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Home > Algebra calculators > Linear Equation in three variables using Inverse matrix method, Gauss-Jordan Elimination method calculator
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Solution
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Solution provided by AtoZmath.com
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Inverse of matrix using Adjoint method calculator
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 1. `[[2,3,1],[0,5,6],[1,1,2]]`
2. `[[2,1,-1],[1,0,-1],[1,1,2]]`
3. `[[2,3],[4,10]]`
4. `[[5,1],[4,2]]`
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Example1. Find Inverse of matrix
`A=[[3,1,1],[-1,2,1],[1,1,1]]`Solution:| `|A|` | = | | `3` | `1` | `1` | | | `-1` | `2` | `1` | | | `1` | `1` | `1` | |
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`=3 xx (2 × 1 - 1 × 1) -1 xx (-1 × 1 - 1 × 1) +1 xx (-1 × 1 - 2 × 1)` `=3 xx (2 -1) -1 xx (-1 -1) +1 xx (-1 -2)` `=3 xx (1) - -1 xx (-2) +1 xx (-3)` `= 3 +2 -3` `=2` | `Adj(A)` | = | | Adj | | `3` | `1` | `1` | | | `-1` | `2` | `1` | | | `1` | `1` | `1` | |
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| = | | `+(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))` | |
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| = | | `+(2 -1)` | `-(-1 -1)` | `+(-1 -2)` | | | `-(1 -1)` | `+(3 -1)` | `-(3 -1)` | | | `+(1 -2)` | `-(3 +1)` | `+(6 +1)` | |
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| = | | `1` | `2` | `-3` | | | `0` | `2` | `-2` | | | `-1` | `-4` | `7` | |
| T |
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| = | | `1` | `0` | `-1` | | | `2` | `2` | `-4` | | | `-3` | `-2` | `7` | |
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`"Now, "A^(-1)=1/|A| × Adj(A)` | = | `1/(2)` × | | `1` | `0` | `-1` | | | `2` | `2` | `-4` | | | `-3` | `-2` | `7` | |
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| = | | `1/2` | `0` | `-1/2` | | | `1` | `1` | `-2` | | | `-3/2` | `-1` | `7/2` | |
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