1. Find `"Cofactor"(A)` ...
`A=[[1,2],[4,5]]`
Solution:
Cofactor of `1=A_(11) = ` | | `=+(5)` | `=5` |
Cofactor of `2=A_(12) = ` | | `=-(4)` | `=-4` |
Cofactor of `4=A_(21) = ` | | `=-(2)` | `=-2` |
Cofactor of `5=A_(22) = ` | | `=+(1)` | `=1` |
The Cofactor matrix of A is `[A_(ij)]` | = | | `A_(11)` | `A_(12)` | | | `A_(21)` | `A_(22)` | |
| = | |
Method-2 : all Cofactors in matrix form
2. Find `"Cofactor"(A)` ...
`A=[[1,2,3],[4,5,6],[7,8,9]]`
Solution:
`COFACTOR(A)` | = | `COFACTOR` | | `1` | `2` | `3` | | | `4` | `5` | `6` | | | `7` | `8` | `9` | |
|
|
Cofactor of `1=A_(11) = ` | | `=+(5 × 9 - 6 × 8)` | `=+(45 -48)` | `=-3` |
Cofactor of `2=A_(12) = ` | | `=-(4 × 9 - 6 × 7)` | `=-(36 -42)` | `=6` |
Cofactor of `3=A_(13) = ` | | `=+(4 × 8 - 5 × 7)` | `=+(32 -35)` | `=-3` |
Cofactor of `4=A_(21) = ` | | `=-(2 × 9 - 3 × 8)` | `=-(18 -24)` | `=6` |
Cofactor of `5=A_(22) = ` | | `=+(1 × 9 - 3 × 7)` | `=+(9 -21)` | `=-12` |
Cofactor of `6=A_(23) = ` | | `=-(1 × 8 - 2 × 7)` | `=-(8 -14)` | `=6` |
Cofactor of `7=A_(31) = ` | | `=+(2 × 6 - 3 × 5)` | `=+(12 -15)` | `=-3` |
Cofactor of `8=A_(32) = ` | | `=-(1 × 6 - 3 × 4)` | `=-(6 -12)` | `=6` |
Cofactor of `9=A_(33) = ` | | `=+(1 × 5 - 2 × 4)` | `=+(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` | |
|
Method-2 : all Cofactors in matrix form
= | | `+(5 × 9 - 6 × 8)` | `-(4 × 9 - 6 × 7)` | `+(4 × 8 - 5 × 7)` | | | `-(2 × 9 - 3 × 8)` | `+(1 × 9 - 3 × 7)` | `-(1 × 8 - 2 × 7)` | | | `+(2 × 6 - 3 × 5)` | `-(1 × 6 - 3 × 4)` | `+(1 × 5 - 2 × 4)` | |
|
= | | `+(45 -48)` | `-(36 -42)` | `+(32 -35)` | | | `-(18 -24)` | `+(9 -21)` | `-(8 -14)` | | | `+(12 -15)` | `-(6 -12)` | `+(5 -8)` | |
|
= | | `-3` | `6` | `-3` | | | `6` | `-12` | `6` | | | `-3` | `6` | `-3` | |
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This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then