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Found in: Page 165

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question: In Exercise 20, find the area of the parallelogram whose vertices are listed.20. $$\left( {0,0} \right),\left( { - {\bf{2}},4} \right),\left( {4, - 5} \right),\left( {2, - 1} \right)$$

The area of the parallelogram is 2 square units.

See the step by step solution

## Step 1: Determine the matrix

The column vectors in the parallelogram are $$\left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 2}\\4\end{array}} \right),\left( {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right),$$ and $$\left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right)$$.

Note that the parallelogram has origin as a vertex and

$$\begin{array}{c}\left( {\begin{array}{*{20}{c}}{ - 2}\\4\end{array}} \right) + \left( {\begin{array}{*{20}{c}}4\\{ - 5}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2 + 4}\\{4 - 5}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right).\end{array}$$

Hence, this parallelogram is determined by the columns of $$A = \left( {\begin{array}{*{20}{c}}{ - 2}&4\\4&{ - 5}\end{array}} \right)$$.

## Step 2: Write the first statement of Theorem 9

The first statement of Theorem 9 states if A is a $$2 \times 2$$ matrix, the area of the parallelogram determined by the columns of A is $$\left| {\det A} \right|$$.

## Step 3: Find the area

By the above statement,

$$\begin{array}{c}\left| {\det A} \right| = \left| {\det \left[ {\begin{array}{*{20}{c}}{ - 2}&4\\4&{ - 5}\end{array}} \right]} \right|\\ = \left| {10 - 8} \right|\\ = \left| 2 \right|\\\left| {\det A} \right| = 2\end{array}$$

Hence, the area of the parallelogram is 2 square units.

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