be represented in the matrix form as AX = b, where b = (8,-10,12,20)^T, X = (x,y,z,w)^T and A =

10

-2

0

4

1

-4

1

-2

0

-2

5

10

6

2

1

3

The augmented matrix of this system is B = [A,b] =

10

-2

0

4

8

1

-4

1

-2

-10

0

-2

5

10

12

6

2

1

3

20

To solve the given system of linear equations, we will reduce B to its RREF as under:

Multiply the 1st row by 1/10; Add -1 times the 1st row to the 2nd row

Add -6 times the 1st row to the 4th row; Multiply the 2nd row by -5/19

Add 2 times the 2nd row to the 3rd row; Add -16/5 times the 2nd row to the 4th row

Multiply the 3rd row by 19/85; Add -35/19 times the 3rd row to the 4th row

Multiply the 4th row by -17/103; Add -214/85 times the 4th row to the 3rd row

Add -12/19 times the 4th row to the 2nd row; Add -2/5 times the 4th row to the 1st row

Add 5/19 times the 3rd row to the 2nd row; Add 1/5 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

0

746/515

0

1

0

0

374/103

0

0

1

0

1784/515

0

0

0

1

20/103

Hence, given system of linear equations is equivalent to x = 746/515, y = 374/103, z = 1784/515 and w = 20/103. The solution is, thus, (x,y,z,w)^T = (746/515, 374/103, 1784/515, 20/103)^T.

10	-2	0	4
1	-4	1	-2
0	-2	5	10
6	2	1	3