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Chapter 3: Q11SE (page 165)

Question: 11. Find the area of the parallelogram determined by the points \(\left( {1,4} \right),\)\(\left( { - 1,5} \right),\)\(\left( {3,9} \right),\) and \(\left( {5,8} \right)\). How can you tell that the quadrilateral determined by the points is actually a parallelogram?

Short Answer

Expert verified

The area of the parallelogram is 12 square units. And the quadrilateral determined by the points is actually a parallelogram since one of the nonzero points can be written as the sum of the other two nonzero points.

Step by step solution

01

Translate the figure

First, translate one of the vertices to the origin. That is, subtract the vertex \(\left( {1,4} \right)\) from all four vertices \(\left( {1,4} \right),\left( { - 1,5} \right),\left( {3,9} \right),\) and \(\left( {5,8} \right)\). The new vertices so obtained are 0,\({v_1} = \left( { - 2,1} \right),\) \({v_2} = \left( {2,5} \right),\) and \({v_3} = \left( {4,4} \right)\).

02

Determine if the translated figure is a parallelogram

\(\begin{array}{c}{v_1} + {v_3} = \left( {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}4\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2 + 4}\\{1 + 4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2\\5\end{array}} \right)\\{v_1} + {v_3} = {v_2}\end{array}\)

Note that the translated figure will be a parallelogram if and only if one of \({v_1},{v_2},\) and \({v_3}\) is the sum of the other two vectors.

Hence, this parallelogram determined by the columns of \(A = \left( {\begin{array}{*{20}{c}}{ - 2}&4\\1&4\end{array}} \right)\).

03

Find the area

\(\begin{array}{c}\left| {\det A} \right| = \left| {\det \left[ {\begin{array}{*{20}{c}}{ - 2}&4\\1&4\end{array}} \right]} \right|\\ = \left| { - 8 - 4} \right|\\ = \left| { - 12} \right|\\\left| {\det A} \right| = 12\end{array}\)

By Theorem 9, the area of the parallelogram is 12 square units.

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Most popular questions from this chapter

Compute the determinants of the elementary matrices given in Exercises 25-30. (See Section 2.2)

\[\left[ {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&k&{\bf{1}}\end{aligned}} \right]\]

Let \(u = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right]\), and \(v = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Compute the area of the parallelogram

determined by u, v, \({\bf{u}} + {\bf{v}}\), and 0, and compute the determinant of \(\left[ {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right]\). How do they compare? Replace the first entry of v by an arbitrary number x, and repeat the problem. Draw a picture and explain what you find.

Let \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\) and let \(k\) be a scalar. Find a formula that relates \(\det kA\) to \(k\) and \(\det A\).

In Exercises 27 and 28, A and B are \[n \times n\] matrices. Mark each statement True or False. Justify each answer.

27. a. A row replacement operation does not affect the determinant of a matrix.

b. The determinant of A is the product of the pivots in any echelon form U of A, multiplied by \({\left( { - {\bf{1}}} \right)^r}\), where r is the number of row interchanges made during row reduction from A to U.

c. If the columns of A are linearly dependent, then \(det\left( A \right) = 0\).

d. \(det\left( {A + B} \right) = det{\rm{ }}A + det{\rm{ }}B\).

Compute the determinants in Exercises 9-14 by cofactor expnasions. At each step, choose a row or column that involves the least amount of computation.

\(\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{ - {\bf{7}}}&{\bf{3}}&{ - {\bf{5}}}\\{\bf{0}}&{\bf{0}}&{\bf{2}}&{\bf{0}}&{\bf{0}}\\{\bf{7}}&{\bf{3}}&{ - {\bf{6}}}&{\bf{4}}&{ - {\bf{8}}}\\{\bf{5}}&{\bf{0}}&{\bf{5}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{0}}&{\bf{9}}&{ - {\bf{1}}}&{\bf{2}}\end{array}} \right|\)
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