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Question
Without expanding determinants, prove that `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)| = |("b"_1, "c"_1, "a"_1),("b"_2, "c"_2, "a"_2),("b"_3, "c"_3, "a"_3)| = |("c"_1, "a"_1, "b"_1),("c"_2, "a"_2, "b"_2),("c"_3, "a"_3, "b"_3)|`
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Solution
Let D = `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)|` ...(i)
Let E = `|("b"_1, "c"_1, "a"_1),("b"_2, "c"_2, "a"_2),("b"_3, "c"_3, "a"_3)|`
Applying C1 ↔ C2, we get
E = `-|("c"_1, "b"_1, "a"_1),("c"_2, "b"_2, "a"_2),("c"_3, "b"_3, "a"_3)|`
Applying C1 ↔ C3, we get
E = `-(- 1) |("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)|`
∴ E = `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)|` ...(ii)
Let F = `|("c"_1, "a"_1, "b"_1),("c"_2, "a"_2, "b"_2),("c"_3, "a"_3, "b"_3)|`
Applying C1 ↔ C2, we get
F = `-|("a"_1, "c"_1, "b"_1),("a"_2, "c"_2, "b"_2),("a"_3, "c"_3, "b"_3)|`
Applying C2 ↔ C3, we get
F = `-(- 1) |("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)|`
∴ F = `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)|` ...(iii)
From (i), (ii) and (iii), we get
`|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)| = |("b"_1, "c"_1, "a"_1),("b"_2, "c"_2, "a"_2),("b"_3, "c"_3, "a"_3)| = |("c"_1, "a"_1, "b"_1),("c"_2, "a"_2, "b"_2),("c"_3, "a"_3, "b"_3)|`
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