`x+y+z+w=1, x-2y+2z+2w=-6 , 2x+y-2z+2w=-5 , 3x-y+3z-3w=-3`

Explanation

We have given:`x+y+z+w=1, x-2y+2z+2w=-6 , 2x+y-2z+2w=-5 , 3x-y+3z-3w=-3`

As we know that the system of linear equation can be written in matrix form as `AX = B`

`implies A=|(1,1,1,1),(1,-2,2,2),(2,1,-2,2),(3,-1,3,-3)|`,`X=[(x),(y),(z)]`,`B=[(1),(-6),(-5),(-3)]`

Now find the determinant of `A`,

`|A|= |(1,1,1,1),(1,-2,2,2),(2,1,-2,2),(3,-1,3,-3)|`

Now apply `C_2->C_2-C_1,C_3->C_3-C_1,C_4->C_4-C_1`

`|A|= |(1,0,0,0),(1,-3,1,1),(2,-1,-4,0),(3,-4,0,-6)|`

Now solve determinant along row1

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