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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
Solution
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.
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