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Found in: Page 277

### Linear Algebra With Applications

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

# Show that the function ${\mathbit{F}}\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& j\end{array}\right]{\mathbf{=}}{\mathbit{b}}{\mathbit{f}}{\mathbit{g}}$is linear in all three columns and in all three rows. See Example 6. Is F alternating on the columns? See Example 4.

Therefore, function F is linear in all rows and columns, but it's not alternating on the columns.

See the step by step solution

## Step 1: Given

Given matrix,

$F\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& j\end{array}\right]=bfg$

## Step 2: To show the function

Given,

$F\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& j\end{array}\right]=bfg$

Contains one factor from each row and from each column, therefore, it's linear in all three rows and all three columns.

On the other hand, we can see that, for example,

role="math" localid="1659505275434" $F\left(\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\right)=1$, but if we switch the second and third column of this matrix, we have

$F\left(\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\right)=1$, which means that F is not alternating on the columns.

Therefore, function F is linear in all rows and columns, but it's not alternating on the