Answered: n ×n matrices A, B, C, D where A is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Given n ×n matrices A, B, C, D where A is invertible and
AC = CA, show that:

det(matrix[

A	B
C	D

]) = det(AD-CB)

Expert Solution

Step 1

Given - A, B, C, D are $n \times n$ matrices where A is invertible

and AC=CA

We have to show that

$d e t [\begin{matrix} A & B \\ C & D \end{matrix}] = d e t [A D - C B]$

$A s A, B, C, D a r e n \times n m a t r i c e s W h e r e A i s i n v e r t i b l e N o w i f A i s i n v e r t i b l e t h e n b y g a u s s e l i m i n a t i o n w e h a v e [\begin{matrix} A & B \\ C & D \end{matrix}] = [\begin{matrix} I & 0 \\ C A^{- 1} & I \end{matrix}] [\begin{matrix} A & B \\ C & D \end{matrix}] w h e r e I i s i d e n t i t y m a t r i x = [\begin{matrix} A - 0 & B - 0 \\ C A^{- 1} A - C & D - C A^{- 1} B \end{matrix}] = [\begin{matrix} A & B \\ 0 & D - C A^{- 1} B \end{matrix}] A i s i n v e r t i b l e s o A^{- 1} A = 1 \Rightarrow [\begin{matrix} A & B \\ C & D \end{matrix}] = [\begin{matrix} A & B \\ 0 & D - C A^{- 1} B \end{matrix}]$