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Find minor [[10,-9,-12],[7,-12,11],[-10,10,3]] [ Calculator, Method and examples ]

Solution:
Your problem `->` minor [[10,-9,-12],[7,-12,11],[-10,10,3]]


`MINOR(A)` = 
`MINOR`
`10``-9``-12`
`7``-12``11`
`-10``10``3`


minor of `10 = A_(11) = `
 `-12`  `11` 
 `10`  `3` 
`=(-12) × 3 - 11 × 10``=-36 -110``=-146`


minor of `-9 = A_(12) = `
 `7`  `11` 
 `-10`  `3` 
`=7 × 3 - 11 × (-10)``=21 +110``=131`


minor of `-12 = A_(13) = `
 `7`  `-12` 
 `-10`  `10` 
`=7 × 10 - (-12) × (-10)``=70 -120``=-50`


minor of `7 = A_(21) = `
 `-9`  `-12` 
 `10`  `3` 
`=(-9) × 3 - (-12) × 10``=-27 +120``=93`


minor of `-12 = A_(22) = `
 `10`  `-12` 
 `-10`  `3` 
`=10 × 3 - (-12) × (-10)``=30 -120``=-90`


minor of `11 = A_(23) = `
 `10`  `-9` 
 `-10`  `10` 
`=10 × 10 - (-9) × (-10)``=100 -90``=10`


minor of `-10 = A_(31) = `
 `-9`  `-12` 
 `-12`  `11` 
`=(-9) × 11 - (-12) × (-12)``=-99 -144``=-243`


minor of `10 = A_(32) = `
 `10`  `-12` 
 `7`  `11` 
`=10 × 11 - (-12) × 7``=110 +84``=194`


minor of `3 = A_(33) = `
 `10`  `-9` 
 `7`  `-12` 
`=10 × (-12) - (-9) × 7``=-120 +63``=-57`


The minor matrix of A is `[A_(ij)]`=
`A_(11)``A_(12)``A_(13)`
`A_(21)``A_(22)``A_(23)`
`A_(31)``A_(32)``A_(33)`
=
`-146``131``-50`
`93``-90``10`
`-243``194``-57`


Method-2 : all minors in matrix form

=
 `-12`  `11` 
 `10`  `3` 
 `7`  `11` 
 `-10`  `3` 
 `7`  `-12` 
 `-10`  `10` 
 `-9`  `-12` 
 `10`  `3` 
 `10`  `-12` 
 `-10`  `3` 
 `10`  `-9` 
 `-10`  `10` 
 `-9`  `-12` 
 `-12`  `11` 
 `10`  `-12` 
 `7`  `11` 
 `10`  `-9` 
 `7`  `-12` 


=
`(-12) × 3 - 11 × 10``7 × 3 - 11 × (-10)``7 × 10 - (-12) × (-10)`
`(-9) × 3 - (-12) × 10``10 × 3 - (-12) × (-10)``10 × 10 - (-9) × (-10)`
`(-9) × 11 - (-12) × (-12)``10 × 11 - (-12) × 7``10 × (-12) - (-9) × 7`


=
`-36 -110``21 +110``70 -120`
`-27 +120``30 -120``100 -90`
`-99 -144``110 +84``-120 +63`


=
`-146``131``-50`
`93``-90``10`
`-243``194``-57`








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