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Question: In Exercise 20, find the area of the parallelogram whose vertices are listed.

20. \(\left( {0,0} \right),\left( { - {\bf{2}},4} \right),\left( {4, - 5} \right),\left( {2, - 1} \right)\)

Short Answer

Expert verified

The area of the parallelogram is 2 square units.

Step by step solution

01

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 \(A = \left( {\begin{array}{*{20}{c}}{ - 2}&4\\4&{ - 5}\end{array}} \right)\).

02

Write the first statement of Theorem 9

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

03

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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Most popular questions from this chapter

Find the determinant in Exercise 15, where \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right| = {\bf{7}}\].

15. \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{{\bf{3g}}}&{{\bf{3h}}}&{{\bf{3i}}}\end{array}} \right|\]

The expansion of a \({\bf{3}} \times {\bf{3}}\) determinant can be remembered by the following device. Write a second type of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals.

Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises 15-18. Warning: This trick does not generalize in any reasonable way to \({\bf{4}} \times {\bf{4}}\) or larger matrices.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{3}}&{\bf{1}}\\{\bf{4}}&{ - {\bf{5}}}&{\bf{0}}\\{\bf{3}}&{\bf{4}}&{\bf{1}}\end{aligned}} \right|\)

In Exercise 33-36, verify that \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\)where E is the elementary matrix shown and \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\).

36. \(\left[ {\begin{aligned}{*{20}{c}}1&0\\0&k\end{aligned}} \right]\)

Find the determinant in Exercise 16, where \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right| = {\bf{7}}\].

16. \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{{\bf{5d}}}&{{\bf{5e}}}&{{\bf{5f}}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right|\]

In Exercise 33-36, verify that \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\)where E is the elementary matrix shown and \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\).

33. \(\left[ {\begin{aligned}{*{20}{c}}1&k\\0&1\end{aligned}} \right]\)

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