First slide

Applications of determinant

Question

$If k \in R, then \det \{a d j (k I_{n})\} is equal to$

Moderate

A

$k^{n - 1}$
B

$k^{n (n - 1)}$
C

$k^{n}$
D

$k$

Solution

$We have | adj A | = | A |^{n - 1}$
$\begin{array}{l} \Rightarrow |adj K I_{n}| = {|K I_{n}|}^{n - 1} \\ = {(K^{n} |I_{n}|)}^{n - 1} (∵ | K A | = K^{n} | A |) \\ = K^{n (n - 1)} (1) (∵ |I_{n}| = 1) \\ = K^{n (n - 1)} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App