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Home > Matrix & Vector calculators > Solving systems of linear equations using Cramer's Rule method example
2. Cramer's Rule method example ( Enter your problem )
( Enter your problem )
  1. Example `2x+5y=21,x+2y=8`
  2. Example `2x+5y=16,3x+y=11`
  3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
  4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`
 		Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method
3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
(Previous example)		3. Gauss-Jordan Elimination method
(Next method)

4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`


Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Cramer's Rule method

Solution:
The equations can be expressed as
`x+y+z-3=0`

`2x-y-z-3=0`

`x-y+z-9=0`

Use Cramer's Rule to find the values of x, y, z.
`(x)/D_x=(-y)/D_y=(z)/D_z=(-1)/D`

`D_x` = 
 `1`  `1`  `-3` 
 `-1`  `-1`  `-3` 
 `-1`  `1`  `-9` 


 =
 `1` × 
 `-1`  `-3` 
 `1`  `-9` 
 `-1` × 
 `-1`  `-3` 
 `-1`  `-9` 
 `-3` × 
 `-1`  `-1` 
 `-1`  `1` 


`=1 xx (-1 × (-9) - (-3) × 1) -1 xx (-1 × (-9) - (-3) × (-1)) -3 xx (-1 × 1 - (-1) × (-1))`

`=1 xx (9 +3) -1 xx (9 -3) -3 xx (-1 -1)`

`=1 xx (12) -1 xx (6) -3 xx (-2)`

`= 12 -6 +6`

`=12`


`D_y` = 
 `1`  `1`  `-3` 
 `2`  `-1`  `-3` 
 `1`  `1`  `-9` 


 =
 `1` × 
 `-1`  `-3` 
 `1`  `-9` 
 `-1` × 
 `2`  `-3` 
 `1`  `-9` 
 `-3` × 
 `2`  `-1` 
 `1`  `1` 


`=1 xx (-1 × (-9) - (-3) × 1) -1 xx (2 × (-9) - (-3) × 1) -3 xx (2 × 1 - (-1) × 1)`

`=1 xx (9 +3) -1 xx (-18 +3) -3 xx (2 +1)`

`=1 xx (12) -1 xx (-15) -3 xx (3)`

`= 12 +15 -9`

`=18`


`D_z` = 
 `1`  `1`  `-3` 
 `2`  `-1`  `-3` 
 `1`  `-1`  `-9` 


 =
 `1` × 
 `-1`  `-3` 
 `-1`  `-9` 
 `-1` × 
 `2`  `-3` 
 `1`  `-9` 
 `-3` × 
 `2`  `-1` 
 `1`  `-1` 


`=1 xx (-1 × (-9) - (-3) × (-1)) -1 xx (2 × (-9) - (-3) × 1) -3 xx (2 × (-1) - (-1) × 1)`

`=1 xx (9 -3) -1 xx (-18 +3) -3 xx (-2 +1)`

`=1 xx (6) -1 xx (-15) -3 xx (-1)`

`= 6 +15 +3`

`=24`


`D` = 
 `1`  `1`  `1` 
 `2`  `-1`  `-1` 
 `1`  `-1`  `1` 


 =
 `1` × 
 `-1`  `-1` 
 `-1`  `1` 
 `-1` × 
 `2`  `-1` 
 `1`  `1` 
 `+1` × 
 `2`  `-1` 
 `1`  `-1` 


`=1 xx (-1 × 1 - (-1) × (-1)) -1 xx (2 × 1 - (-1) × 1) +1 xx (2 × (-1) - (-1) × 1)`

`=1 xx (-1 -1) -1 xx (2 +1) +1 xx (-2 +1)`

`=1 xx (-2) -1 xx (3) +1 xx (-1)`

`= -2 -3 -1`

`=-6`


`(x)/D_x=(-y)/D_y=(z)/D_z=(-1)/D`

`:.(x)/12=(-y)/18=(z)/24=(-1)/-6`

`:.(x)/12=(-1)/-6,(-y)/18=(-1)/-6,(z)/24=(-1)/-6`

`:.x=(-12)/(-6),y=(18)/(-6),z=(-24)/(-6)`

`:.x=2,y=-3,z=4`



This material is intended as a summary. Use your textbook for detail explanation.
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3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
(Previous example)		3. Gauss-Jordan Elimination method
(Next method)


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