|
Question 827186: I Want to know the solution.
solve by Cramer's Rule, equations are
2x-2y+3z=0
x+y+z=0
x+3y+z=0
of 3X3 Matrix
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let A be the 3 by 3 matrix
| a b c |
| d e f |
| g h i |
then the determinant of A is written |A| = D and
D = a(ei-fh) -b(di-fg) +c(dh-eg)
we are given a system of linear equations whose 3 by 3 coefficient matrix is the following and the solution column is three 0's
| 2 -2 3|
| 1 1 1|
| 1 3 1|
let's calculate the determinant of the coefficient matrix
D = 2(1-3) -(-2)(1-1) +3(3-1) = 2
now the determinant for x is calculated from the coefficient matrix with the x column replaced with the solution column
| 0 -2 3|
| 0 1 1|
| 0 3 1|
Dx = 0 -(-2)(0-0) +3(0-0) = 0
now the determinant for y is calculated from the coefficient matrix with the y column replaced with the solution column
| 2 0 3|
| 1 0 1|
| 1 0 1|
Dy = 2(0-0) - 0(1-1) +3(0-0) = 0
now the determinant for z is calculated from the coefficient matrix with the z column replaced with the solution column
| 2 -2 0|
| 1 1 0|
| 1 3 0|
Dz = 2(0-0) -(-2)(0-0) +0(3-1) = 0
now Cramer's rule tells us that x = Dx/D, y = Dy/D, z = Dz/D, therefore
x = 0/2 = 0
y = 0/2 = 0
z = 0/2 = 0
|
|
|
| |
