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)|` 

Answer 1

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)|`