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Question
Answer the following question:
If `|("a", 1, 1),(1, "b", 1),(1, 1, "c")|` = 0 then show that `1/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
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Solution
`|("a", 1, 1),(1, "b", 1),(1, 1, "c")|` = 0
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
`|("a", 1, 1),(1- "a", "b" - 1, 0),(1- "a", 0, "c" - 1)|` = 0
∴ a[(b – 1) (c – 1) – 0] – 1[(1 – a) (c – 1) – 0] + 1[0 – (b – 1) (1 – a)] = 0
∴ a(1 – b) (1 – c) + (1 – a) (1 – c) + (1 – b) (1 – a) = 0
Dividing throughout by (1 – a) (1 – b) (1 – c), we get
`"a"/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 0
Adding 1 on both the sides, we get
`1 + "a"/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
∴ `(1 - "a" + "a")/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
∴ `1/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
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