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Found in: Page 103

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Consider a block matrix ${\mathbf{A}}{\mathbf{=}}\left[\begin{array}{cc}{A}_{11}& 0\\ 0& {A}_{22}^{}\end{array}\right]$, where role="math" localid="1660371822794" ${{\mathbit{A}}}_{{\mathbf{11}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbit{A}}}_{{\mathbf{12}}}$are square matrices. For which choices of ${{\mathbit{A}}}_{{\mathbf{11}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbit{A}}}_{{\mathbf{22}}}$ is A invertible? In these cases, what is ?

The matrix A is invertible when ${\mathrm{A}}_{11}\mathrm{and}{\mathrm{A}}_{22}$are invertible.

The invertible matrix is, $A=\left[\begin{array}{cc}{A}_{11}^{-1}& 0\\ 0& {A}_{22}^{-1}\end{array}\right]$.

See the step by step solution

## Step 1: Reducing A to identity theoretically

Consider $A=\left[\begin{array}{cc}{A}_{11}& 0\\ 0& {A}_{22}\end{array}\right]$where role="math" localid="1660372094980" ${\mathrm{A}}_{11}\mathrm{and}{\mathrm{A}}_{22}$are square matrices. We want to find what choices of ${\mathrm{A}}_{11}\mathrm{and}{\mathrm{A}}_{22}$make A invertible and what is for these choices.

Obviously, A will be invertible only if ${\mathrm{A}}_{11}\mathrm{and}{\mathrm{A}}_{22}$are invertible.This is because if ${\mathrm{A}}_{11}\mathrm{and}{\mathrm{A}}_{22}$are invertible they can reduce to the identity which means that A can reduce to the identity as well.

## Step 2:Solving for A-1

Logically this means that, $\mathrm{A}=\left[\begin{array}{cc}{\mathrm{A}}_{11}^{-1}& 0\\ 0& {\mathrm{A}}_{22}^{-1}\end{array}\right]$.

Now, check it by finding the product of ${\mathrm{AA}}^{-1}$.

$A{A}^{-1}=\left[\begin{array}{cc}{\mathrm{A}}_{11}^{}& 0\\ 0& {\mathrm{A}}_{22}^{}\end{array}\right]\left[\begin{array}{cc}{\mathrm{A}}_{11}^{-1}& 0\\ 0& {\mathrm{A}}_{22}^{-1}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}{\mathrm{A}}_{11}{\mathrm{A}}_{11}^{-1}& 0\\ 0& {\mathrm{A}}_{22}{\mathrm{A}}_{22}^{-1}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}{\mathrm{l}}_{\mathrm{n}}& 0\\ 0& {\mathrm{l}}_{\mathrm{n}}\end{array}\right]$

Where the product is identity matrix.

Thus, ${A}^{-1}=\left[\begin{array}{cc}{\mathrm{A}}_{11}^{-1}& 0\\ 0& {\mathrm{A}}_{22}^{-1}\end{array}\right]$.