If the determinant of a $3^{\text {rd }}$ order matrix A is K , then the sum of the determinants of the…
If the determinant of a $3^{\text {rd }}$ order matrix A is K , then the sum of the determinants of the matrices $\left(A A^T\right)$ and $\left(A-A^T\right)$ is
$\mathrm{2K}$
$0$
$\mathrm{K}^2$
$\mathrm{K}$
Solution
Given, A is $3 \times 3$ matrix and $|\mathrm{A}|=\mathrm{k}$
Now, $\mathrm{A}-\mathrm{A}^{\mathrm{T}}$ is always a skew symmetric matrix of order 3
So, $\left|\mathrm{A}-\mathrm{A}^{\mathrm{T}}\right|=0$ and $\left|\mathrm{AA}^{\mathrm{T}}\right|=|\mathrm{A}|\left|\mathrm{A}^{\mathrm{T}}\right|=\mathrm{K} \cdot \mathrm{K}=\mathrm{K}^2$
So, $\left|\mathrm{AA}^{\mathrm{T}}\right|+\left|\mathrm{A}-\mathrm{A}^{\mathrm{T}}\right|=\mathrm{O}+\mathrm{K}^2=\mathrm{K}^2$