Chapter 1, Problem 27SE

In the special case where matrix A21 in Exercise 26 is zero, the matrix A simplifies to

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

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

$A^{-1}=\left[\begin{array}{l}{A}_{11}^{-1}-A_{11}^{-1}{A}_{12}{A}_{22}^{-1}\\ 0A_{22}^{-1}\end{array}\right]$