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Q83E

Expert-verifiedFound in: Page 102

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.

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

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 ${I}_{n}$.

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

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

Then we have

${E}^{-1}=\left[\begin{array}{cccc}1& 0& -& 0\\ 0& 1/k& -& 0\\ |& |& \backslash & 0\\ 0& 0& -& 1\end{array}\right]$Notice that ${E}^{-1}E={I}_{n}=E{E}^{-1}$

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

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

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

Then we have

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

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

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

Suppose that an elementary matrix ${E}_{n\times n}$ 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

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

and

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

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 $({R}_{i}-k{R}_{j}\to {R}_{i})$. Hence *E* is invertible and ${E}^{-1}$ is also an elementary matrix.

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

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