If K R_0 then det. | adj (KI_n) | is equal to - Vidyadip

MCQ

If $K \in R_0$ then det. $ |{adj (KI_n)}|$ is equal to

A
$K^{n - 1}$
✓
$K^{n(n - 1)}$
C
$K^n$
D
$K$

✓

Answer

Correct option: B.

$K^{n(n - 1)}$

b
$(KI_n)$ $adj(KI_n) = | KI_n | I_n$

[Using $A (adj A) = | A | I$]

$adj\, (KI_n) = K^{n - 1} I_n$

$| adj\, (KI_n) | = K^{n (n - 1)}$

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