First slide

Applications of determinant

Question

If $n^{t h}$ -order square matrix A is a orthogonal , then, $| a d j (a d j A) |$ is

Moderate

A

always -1 if n is even
B

always 1 if n is odd
C

always 1
D

none of these

Solution

Since A is orthogonal , hence
$\begin{aligned} A A^{T} = I \\ \Rightarrow |A A^{T}| = 1 \\ |A ∥ A^{T}| = 1 \end{aligned}$
$\begin{aligned} \Rightarrow |A^{2}| = 1 \\ \Rightarrow | A | = \pm 1 \end{aligned}$
$Now, | adj (adj A) | = | A |^{(n - 1)^{2}}$
$= {(\pm 1)}^{{(n - 1)}^{2}}$

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