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Home > Matrix & Vector calculators > Adjoint of a matrix example
7. Adjoint matrix or adjugate matrix example ( Enter your problem )
( Enter your problem )
  1. Definition and Examples
  2. Example-2
 		Other related methods
  1. Addition of two matrix
  2. Multiplication of two matrix
  3. Division of two matrix
  4. Power of a matrix
  5. Transpose of a matrix
  6. Determinant of a matrix
  7. Adjoint of a matrix
  8. Inverse of a matrix
  9. Prove that any two matrix expression is equal or not
  10. Minor of a matrix
  11. Cofactor of a matrix
  12. Trace of a matrix
6. Determinant of a matrix
(Previous method)		2. Example-2
(Next example)

1. Definition and Examples


Adjoint of a matrix
Let `A` be a square matrix then Adjoint of matrix A is transpose of Cofactor matrix of A. It is denoted by adj`A` Adjoint matrix is also called adjugate matrix.

`A=[[1,2,3],[4,5,6],[7,8,9]]`
To find adjoint of a matrix, we have to first find Cofactor matrix of A and then of transpose of this Cofactor matrix A.

Cofactor of `1 = A_11 = +|[5,6],[8,9]| = +(45-48) = -3`
Cofactor of `2 = A_12 = -|[4,6],[7,9]| = -(36-42) = 6`
Cofactor of `3 = A_13 = +|[4,5],[7,8]| = +(32-35) = -3`

Cofactor of `4 = A_21 = -|[2,3],[8,9]| = -(18-24) = 6`
Cofactor of `5 = A_22 = +|[1,3],[7,9]| = +(9-21) = -12`
Cofactor of `6 = A_23 = -|[1,2],[7,8]| = -(8-14) = 6`

Cofactor of `7 = A_31 = +|[2,3],[5,6]| = +(12-15) = -3`
Cofactor of `8 = A_32 = -|[1,3],[4,6]| = -(6-12) = 6`
Cofactor of `9 = A_33 = +|[1,2],[4,5]| = +(5-8) = -3`
The Cofactor matrix of A is `[A_(ij)]=[[A_11,A_12,A_13],[A_21,A_22,A_23],[A_31,A_32,A_33]] =[[-3,6,-3],[6,-12,6],[-3,6,-3]]`

Now find transpose of `A_ij`
Adj`A=[[-3,6,-3],[6,-12,6],[-3,6,-3]]^T` `=[[-3,6,-3],[6,-12,6],[-3,6,-3]]^T`
Examples
1. Find `Adj(A)` ...
`A=[[3,1,1],[-1,2,1],[1,1,1]]`

Solution:
`Adj(A)` = 
Adj
`3``1``1`
`-1``2``1`
`1``1``1`


 = 
 + 
 `2`  `1` 
 `1`  `1` 
 - 
 `-1`  `1` 
 `1`  `1` 
 + 
 `-1`  `2` 
 `1`  `1` 
 - 
 `1`  `1` 
 `1`  `1` 
 + 
 `3`  `1` 
 `1`  `1` 
 - 
 `3`  `1` 
 `1`  `1` 
 + 
 `1`  `1` 
 `2`  `1` 
 - 
 `3`  `1` 
 `-1`  `1` 
 + 
 `3`  `1` 
 `-1`  `2` 
T


 = 
`+(2 × 1 - 1 × 1)``-(-1 × 1 - 1 × 1)``+(-1 × 1 - 2 × 1)`
`-(1 × 1 - 1 × 1)``+(3 × 1 - 1 × 1)``-(3 × 1 - 1 × 1)`
`+(1 × 1 - 1 × 2)``-(3 × 1 - 1 × (-1))``+(3 × 2 - 1 × (-1))`
T


 = 
`+(2 -1)``-(-1 -1)``+(-1 -2)`
`-(1 -1)``+(3 -1)``-(3 -1)`
`+(1 -2)``-(3 +1)``+(6 +1)`
T


 = 
`1``2``-3`
`0``2``-2`
`-1``-4``7`
T


 = 
`1``0``-1`
`2``2``-4`
`-3``-2``7`


2. Find `Adj(A)` ...
`A=[[2,3,1],[0,5,6],[1,1,2]]`

Solution:
`Adj(A)` = 
Adj
`2``3``1`
`0``5``6`
`1``1``2`


 = 
 + 
 `5`  `6` 
 `1`  `2` 
 - 
 `0`  `6` 
 `1`  `2` 
 + 
 `0`  `5` 
 `1`  `1` 
 - 
 `3`  `1` 
 `1`  `2` 
 + 
 `2`  `1` 
 `1`  `2` 
 - 
 `2`  `3` 
 `1`  `1` 
 + 
 `3`  `1` 
 `5`  `6` 
 - 
 `2`  `1` 
 `0`  `6` 
 + 
 `2`  `3` 
 `0`  `5` 
T


 = 
`+(5 × 2 - 6 × 1)``-(0 × 2 - 6 × 1)``+(0 × 1 - 5 × 1)`
`-(3 × 2 - 1 × 1)``+(2 × 2 - 1 × 1)``-(2 × 1 - 3 × 1)`
`+(3 × 6 - 1 × 5)``-(2 × 6 - 1 × 0)``+(2 × 5 - 3 × 0)`
T


 = 
`+(10 -6)``-(0 -6)``+(0 -5)`
`-(6 -1)``+(4 -1)``-(2 -3)`
`+(18 -5)``-(12 +0)``+(10 +0)`
T


 = 
`4``6``-5`
`-5``3``1`
`13``-12``10`
T


 = 
`4``-5``13`
`6``3``-12`
`-5``1``10`




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6. Determinant of a matrix
(Previous method)		2. Example-2
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