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प्रश्न
Let Δ = `|("a", "p", x),("b", "q", y),("c", "r", z)|` = 16, then Δ1 = `|("p" + x, "a" + x, "a" + "p"),("q" + y, "b" + y, "b" + "q"),("r" + z, "c" + z, "c" + "r")|` = 32.
पर्याय
True
False
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उत्तर
This statement is True.
Explanation:
Given that Δ = `|("a", "p", x),("b", "q", y),("c", "r", z)|` = 16
L.H.S. Δ1 = `|("p" + x, "a" + x, "a" + "p"),("q" + y, "b" + y, "b" + "q"),("r" + z, "c" + z, "c" + "r")|`
C1 → C1 + C2 + C3
= `|("2p" + 2x + 2"a", "a" + x, "a" + "p"),(2"q" +2y + 2"b", "b" + y, "b" + "q"),(2"r" + 2z + 2"c", "c" + z, "c" + "r")|`
= `2|("p" + x + "a", "a" + x, "a" + "p"),("q" +y + "b", "b" + y, "b" + "q"),("r" + z + "c", "c" + z, "c" + "r")|` ......[Taking 2 common from C1]
C1 → C1 – C2 = `2|("p", "a" + x, "a" + "p"),("q", "b" + y, "b" + "q"),("r", "c" + z, "c" + "r")|`
C3 → C3 – C2d = `2|("p", "a" + x, "a"),("q", "b" + y, "b"),("r", "c" + z, "c")|`
Splitting up C2
= `2|("p", "a", "a"),("q", "b", "b"),("r", "c", "c")| + 2|("p", x, "a"),("q", "y", "b"),("r", "z", "c")|`
= `2(0) + 2|("p", x, "a"),("q", y, "b"),("r", z, "c")|`
= `2|("p", x, "a"),("q", y, "b"),("r", z, "c")|`
⇒ `2|("a", "p", x),("b", "q", y),("c", "r", z)|` ......(C1 ↔ C3 and C2 ↔ C3)
= 2 × 16
= 32
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