Question

If the determinant `|(x + "a", "p" + "u", "l" + "f"),("y" + "b", "q" + "v", "m" + "g"),("z" + "c", "r" + "w", "n" + "h")|` splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

Options

• True

• False

Answer 1

This statement is True.

Explanation:

Let Δ = `|(x + "a", "p" + "u", "l" + "f"),("y" + "b", "q" + "v", "m" + "g"),("z" + "c", "r" + "w", "n" + "h")|`

Splitting up C₁

⇒ `|(x, "p" + "u", "l" + "f"),(y, "q" + "v", "m" + "g"),(z, "r" + "w", "n" + "h")| + |("a", "p" + "u", "l" + "f"),("b", "q" + "v", "m" + "g"),("c", "r" + "w", "n" + "h")|`

Splitting up C₂ in both determinants

⇒ `|(x, "p", "l" + "f"),(y, "q", "m" + "g"),(z, "r", "n" + "h")| + |(x, "u", "l" + "f"),(y, "v", "m" + "g"),(z, "w", "n" + "h")| + |("a", "p", "l" + "f"),("b", "q", "m" + "g"),("c", "r", "n" + "h")| + |("a", "u", "l" + "f"),("b", "v", "m" + "g"),("c", "w", "n" + "h")|`

Similarly by splitting C₃ in each determinant, we will get 8 determinants.