Solve Equations 2x+3y-z=5,3x+2y+z=10,x-5y+3z=0 using Inverse Matrix method

Solution:
Here `2x+3y-z=5`
`3x+2y+z=10`
`x-5y+3z=0`

Now converting given equations into matrix form
`[[2,3,-1],[3,2,1],[1,-5,3]] [[x],[y],[z]]=[[5],[10],[0]]`

Now, A = `[[2,3,-1],[3,2,1],[1,-5,3]]`, X = `[[x],[y],[z]]` and B = `[[5],[10],[0]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`2`   `3`   `-1`   `3`   `2`   `1`   `1`   `-5`   `3`

=

`2`   ×    `2`   `1`   `-5`   `3`

`-3`   ×    `3`   `1`   `1`   `3`

`-1`   ×    `3`   `2`   `1`   `-5`

`=2 xx (2 × 3 - 1 × (-5)) -3 xx (3 × 3 - 1 × 1) -1 xx (3 × (-5) - 2 × 1)`

`=2 xx (6 +5) -3 xx (9 -1) -1 xx (-15 -2)`

`=2 xx (11) -3 xx (8) -1 xx (-17)`

`= 22 -24 +17`

`=15`


`"Here, " |A| = 15 != 0`

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

`Adj(A)`

=

Adj `2` `3` `-1` `3` `2` `1` `1` `-5` `3`

=	+   `2`   `1`   `-5`   `3`   -   `3`   `1`   `1`   `3`   +   `3`   `2`   `1`   `-5`   -   `3`   `-1`   `-5`   `3`   +   `2`   `-1`   `1`   `3`   -   `2`   `3`   `1`   `-5`   +   `3`   `-1`   `2`   `1`   -   `2`   `-1`   `3`   `1`   +   `2`   `3`   `3`   `2`  T

=	`+(2 × 3 - 1 × (-5))` `-(3 × 3 - 1 × 1)` `+(3 × (-5) - 2 × 1)` `-(3 × 3 - (-1) × (-5))` `+(2 × 3 - (-1) × 1)` `-(2 × (-5) - 3 × 1)` `+(3 × 1 - (-1) × 2)` `-(2 × 1 - (-1) × 3)` `+(2 × 2 - 3 × 3)` T

=	`+(6 +5)` `-(9 -1)` `+(-15 -2)` `-(9 -5)` `+(6 +1)` `-(-10 -3)` `+(3 +2)` `-(2 +3)` `+(4 -9)` T

=	`11` `-8` `-17` `-4` `7` `13` `5` `-5` `-5` T

=	`11` `-4` `5` `-8` `7` `-5` `-17` `13` `-5`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(15)` ×

`11` `-4` `5` `-8` `7` `-5` `-17` `13` `-5`

×

`5` `10` `0`

=

`1/15` ×

`11×5-4×10+5×0` `-8×5+7×10-5×0` `-17×5+13×10-5×0`

=

`1/15` ×

`15` `30` `45`

=	`1` `2` `3`

`:.[[x],[y],[z]]=[[1],[2],[3]]`

`:.x=1,y=2,z=3`

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