Solve Equations 2x+3y-z=5,3x+2y+z=10,x-5y+3z=0 using Cramer's Rule method

Solution:
The equations can be expressed as
`2x+3y-z-5=0`

`3x+2y+z-10=0`

`x-5y+3z+0=0`

Use Cramer's Rule to find the values of x, y, z.
`(x)/D_x=(-y)/D_y=(z)/D_z=(-1)/D`

`D_x`

=

`3`   `-1`   `-5`   `2`   `1`   `-10`   `-5`   `3`   `0`

=

`3`   ×    `1`   `-10`   `3`   `0`

`+1`   ×    `2`   `-10`   `-5`   `0`

`-5`   ×    `2`   `1`   `-5`   `3`

`=3 xx (1 × 0 - (-10) × 3) +1 xx (2 × 0 - (-10) × (-5)) -5 xx (2 × 3 - 1 × (-5))`

`=3 xx (0 +30) +1 xx (0 -50) -5 xx (6 +5)`

`=3 xx (30) +1 xx (-50) -5 xx (11)`

`= 90 -50 -55`

`=-15`

`D_y`

=

`2`   `-1`   `-5`   `3`   `1`   `-10`   `1`   `3`   `0`

=

`2`   ×    `1`   `-10`   `3`   `0`

`+1`   ×    `3`   `-10`   `1`   `0`

`-5`   ×    `3`   `1`   `1`   `3`

`=2 xx (1 × 0 - (-10) × 3) +1 xx (3 × 0 - (-10) × 1) -5 xx (3 × 3 - 1 × 1)`

`=2 xx (0 +30) +1 xx (0 +10) -5 xx (9 -1)`

`=2 xx (30) +1 xx (10) -5 xx (8)`

`= 60 +10 -40`

`=30`

`D_z`

=

`2`   `3`   `-5`   `3`   `2`   `-10`   `1`   `-5`   `0`

=

`2`   ×    `2`   `-10`   `-5`   `0`

`-3`   ×    `3`   `-10`   `1`   `0`

`-5`   ×    `3`   `2`   `1`   `-5`

`=2 xx (2 × 0 - (-10) × (-5)) -3 xx (3 × 0 - (-10) × 1) -5 xx (3 × (-5) - 2 × 1)`

`=2 xx (0 -50) -3 xx (0 +10) -5 xx (-15 -2)`

`=2 xx (-50) -3 xx (10) -5 xx (-17)`

`= -100 -30 +85`

`=-45`

`D`

=

`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`


`(x)/D_x=(-y)/D_y=(z)/D_z=(-1)/D`

`:.(x)/-15=(-y)/30=(z)/-45=(-1)/15`

`:.(x)/-15=(-1)/15,(-y)/30=(-1)/15,(z)/-45=(-1)/15`

`:.x=(15)/(15),y=(30)/(15),z=(45)/(15)`

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

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