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Home > Algebra calculators > Inverse Matrix method example
6. Inverse matrix Method example ( Enter your problem )
( Enter your problem )
  1. Examples
 		Other related methods
  1. Substitution method
  2. Elimination method
  3. Cross Multiplication method
  4. Addition-Substraction method
  5. Graphical method
  6. Inverse matrix method
  7. Cramer's Rule method
5. Graphical method
(Previous method)		7. Cramer's Rule method
(Next method)

1. Examples


1. Solve linear equations x+y=2 and 2x+3y=4 using Inverse matrix Method

Solution:
Here `x+y=2`
`2x+3y=4`

Now converting given equations into matrix form
`[[1,1],[2,3]] [[x],[y]]=[[2],[4]]`

Now, A = `[[1,1],[2,3]]`, X = `[[x],[y]]` and B = `[[2],[4]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|` = 
 `1`  `1` 
 `2`  `3` 


`=1 × 3 - 1 × 2`

`=3 -2`

`=1`


`"Here, " |A| = 1 != 0`

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

`Adj(A)` = 
Adj
`1``1`
`2``3`


 = 
`+(3)``-(2)`
`-(1)``+(1)`
T


 = 
`3``-2`
`-1``1`
T


 = 
`3``-1`
`-2``1`



`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

 = `1/(1)` ×
`3``-1`
`-2``1`
 ×
`2`
`4`


 = 
`3×2-1×4`
`-2×2+1×4`


 = 
`2`
`0`


`:.[[x],[y]]=[[2],[0]]`

`:.x=2,y=0`

2. Solve linear equations 2x+7y-11=0 and 3x-y-5=0 using Inverse matrix Method

Solution:
Here `2x+7y-11=0`
`3x-y-5=0`

Now converting given equations into matrix form
`[[2,7],[3,-1]] [[x],[y]]=[[11],[5]]`

Now, A = `[[2,7],[3,-1]]`, X = `[[x],[y]]` and B = `[[11],[5]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|` = 
 `2`  `7` 
 `3`  `-1` 


`=2 × (-1) - 7 × 3`

`=-2 -21`

`=-23`


`"Here, " |A| = -23 != 0`

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

`Adj(A)` = 
Adj
`2``7`
`3``-1`


 = 
`+(-1)``-(3)`
`-(7)``+(2)`
T


 = 
`-1``-3`
`-7``2`
T


 = 
`-1``-7`
`-3``2`



`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

 = `1/(-23)` ×
`-1``-7`
`-3``2`
 ×
`11`
`5`


 = `-1/23` ×
`-1×11-7×5`
`-3×11+2×5`


 = `-1/23` ×
`-46`
`-23`


 = 
`2`
`1`


`:.[[x],[y]]=[[2],[1]]`

`:.x=2,y=1`

3. Solve linear equations 3x-y=3 and 7x+2y=20 using Inverse matrix Method

Solution:
Here `3x-y=3`
`7x+2y=20`

Now converting given equations into matrix form
`[[3,-1],[7,2]] [[x],[y]]=[[3],[20]]`

Now, A = `[[3,-1],[7,2]]`, X = `[[x],[y]]` and B = `[[3],[20]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|` = 
 `3`  `-1` 
 `7`  `2` 


`=3 × 2 - (-1) × 7`

`=6 +7`

`=13`


`"Here, " |A| = 13 != 0`

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

`Adj(A)` = 
Adj
`3``-1`
`7``2`


 = 
`+(2)``-(7)`
`-(-1)``+(3)`
T


 = 
`2``-7`
`1``3`
T


 = 
`2``1`
`-7``3`



`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

 = `1/(13)` ×
`2``1`
`-7``3`
 ×
`3`
`20`


 = `1/13` ×
`2×3+1×20`
`-7×3+3×20`


 = `1/13` ×
`26`
`39`


 = 
`2`
`3`


`:.[[x],[y]]=[[2],[3]]`

`:.x=2,y=3`

4. Solve linear equations 2x-y=11 and 5x+4y=1 using Inverse matrix Method

Solution:
Here `2x-y=11`
`5x+4y=1`

Now converting given equations into matrix form
`[[2,-1],[5,4]] [[x],[y]]=[[11],[1]]`

Now, A = `[[2,-1],[5,4]]`, X = `[[x],[y]]` and B = `[[11],[1]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|` = 
 `2`  `-1` 
 `5`  `4` 


`=2 × 4 - (-1) × 5`

`=8 +5`

`=13`


`"Here, " |A| = 13 != 0`

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

`Adj(A)` = 
Adj
`2``-1`
`5``4`


 = 
`+(4)``-(5)`
`-(-1)``+(2)`
T


 = 
`4``-5`
`1``2`
T


 = 
`4``1`
`-5``2`



`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

 = `1/(13)` ×
`4``1`
`-5``2`
 ×
`11`
`1`


 = `1/13` ×
`4×11+1×1`
`-5×11+2×1`


 = `1/13` ×
`45`
`-53`


 = 
`45/13`
`-53/13`


`:.[[x],[y]]=[[45/13],[-53/13]]`

`:.x=45/13,y=-53/13`



This material is intended as a summary. Use your textbook for detail explanation.
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5. Graphical method
(Previous method)		7. Cramer's Rule method
(Next method)


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