General expression for determinant of a block-diagonal matrix
First write $$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] $$ Also, $$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &&&& \\[-3pt] & \ddots \hspace{-5pt} &&& \\ && A_j \hspace{-5pt} && \\[-3pt] &&& \ddots \hspace{-5pt} & \\ &&&& \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$ which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1
$$ \implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$
The determinant of a block diagonal matrix is equal to the product of the determinants of the diagonal blocks. In your case, you have a block diagonal matrix of the form $$A=\left(\begin{array}{ccc} A_1 & 0 & 0\\ 0 & A_2 & 0\\ 0 & 0 & A_3 \end{array}\right)$$ so $\det(A) = \det(A_1)\det(A_2)\det(A_3)$.
You can, though it is a bit ad-hoc. For example, note that if we let $$T_1 = \left(\begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$$ then $T_1^tA_1T_1$ is the block diagonal matrix $$\left(\begin{array}{ccc} A_1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$$ Likewise, if we let $$\begin{align*} T_2 &=\left(\begin{array}{ccccccccc} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{array}\right)\\ T_3 &= \left(\begin{array}{ccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right) \end{align*}$$ then the resulting matrix is $$f(A_1,A_2,A_3) = T_1^tA_1T_1 + T_2^tA_2T_2 + T_3^t A_3 T_3.$$
As stated as a comment, the result is here. It really makes block diagonal matrices wonderful, hence finding canonical forms important.