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A First Course in Linear Algebra, 2017a

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82 Matrices<br />

You can see that B is the matrix obta<strong>in</strong>ed by add<strong>in</strong>g 2 times the first row of A to the third row.<br />

♠<br />

Suppose we have applied a row operation to a matrix A. Consider the row operation required to return<br />

A to its orig<strong>in</strong>al form, to undo the row operation. It turns out that this action is how we f<strong>in</strong>d the <strong>in</strong>verse of<br />

an elementary matrix E.<br />

Consider the follow<strong>in</strong>g theorem.<br />

Theorem 2.51: Elementary Matrices and Inverses<br />

Every elementary matrix is <strong>in</strong>vertible and its <strong>in</strong>verse is also an elementary matrix.<br />

In fact, the <strong>in</strong>verse of an elementary matrix is constructed by do<strong>in</strong>g the reverse row operation on I.<br />

E −1 will be obta<strong>in</strong>ed by perform<strong>in</strong>g the row operation which would carry E back to I.<br />

•IfE is obta<strong>in</strong>ed by switch<strong>in</strong>g rows i and j, thenE −1 is also obta<strong>in</strong>ed by switch<strong>in</strong>g rows i and j.<br />

•IfE is obta<strong>in</strong>ed by multiply<strong>in</strong>g row i by the scalar k, thenE −1 is obta<strong>in</strong>ed by multiply<strong>in</strong>g row i by<br />

the scalar 1 k .<br />

•IfE is obta<strong>in</strong>ed by add<strong>in</strong>g k times row i to row j, thenE −1 is obta<strong>in</strong>ed by subtract<strong>in</strong>g k times row i<br />

from row j.<br />

Consider the follow<strong>in</strong>g example.<br />

Example 2.52: Inverse of an Elementary Matrix<br />

Let<br />

[ ] 1 0<br />

E =<br />

0 2<br />

F<strong>in</strong>d E −1 .<br />

Solution. Consider the elementary matrix E given by<br />

[ ]<br />

1 0<br />

E =<br />

0 2<br />

Here, E is obta<strong>in</strong>ed from the 2 × 2 identity matrix by multiply<strong>in</strong>g the second row by 2. In order to carry E<br />

back to the identity, we need to multiply the second row of E by 1 2 . Hence, E−1 is given by<br />

[ ]<br />

1 0<br />

E −1 =<br />

0 1 2<br />

We can verify that EE −1 = I. Take the product EE −1 ,givenby<br />

[ ] [ ] [ ]<br />

EE −1 1 0 1 0 1 0<br />

=<br />

=<br />

0 2<br />

0 1<br />

0 1 2

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