Looks like
[tex]B=\begin{bmatrix}11&0&0&13&4\\12&5&0&17&19\\0&0&0&1&0\\21&23&3&29&37\\2&0&0&35&0\end{bmatrix}[/tex]
Recall that for any two matrices [tex]A,B[/tex],
[tex]\det(AB)=\det A\det B[/tex]
which means
[tex]\det(B^{100})=(\det B)^{100}[/tex]
Laplace expansion along the third row gives us
[tex]\det B=-\det\begin{bmatrix}11&0&0&4\\12&5&0&19\\21&23&3&37\\2&0&0&0\end{bmatrix}[/tex]
Expansion along the last row gives
[tex]\det B=-\left(-2\det\begin{bmatrix}0&0&4\\5&0&19\\23&3&37\end{bmatrix}\right)=2\det\begin{bmatrix}0&0&4\\5&0&19\\23&3&37\end{bmatrix}[/tex]
Expansion along the first row gives
[tex]\det B=2\left(4\det\begin{bmatrix}5&0\\23&3\end{bmatrix}\right)=8\det\begin{bmatrix}5&0\\23&3\end{bmatrix}[/tex]
and finally
[tex]\det\begin{bmatrix}5&0\\23&3\end{bmatrix}=15[/tex]
so that
[tex]\det B=8\cdot15=120[/tex]
Then
[tex]\boxed{\det(B^{100})=120^{100}}[/tex]
