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Expert-verified Found in: Page 267 ### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384 # Let G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right). Use formula $$\left( {\bf{1}} \right)$$ for the determinant in section $${\bf{5}}{\bf{.2}}$$ to explain why $$\det G = \left( {\det A} \right)\left( {\det B} \right)$$. From this, deduce that the characteristic polynomial of $$G$$ is the product of the characteristic polynomials of $$A$$ and $$B$$.

For any scalar $$\lambda$$, the matrix $$G - \lambda I$$ has the same partitioned form as $$G$$, with $$A - \lambda I$$ , and $$B - \lambda I$$ as its diagonal blocks.

See the step by step solution

## Step 1: Write the echelon form

Consider G = \left( {\begin{aligned}{*{20}{c}}A&X\\0&B\end{aligned}} \right).

Assume $$U$$ and $$V$$ be echelon forms of $$A$$ and $$B$$ obtained by $$r$$ and $$s$$ row interchanges.

\begin{aligned}{l}\det A &= {\left( { - 1} \right)^r}\det U\\\det B &= {\left( { - 1} \right)^s}\det V\end{aligned}

## Step 2: Explain why $$\det G = \left( {\det A} \right)\left( {\det B} \right)$$

Using row operations, when $$A$$ is reduced to $$U$$, then $$G$$ is reduced to G' = \left( {\begin{aligned}{*{20}{c}}U&Y\\0&B\end{aligned}} \right).

Using row operations, when $$B$$ is reduced to $$V$$, then $$G$$ is reduced to G'' = \left( {\begin{aligned}{*{20}{c}}U&Y\\0&V\end{aligned}} \right).

Since there are $$r + s$$ row operations then we get,

\begin{aligned}{c}\det G &= \det \left( {\begin{aligned}{*{20}{c}}A&X\\0&B\end{aligned}} \right)\\ &= {\left( { - 1} \right)^{r + s}}\det \left( {\begin{aligned}{*{20}{c}}U&Y\\0&V\end{aligned}} \right)\\ &= {\left( { - 1} \right)^{r + s}}\left( {\det U} \right)\left( {\det V} \right)\\ &= \left( {\det A} \right)\left( {\det B} \right)\end{aligned}

Thus, $$\det \left( {G - \lambda I} \right) = \det \left( {A - \lambda I} \right)\det \left( {B - \lambda I} \right)$$.

For any scalar $$\lambda$$, the matrix $$G - \lambda I$$ has the same partitioned form as $$G$$, with $$A - \lambda I$$ and $$B - \lambda I$$ as its diagonal blocks, that is, $$\det \left( {G - \lambda I} \right) = \det \left( {A - \lambda I} \right)\det \left( {B - \lambda I} \right)$$. ### Want to see more solutions like these? 