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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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