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Solution:

Given: $f(x)=x^{3}+4 x^{2}-x$

$f(A)=A^{3}+4 A^{2}-A$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left|\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right|\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{lll}0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9-0 & 4-0+0 \\ 0-2+0 & 1+3-0 & 2-0+0\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]$

$A^{3}=A^{2} A$

$\Rightarrow A^{3}=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{ccc}0-10+0 & 4+15-0 & 8-0+0 \\ 0+22+4 & -6-33-4 & -12+0+0 \\ 0+8+2 & -2-12-2 & -4+0+0\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]$

$f(A)=A^{3}+4 A^{2}-A$

$\Rightarrow f(A)=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+4\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]$

$\Rightarrow f(A)=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+\left[\begin{array}{ccc}16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]$

$\Rightarrow f(A)=\left[\begin{array}{cc}-10+16-0 & 19-20-1 & 8+0-2 \\ 26-24-2 & -43+44+3 & -12+16-0 \\ 10-8-1 & -16+16+1 & -4+8+0\end{array}\right]$

$\Rightarrow f(A)=\left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]$