?In the special case where matrix $A_{21}$ in Exercise 26 is zero, the matrix

Chapter 1, Problem 27

(choose chapter or problem)

In the special case where matrix $A_{21}$ in Exercise 26 is zero, the matrix $A$ simplifies to

$A=\left[\begin{array}{lll} A_{11} & A_{12} & 0 & A_{22} \end{array}\right] $

which is said to be in block upper triangular form. Use the result of Exercise 26 to show that in this case

$A^{-1}=\left[A_{11}^{-1}-A_{11}^{-1} A_{12} A_{22}^{-1} 0 A_{22}^{-1}\right]$

Equation Transcription:

${A}_{21}$

$A=\left[{{A}_{11}\ {A}_{12}\ 0\ {A}_{22}\ }\right]$

${A}^{-1}=\left[{{A}_{11}^{-1}\ -{A}_{11}^{-1}{A}_{12}{A}_{22}^{-1}\ 0\ {A}_{22}^{-1}\ }\right]$

Text Transcription:

A_21

A=[A_11 A_12 0 A_22 ]

A^-1=[A_11^-1 -A_11^-1A_12 A_22^-1 0 A_22^-1]

Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.

?In the special case where matrix \(A_{21}\) in Exercise 26 is zero, the matrix