Solve the following equations by the invers method :
` x + y + z = - 1, x - y + z = 2, x + y - z = 3`.

Explanation

The given equations can be written in the matrix form as :
`{:((1,1,1),(1,-1,1),(1,1,-1)):}{:((x),(y),(z)):}= {:((-1),(2),(3)):}`
This is of the form AX = B, where
`{:((1,1,1),(1,-1,1),(1,1,-1)):}, X = {:((x),(y),(z)):}and B= {:((-1),(2),(3)):}`
Let us find `A^(-1)`
`|A| = |(1,1,1),(1,-1,1),(1,1,-1)|`
` = 1 (1-1) -1(-1-1)+1(1+1)`
` = 0 + 2 + 2 = 4 ne 0`
` :. A^(-1)` exists.
Consider `A A^(-1) = I`
`:. {:((1,1,1),(1,-1,1),(1,1,-1)):} A^(-1) = {:((1,0,0),(0,1,0),(0,0,1)):}`
By `R_(2) - R_(1) and R_(3) - R_(1) `, we get,
` {:((1,1,1),(0,-2,0),(0,0,-2)):} A^(-1) = {:((1,0,0),(-1,1,0),(-1,0,1)):}`
By `(-1/2) R_(2) and (-1/2) R_(3)`, we get ,
` {:((1,1,1),(0,1,0),(0,0,1)):} A^(-1) = {:((1,0,0),(1//2,-1//2,0),(1//2,0,-1//2)):}`
By `R_(1) - R_(2)`, we get ,
` {:((1,0,1),(0,1,0),(0,0,1)):} A^(-1) = {:((1//2,1//2,0),(1//2,-1//2,0),(1//2,0,-1//2)):}`
By `R_(1) - R_(3)`, we get ,
` {:((1,0,0),(0,1,0),(0,0,1)):} A^(-1) = {:((0,1//2,1//2),(1//2,-1//2,0),(1//2,0,-1//2)):}`
`:. A^(-1)= {:((0,1//2,1//2),(1//2,-1//2,0),(1//2,0,-1//2)):}`
Now , premultiply AX = B by `A^(-1)`, we get,
`A^(-1) (AX) = A^(-1) B`
` :. (A^(-1)A)X = A^(-1)B`
` :. IX = A^(-1)B`
` :. X = {:((0,1//2,1//2),(1//2,-1//2,0),(1//2,0,-1//2)):}{:((-1),(2),(3)):}`
` {:((0+1+3//2),(-1//2-1+0),(-1//2+0-3//2)):}`
`:. {:((x),(y),(z)):}={:((5//2),(-3//2),(-2)):}`
`:. ` by equality of the matrices , `x = 5/2 , y = - 3/2 , z = - 2` is the required solution.