Let us consider a frame of reference with three mutually perpendicular axes meeting at the origin O. This is fixed in a rotating body so it rotates with the body in such a way that the product of inertia about them is zero. These mutually perpendicular axes are called the principal axes of inertia of the body. When a rigid body rotates about the principal axis, the direction of angular momentum is the same as that of the angular velocity.

We know that,

\( \vec{L}=I\vec{\omega} \)

Therefore,

\( I_{xx}\omega_{x}+I_{xy}\omega{y}+I_{xz}\omega{z}=I\omega_{x} \)

\( I_{yx}\omega_{x}+I_{yy}\omega{y}+I_{yz}\omega{z}=I\omega_{y} \)

\( I_{zx}\omega_{x}+I_{zy}\omega{y}+I_{zz}\omega{z}=I\omega_{z} \)

Or we can write,

\( (I_{xx}-I)\omega_{x}+I_{xy}\omega{y}+I_{xz}\omega{z}=0 \)

\( I_{xx}\omega_{x}+(I_{xy}-I)\omega{y}+I_{xz}\omega{z}=0 \)

\( I_{xx}\omega_{x}+I_{xy}\omega{y}+(I_{xz}-I)\omega{z}=0 \)

The Principal moment of inertia are found by setting the determinant of the co-efficients of \( \omega_x \), \( \omega_y \), \( \omega_z \) equal to zero.

\( \begin{vmatrix} (I_{xx}-I) & I_{xy} & I_{xz} \\ I_{yx} & (I_{yy}-I) & I_{yz} \\ I_{zx} & I_{zy} & (I_{zz}-I)  \end{vmatrix} =0 \)

This leads to a cubic equation in \( I \). It has three real roots \( I_1 \), \( I_2 \), \( I_3 \). These are called the principal moment of inertia.