Using matrices, how do you solve

Question

2x+3y+4z=28

2x+y-z=0

x+y+3z=16

Answer 1

There are two ways of solving them, the Gauss Jordan elimination or the Cramer's rule I adopted the second.

2x+3y+4z=28
2x+y-z=0
x+y+3z=16

The original coefficient matrix to obtain the determinant is:

¦2  3   4¦

¦2  1  -1¦

¦1  1   3 ¦ = 2x1x3 + 3x-1x1 + 4x2x1 - 4x1x1 - 3x2x3 - 2x-1x1 = -9

now we make the x replacement and substitutional matrix to calculate the determinant:

28  3   4

0    1  -1

16  1   3  = 28x1x3 + 3x-1x16 + 4x0x1 - 4x1x16 - 3x0x3 - 28x-1x1= 0

Then x = 0/-9 = 0

for y the matrix will be

2   28    4    

2    0    -1

1   16    3  =  0 - 28 + 128 - 0 - 168 + 32 = -36

Then Y= -36/-9 = 4

Then Z obtained in the same way is -36/-9 = 4

 

 

Answer 2

Make a three by three matrix of the coefficients (I call this A) and the right hand side (b). Then assign x to be the variables x1 x2 x3 in a three by one matrix (b is threE by one as well.)

 

To solve, use excel or handy transpose and special tricks to find the inverse of A. On the left hand side A multiplied by its in verse is the identity which is good because it isolates out the variables.  Thenmultiply b by the inverse of A.