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If `S=[(0,1,1),(1,0,1),(1,1,0)]` and `A=[(b+c,c-a,b-a),(c-b,c+b,a-b),(b-c,a-c,a+b)]` `(a, b, c ne 0)`, then `SAS^(-1)` isA. symmetric matrixB. diagonal matrixC. invertible matrixD. singular matrix

Answer» `S[(0,1,1),(1,0,1),(1,1,0)]` or `S^(-1)=1/2 [(-1,1,1),(1,-1,1),(1,1,-1)]`
We have,
`SA=[(0,1,1),(1,0,1),(1,1,0)][(b+c,c-a,b-a),(c-b,c+a,a-b),(b-c,a-c,a+b)]`
`=[(0,2a,2a),(2b,0,2b),(2c,2c,0)]`
`:. SAS^(-1)=[(0,2a,2a),(2b,0,2b),(2c,2c,0)] 1/2 [(-1,1,1),(1,-1,1),(1,1,-1)]`
`=[(0,a,a),(b,0,b),(c,c,0)][(-1,1,1),(1,-1,1),(1,1,-1)]`
`=[(2a,0,0),(0,2b,0),(0,0,2c)]`
`=` diag `(2a, 2b, 2c)`


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