Answer
True
Work Step by Step
Given $\begin{vmatrix}
a_1 & a_2 &0 & 0\\
a_3 & a_4 & 0 & 0\\
0 & 0 & b_1 & b_2\\
0 & 0 & b_3 & b_4
\end{vmatrix}$
We have $\begin{vmatrix}
a_1 & a_2 &0 & 0\\
a_3 & a_4 & 0 & 0\\
0 & 0 & b_1 & b_2\\
0 & 0 & b_3 & b_4
\end{vmatrix}\\
=a_1(a_4b_1b_4+0.b_2.0+0.0.b_3-(0.b_1.0+a_4b_2b_3+0.0.b_4))-a_2(a_3b_1b_4+0.b_2.0+0.0.b_3-(0.b_1.0+a_3b_2b_3+0.0.b_4))\\
=a_1(a_4b_1b_4-a_4b_2b_3)-a_2(a_3b_1b_4-a_3b_2b_3)\\
=a_1a_4b_1b_4-a_1a_4b_2b_3-a_2a_3b_1b_4+a_2a_3b_2b_3$
then $\begin{vmatrix}
a_1 & a_2 \\ a_3 & a_4
\end{vmatrix}=a_1a_4-a_2a_3$
and $\begin{vmatrix}
b_1 & b_2 \\ b_3 & b_4
\end{vmatrix}=b_1b_4-b_2b_3\\
\rightarrow \begin{vmatrix}
a_1 & a_2 \\ a_3 & a_4
\end{vmatrix}\begin{vmatrix}
b_1 & b_2 \\ b_3 & b_4
\end{vmatrix}\\=(a_1a_4-a_2a_3)(b_1b_2-b_2b_3)\\
=a_1a_4b_1b_4-a_1a_4b_2b_3-a_2a_3b_1b_4+a_2a_3b_2b_3$
Hence, the statement is true.