`2x-3z+w=1x-y+2w=1-3y+z+w=1x+y+z=1`

Explanation

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

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

`implies A=|(2,0,-3,1),(1,-1,0,2),(0,-3,1,1),(1,1,1,0)|`,`X=[(x),(y),(z)]`,`B=[(1),(1),(1),(1)]`

Now find the determinant of `A`,

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

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

`|A|= |(2,-2,-5,1),(1,-2,-1,2),(0,-3,1,1),(1,0,0,0)|`

Now solve determinant along row1

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