Solution:

$A=\left[\begin{array}{ccc}{0} & {1} & {-1} \\ {4} & {-3} & {4} \\ {3} & {-3} & {4}\end{array}\right]$

$A^{2} = AA$

$=\left[\begin{array}{ccc}{0} & {1} & {-1} \\ {4} & {-3} & {4} \\ {3} & {-3} & {4}\end{array}\right]\left[\begin{array}{ccc}{0} & {1} & {-1} \\ {4} & {-3} & {4} \\ {3} & {-3} & {4}\end{array}\right]$

$=\left[\begin{array}{ccc}{0+4-3} & {0-3+3} & {0+4-4} \\ {0-12+12} & {4+9-12} & {-4-12+16} \\ {0-12+12} & {3+9-12} & {-3-12+16}\end{array}\right]$

$=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$

$=I$

$\therefore A^{2}=I$