Question

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

Answer 1

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 – C₂^d = `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