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Let [11 0 0 13 4] 12 5 0 17 19 B= | 0 0 0 1 0 21 23 3 29 37 | 2 0 0 35 0 Compute det(B100)

User Prakhar
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1 Answer

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Looks like


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}

Recall that for any two matrices
A,B,


\det(AB)=\det A\det B

which means


\det(B^(100))=(\det B)^(100)

Laplace expansion along the third row gives us


\det B=-\det\begin{bmatrix}11&0&0&4\\12&5&0&19\\21&23&3&37\\2&0&0&0\end{bmatrix}

Expansion along the last row gives


\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}

Expansion along the first row gives


\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}

and finally


\det\begin{bmatrix}5&0\\23&3\end{bmatrix}=15

so that


\det B=8\cdot15=120

Then


\boxed{\det(B^(100))=120^(100)}

User Alexblum
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