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Question
Without expanding the determinant, prove that `|("s", "a"^2, "b"^2 + "c"^2),("s", "b"^2, "c"^2 + "a"^2),("s", "c"^2, "a"^2 + "b"^2)|` = 0
Solution
`|("s", "a"^2, "b"^2 + "c"^2),("s", "b"^2, "c"^2 + "a"^2),("s", "c"^2, "a"^2 + "b"^2)| = "s"|(1, "a"^2, "b"^2 + "c"^2),(1, "b"^2, "c"^2 + "a"^2),(1, "c"^2, "a"^2 + "b"^2)|`
= `"s"|(1, "a"^2, "a"^2 + "b"^2 + "c"^2),(1, "b"^2, "a"^2 + "b"^2 + "c"^2),(1, "c"^2, "a"^2 + "b"^2 + "c"^2)| "C"_3 -> "C"_3 + "C"_2`
= `"s"("a"^2 + "b"^2 + "c"^2)|(1, "a"^2, 1),(1, "b"^2, 1),(1, "c"^2, 1)|`
= s (a2 + b2 + c2) × 0
Since two columns are equal.
= 0
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