2. Find `A/B` ...
`A=[[2,3,1],[0,5,6],[1,1,2]]`,`B=[[2,1,-1],[1,0,-1],[1,1,2]]`
Solution:
`|B|` | = | | `2` | `1` | `-1` | | | `1` | `0` | `-1` | | | `1` | `1` | `2` | |
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`=2 xx (0 × 2 - (-1) × 1) -1 xx (1 × 2 - (-1) × 1) -1 xx (1 × 1 - 0 × 1)`
`=2 xx (0 +1) -1 xx (2 +1) -1 xx (1 +0)`
`=2 xx (1) -1 xx (3) -1 xx (1)`
`= 2 -3 -1`
`=-2`
`Adj(B)` | = | Adj | | `2` | `1` | `-1` | | | `1` | `0` | `-1` | | | `1` | `1` | `2` | |
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= | | `+(0 × 2 - (-1) × 1)` | `-(1 × 2 - (-1) × 1)` | `+(1 × 1 - 0 × 1)` | | | `-(1 × 2 - (-1) × 1)` | `+(2 × 2 - (-1) × 1)` | `-(2 × 1 - 1 × 1)` | | | `+(1 × (-1) - (-1) × 0)` | `-(2 × (-1) - (-1) × 1)` | `+(2 × 0 - 1 × 1)` | |
| T |
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= | | `+(0 +1)` | `-(2 +1)` | `+(1 +0)` | | | `-(2 +1)` | `+(4 +1)` | `-(2 -1)` | | | `+(-1 +0)` | `-(-2 +1)` | `+(0 -1)` | |
| T |
|
= | | `1` | `-3` | `1` | | | `-3` | `5` | `-1` | | | `-1` | `1` | `-1` | |
| T |
|
= | | `1` | `-3` | `-1` | | | `-3` | `5` | `1` | | | `1` | `-1` | `-1` | |
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`"Now, "B^(-1)=1/|B| × Adj(B)`
= | `1/(-2)` × | | `1` | `-3` | `-1` | | | `-3` | `5` | `1` | | | `1` | `-1` | `-1` | |
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= | | `-0.5` | `1.5` | `0.5` | | | `1.5` | `-2.5` | `-0.5` | | | `-0.5` | `0.5` | `0.5` | |
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`A×(B^(-1))` | = | | `2` | `3` | `1` | | | `0` | `5` | `6` | | | `1` | `1` | `2` | |
| × | | `-0.5` | `1.5` | `0.5` | | | `1.5` | `-2.5` | `-0.5` | | | `-0.5` | `0.5` | `0.5` | |
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= | | `2×-0.5+3×1.5+1×-0.5` | `2×1.5+3×-2.5+1×0.5` | `2×0.5+3×-0.5+1×0.5` | | | `0×-0.5+5×1.5+6×-0.5` | `0×1.5+5×-2.5+6×0.5` | `0×0.5+5×-0.5+6×0.5` | | | `1×-0.5+1×1.5+2×-0.5` | `1×1.5+1×-2.5+2×0.5` | `1×0.5+1×-0.5+2×0.5` | |
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= | | `-1+4.5-0.5` | `3-7.5+0.5` | `1-1.5+0.5` | | | `0+7.5-3` | `0-12.5+3` | `0-2.5+3` | | | `-0.5+1.5-1` | `1.5-2.5+1` | `0.5-0.5+1` | |
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= | | `3` | `-4` | `0` | | | `4.5` | `-9.5` | `0.5` | | | `0` | `0` | `1` | |
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This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then