If A is a square matrix satisfying

Let $A=\left[a_{i j}\right]$ be a square matrix of order $\mathrm{n}$.

Here,

$|A|=\left|A^{T}\right|$

[By property of determinants]

Given: $\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I}$

$\Rightarrow\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}\right|=1$

Then,

$\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}\right|=\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{A}| \quad$ [Since the determinants are of the same order]

$\Rightarrow\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{A}|=1$

$\Rightarrow|\mathrm{A}|=\frac{1}{\left|\mathrm{~A}^{\mathrm{T}}\right|}$

$\Rightarrow|\mathrm{A}|=\frac{1}{|\mathrm{~A}|}$      $\left[\therefore|A|=\left|A^{T}\right|\right]$

$\Rightarrow|\mathrm{A}|^{2}=1$

$\Rightarrow|A|=+1$