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Home > Matrix & Vector calculators > Solving systems of linear equations using Inverse Matrix method example
1. Inverse Matrix 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)		2. Cramer's Rule 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 Inverse Matrix method

Solution:
Here `x+y+z=3`
`2x-y-z=3`
`x-y+z=9`

Now converting given equations into matrix form
`[[1,1,1],[2,-1,-1],[1,-1,1]] [[x],[y],[z]]=[[3],[3],[9]]`

Now, A = `[[1,1,1],[2,-1,-1],[1,-1,1]]`, X = `[[x],[y],[z]]` and B = `[[3],[3],[9]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|` = 
 `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`


`"Here, " |A| = -6 != 0`

`:. A^(-1) " is possible."`

`Adj(A)` = 
Adj
`1``1``1`
`2``-1``-1`
`1``-1``1`


 = 
 + 
 `-1`  `-1` 
 `-1`  `1` 
 - 
 `2`  `-1` 
 `1`  `1` 
 + 
 `2`  `-1` 
 `1`  `-1` 
 - 
 `1`  `1` 
 `-1`  `1` 
 + 
 `1`  `1` 
 `1`  `1` 
 - 
 `1`  `1` 
 `1`  `-1` 
 + 
 `1`  `1` 
 `-1`  `-1` 
 - 
 `1`  `1` 
 `2`  `-1` 
 + 
 `1`  `1` 
 `2`  `-1` 
T


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


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


 = 
`-2``-3``-1`
`-2``0``2`
`0``3``-3`
T


 = 
`-2``-2``0`
`-3``0``3`
`-1``2``-3`



`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

 = `1/(-6)` ×
`-2``-2``0`
`-3``0``3`
`-1``2``-3`
 ×
`3`
`3`
`9`


 = `-1/6` ×
`-2×3-2×3+0×9`
`-3×3+0×3+3×9`
`-1×3+2×3-3×9`


 = `-1/6` ×
`-12`
`18`
`-24`


 = 
`2`
`-3`
`4`


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

`:.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)		2. Cramer's Rule method
(Next method)


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