Question

write the general form of a 3x3 skew symmetric matrix and prove its determinant is zero​

Answer 1

Answer:

$\left[\begin{array}{ccc}0&a&b\\-a&0&d\\-b&-d&0\end{array}\right]$

Step-by-step explanation:

The general form of a 3 × 3 skey symmetric matrix is:

$\left[\begin{array}{ccc}0&a&b\\-a&0&d\\-b&-d&0\end{array}\right]$

Let's evaluate its determinant.

$det \left[\begin{array}{ccc}0&a&b\\-a&0&d\\-b&-d&0\end{array}\right]$

= 0(0 + $d^{2}$) - a(0 + bd) + b(ad - 0)

= - abd + abd

= 0