Solution:

$A=\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$

$|A-\lambda I|=0$

$A=\left |\begin{matrix} 0-\lambda &c & -b \\ -c & 0-\lambda & a \\ b & -a & 0- \lambda \end{matrix} \right | $=0

$=-\lambda [\lambda ^2+a^2]-c[ c\lambda -ab]-b[ac+b\lambda]=0$

$=-\lambda^3-\lambda a^2-c^2 \lambda +abc-abc-b^2\lambda=0$

$=-\lambda^3+(-\lambda a^2-b^2 \lambda -c^2\lambda=0$

$=-\lambda^3-(a^2-b^2-c^2)\lambda=0$

$=\lambda^3+(a^2+b^2+c^2)\lambda=0$-------(1)

$\lambda \ in \ terms \ of \ A$

$\therefore A^3+(a^2+b^2+c^2)A=0$-------(2)

$A=\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$-------(3)

$A^2=\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}.\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$

$A^2=\begin{bmatrix} -c-b^2&ab & ac \\ ab & -c^2- a^2& bc \\ ac & bc & -b^2-a^2 \end{bmatrix}$------(4)

$A^3=A^2.A = \begin{bmatrix} -c-b^2&ab & ac \\ ab & -c^2 -a^2& bc \\ ac & bc & -b^2-a^2 \end{bmatrix} \begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$

$A^3=\begin{bmatrix} 0 &-c^3-cb^2-ca^2 &b^3+bc^2+ba^2 \\ c^3+ca^2+cb^2& 0& -ab^2-ac^2-a^3 \\ -bc^2-b^3-a^2b& ac^2+ab^2+a^3& 0 \end{bmatrix}$

$A^3= -(a^2+b^2+c^2)\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$

$A^3=-(a^2+b^2+c^2)A$------(5)

from equation (5)

$A^3+(a^2+b^2+c^2)A=0$-----------(6)

From equation (2) and (6)

Hence, A satisfies eqn (1)

A satisfies the cayley hamilton theorem

Now determinant of matrix

$|A|=\begin{bmatrix} 0 &c & -b \\ -c & 0& a \\ b & -a & 0 \end{bmatrix}$

=$0(0+a^2)-c(0-ab)-b(ac-0)$

=$0+abc-abc$

$|A|=0$

$A^{-1}$ does not exhist