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प्रश्न
Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.
उत्तर
L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`
Applying C1 → C1 – (C2 + C3), we get
L.H.S. = `|(0, "a", "b"),(-2"c", "a" + "c", "c"),(-2"c", "c", "b" + "c")|`
Taking (– 2) common from C1, we get
L.H.S. = ` - 2|(0, "a", "b"),("c", "a" + "c", "c"),("c", "c", "b" + "c")|`
Applying C2 → C2 – C1 and C3 → C3 – C1, we get
L.H.S. = `-2|(0, "a", "b"),("c", "a", 0),("c", 0, "b")|`
= - 2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= – 2(0 – abc – abc)
= – 2(– 2abc)
= 4abc
= R.H.S.
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