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If `A`is a square matrix such that `A^2=I`, then find the simplified value of `(A-I)^3+(A+I)^3-7Adot` |
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Answer» Here, we are given, `A^2 = I`. Now, `(A+I)^3 = (A+I)(A+I)(A+I)` `= (A^2+IA+AI+I^2)(A+I)` `=(I+A+A+I)(I+A)` (As `A^2 = I`) `=(2I+2A)(I+A)` `=2(I+A)(I+A)` `=2(2(I+A)` `=4I+4A` `:. (A+I)^3 = 4I+4A` Now, `(A-I)^3 = (A-I)(A-I)(A-I)` `=(A^2-IA-IA+I^2)(A-I)` `=(I-2A+I)(A-I)` `=(2I-2A)(A-I)` `=-2(A-I)(A-I)` `=-2(-2(A-I))` `=4A-4I` `:. (A-I)^3+(A+I)^3- 7A = 4A-4I+4I+4A-7A = 8A-7A = A` `=>(A-I)^3+(A+I)^3- 7A = A` |
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