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Expert-verified Found in: Page 102 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Are elementary matrices invertible? If so, is the inverse of an elementary matrix elementary as well? Explain the significance of your answers in terms of elementary row operations.

The inverse of an elementary matrix is an elementary matrix of the same type.

See the step by step solution

## Step 1: Definition of the invertible matrix

An invertible matrix is defined as a matrix A of the dimension ${n}{×}{n}$ and is called invertible if and only if there exists another matrix B of the same dimensions.

## Step 2: Some constants that the matrix constructed are noninvertible

We prove that “The inverse of an elementary matrix is again an elementary matrix of the same type."

An elementary matrix of order n is obtained by performing exactly one elementary row operation on the identity matrix ${I}_{n}$.

## Step 3: Type 1

Suppose that an elementary matrix ${E}_{n×n}$ is obtained from ${I}_{n}$ by multiplying a row by a constant $k\in \mathrm{ℝ},k\ne 0$ . Without loss of generality, suppose we multiply the second row of ${I}_{n}$ by k. Then we have

$E=\left[\begin{array}{cccc}1& 0& -& 0\\ 0& k& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]$

Then we have

${E}^{-1}=\left[\begin{array}{cccc}1& 0& -& 0\\ 0& 1/k& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]$

Notice that ${E}^{-1}E={I}_{n}=E{E}^{-1}$

Therefore ${E}^{-1}$ is obtained from by multiplying a row by $\frac{1}{k}\left(\frac{1}{k}{R}_{a}↔{R}_{a}\right)$. Hence is invertible and ${E}^{-1}$ is also an elementary matrix.

## Step 4: Type 2

Suppose that an elementary matrix ${E}_{n×n}$ is obtained from ${I}_{n}$ by interchanging two rows $\left({R}_{i}↔{R}_{j}\right)$. Without loss of generality, suppose we interchange the first and second row of ${I}_{n}$ . We have

$E=\left[\begin{array}{cccc}1& 0& -& 0\\ 0& 0& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]$

Then we have

${E}^{-1}=\left[\begin{array}{cccc}0& 1& -& 0\\ 1& 0& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]=E$

Notice that ${E}^{-1}E={I}_{n}=E{E}^{-1}$

Therefore ${E}^{-1}$ is obtained from ${I}_{n}$ by interchanging same rows of ${I}_{n}\left({R}_{i}-{R}_{j}\right)$ . Hence E is invertible and ${E}^{-1}$ is also an elementary matrix.

## Step 5: Type 3

Suppose that an elementary matrix ${E}_{n×n}$ is obtained from ${I}_{n}$ by adding a multiple of one row to another $\left({R}_{i}+k{R}_{j}\to {R}_{i}\right){R}_{i}+k$ . Without loss of generality, suppose we add k times second row of ${I}_{n}$ in first row ${I}_{n}$. of Then we have

$E=\left[\begin{array}{cccc}1& k& -& 0\\ 0& 1& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]$

and

${E}^{-1}=\left[\begin{array}{cccc}1& -k& -& 0\\ 0& 1& -& 0\\ |& |& \& 0\\ 0& 0& -& 1\end{array}\right]$

Notice that ${E}^{-1}E={I}_{n}=E{E}^{-1}$

Therefore ${E}^{-1}$ is obtained from ${I}_{n}$ by subtracting a multiple of one row to another $\left({R}_{i}-k{R}_{j}\to {R}_{i}\right)$. Hence E is invertible and ${E}^{-1}$ is also an elementary matrix.

## Step 6: The final answer

The inverse of an elementary matrix is an elementary matrix of the same type. ### Want to see more solutions like these? 