$$\begin{bmatrix}1&0 & 0 \\ 0 &0&1\\ 0 & 1 & 0 \end{bmatrix}$$, then the absolute value of the determinant of $$(A^9+A^6+A^3+A)$$ is ________.

Given, $$A=$$ $$\begin{bmatrix}1&0 & 0 \\ 0 &0&1\\ 0 & 1 & 0 \end{bmatrix}$$

Now, $$A^2=A\cdot A$$ = $$\begin{bmatrix}1&0 & 0 \\ 0 &0&1\\ 0 & 1 & 0 \end{bmatrix}$$ $$\times\ $$ $$\begin{bmatrix}1&0 & 0 \\ 0 &0&1\\ 0 & 1 & 0 \end{bmatrix}$$

Upon performing matrix multiplication, $$A^2$$= $$\begin{bmatrix}1&0 & 0 \\ 0 &1&0\\ 0 & 0 & 1 \end{bmatrix}$$

So, $$A^2$$ is identity matrix $$I$$

So, $$A^3=A^2\cdot A=I\cdot A=A$$

$$A^6=(A^2)^3=I^3=I$$

$$A^9=(A^2)^4\cdot A=(I)^4\cdot A=I\cdot A=A$$

So, $$A^9+A^6+A^3+A=A+I+A+A=3A+I$$

Now, $$3A+I$$ = $$3\cdot$$ $$\begin{bmatrix}1&0 & 0 \\ 0 &0&1\\ 0 & 1 & 0 \end{bmatrix}$$ + $$\begin{bmatrix}1&0 & 0 \\ 0 &1&0\\ 0 & 0 & 1 \end{bmatrix}$$

=$$\begin{bmatrix}3&0 & 0 \\ 0 &0&3\\ 0 & 3 & 0 \end{bmatrix}$$ + $$\begin{bmatrix}1&0 & 0 \\ 0 &1&0\\ 0 & 0 & 1 \end{bmatrix}$$

=$$\begin{bmatrix}4&0 & 0 \\ 0 &1&3\\ 0 & 3 & 1 \end{bmatrix}$$

Now, determinant of $$(A^9+A^6+A^3+A)$$ = $$\begin{vmatrix}4&0 & 0 \\ 0 &1&3\\ 0 & 3 & 1 \end{vmatrix}$$

Breaking along $$R1$$,

=$$4\cdot$$ $$\begin{vmatrix}1&3 \\ 3&1\end{vmatrix}$$

=$$4(1\times\ 1-3\times\ 3)$$

$$=4(1-9)$$

$$=-32$$

So, absolute value of deteminant is 32.