Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Inverse Matrix method

Solution:
Here `x+y+z=3`
`2x-y-z=3`
`x-y+z=9`

Now converting given equations into matrix form
`[[1,1,1],[2,-1,-1],[1,-1,1]] [[x],[y],[z]]=[[3],[3],[9]]`

Now, A = `[[1,1,1],[2,-1,-1],[1,-1,1]]`, X = `[[x],[y],[z]]` and B = `[[3],[3],[9]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`1`   `1`   `1`   `2`   `-1`   `-1`   `1`   `-1`   `1`

=

`1`   ×    `-1`   `-1`   `-1`   `1`

`-1`   ×    `2`   `-1`   `1`   `1`

`+1`   ×    `2`   `-1`   `1`   `-1`

`=1 xx (-1 × 1 - (-1) × (-1)) -1 xx (2 × 1 - (-1) × 1) +1 xx (2 × (-1) - (-1) × 1)`

`=1 xx (-1 -1) -1 xx (2 +1) +1 xx (-2 +1)`

`=1 xx (-2) -1 xx (3) +1 xx (-1)`

`= -2 -3 -1`

`=-6`


`"Here, " |A| = -6 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

=

Adj `1` `1` `1` `2` `-1` `-1` `1` `-1` `1`

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

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

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

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

=	`-2` `-2` `0` `-3` `0` `3` `-1` `2` `-3`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(-6)` ×

`-2` `-2` `0` `-3` `0` `3` `-1` `2` `-3`

×

`3` `3` `9`

=

`-1/6` ×

`-2×3-2×3+0×9` `-3×3+0×3+3×9` `-1×3+2×3-3×9`

=

`-1/6` ×

`-12` `18` `-24`

=	`2` `-3` `4`

`:.[[x],[y],[z]]=[[2],[-3],[4]]`

`:.x=2,y=-3,z=4`

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