By the method of matrix inversion, solve the system.
`[(1,1,1),(2,5,7),(2,1,-1)][(x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3))]=[(9,2),(52,15),(0,-1)]`

Answer

`x- 1, u =-1, y = 3, v= 2, z=5, w = 1`

Explanation

We have, `[[1,1,1],[2,5,7],[2,1,-7]][[x,u],[y,v],[x,omega]][[9,2],[52,15],[0,-1]]`
or `AX = B`
or ` X = a^(-1) B` …(i)
Where, `A= [[1,1,1],[2,5,7],[2,1,-7]],X=[[x,u],[y,v],[x,omega]]and B=[[9,2],[52,15],[0,-1]]`
`therefore abs(A) = (-5-7)-1(-2-14)+1(2-10)`
` = -12+16-8=-4ne0`
Let C be the matrix of cofactors of elements of `abs(A)`.
`therefore C=[[C_(11),C_(12),C_(13)],[C_(21),C_(22),C_(23)],[C_(31),C_(32),C_(33)]]`
`=[[abs([5,7],[1,-1]),-abs((2,7),(2,-1)),abs((2,5),(2,1))],[-abs((1,1),(1,-1)),abs((1,1),(2,-1)),-abs((1,1),(2,1))],[abs((1,1),(5,7)),-abs((1,1),(2,7)),abs((1,1),(2,5))]]`
`= [[-12,2,2],[16,-3,-5],[-8,1,3]]`
`therefore" adj "A = C'= [[-12,2,2],[16,-3,-5],[-8,1,3]]`
`therefore" "A^(-1)=("adj "A)/abs(A) =-1/4 [[-12,2,2],[16,-3,-5],[-8,1,3]]`
Now, `therefore" "A^(-1)B =-1/4 [[-12,2,2],[16,-3,-5],[-8,1,3]]xx[[9,2],[52,15],[0,-1]]`
`= -1/4 [[-4,4],[-12,-8],[-20,-4]]= [[1,-1],[3,2],[5,1]]`
From Eq. (i) `X=A^(-1) B`
`rArr [[x,u],[y,v],[z,w]]=[[1,-1],[3,2],[5,1]]`
on equating the corresponding elements, we have
`x =1, u = -1`
`y = 3, v=2`
` z = 5, w= 1`