3.3soln - Solutions to even problems from section 3.3 2....

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Unformatted text preview: Solutions to even problems from section 3.3 2. Use Cramer’s rule to compute the solutions of the linear system 4x1 + x2 = 6 5x1 + 2x2 = 7 41 . Then det A = 4(2) − 5(1) = 3 = 0 so A is invertible. Since A 52 6 ( for i = 1, 2. is invertible we can use Cramer’s rule which says xi = det AiAb) where b = det 7 Thus 161 6(2) − 7(1) 5 x1 = = =, 372 3 3 Solution: Let A = x2 = 1 3 46 57 = 4(7) − 6(5) 2 =− . 3 3 4. Use Cramer’s rule to compute the solutions of the linear system −5x1 + 3x2 = 9 3x1 − x2 = −5 −5 3 . Then det A = −5(−1) − 3(3) = −4 = 0 so A is invertible. 3 −1 Since A is invertible we can use Cramer’s rule: 9(−1) − 3(−5) 3 1 9 3 = =− , x1 = − 4 −5 −1 −4 2 Solution: Let A = x2 = − 1 4 −5 9 3 −5 = −5(−5) − 9(3) 1 =. −4 2 18. Suppose all entries in A are integers and det A = 1. Explain why all entries in A−1 are integers. Solution: The inverse formula (theorem 8 on page 203) tells us that if A is invertible (which it is because det A = 0) then 1 A−1 = adj A det A Since det A = 1 we have A−1 = adj A. Now the (i, j )-entry in adj A is the (j, i)-cofactor Cji of A. In other words the(i, j )-entry in adj A is (−1)i+j det Aji . Since all entries in A are integers we know all entries in Aji are integers which tells us det Aji ∈ Z for each i, j . Thus the entries in adj A = A−1 are integers. 20. Find the area of the parallelogram whose vertices are (0, 0), Solution: Here’s a picture: (−1, 3), (4, −5), (3, −2). Thus the area of the parallelogram is det −1 4 3 −5 = |(−1)(−5) − 4(3)| = 7. 22. Find the area of the parallelogram whose vertices are (0, −2), (6, −1), (−3, 1), (3, 2). Solution: First let’s shift the parallelogram so that one of the vertices is at the origin. Here’s a picture: Thus the area of the parallelogram is det −3 6 31 = |(−3)(1) − 6(3)| = 21. 24. Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 4, 0), (−2, −5, 2), (−1, 2, −1). Solution: Using theorem 9 we see that the area of the parallelepiped is 1 −2 −1 −5 2 −2 −1 det 4 −5 2 = det = |(5 − 4) − 4(2 + 2)| = 15. − 4 det 2 −1 2 −1 0 2 −1 30. Let R be the triangle with vertices (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ). Show that x1 y1 1 1 (Area of R) = det x2 y2 1 . 2 x3 y3 1 Solution: First we shift R so that one of the vertices is at the origin using the map a b → a − x1 b − y1 Here’s where all the vertices are mapped to under this shift: x1 y1 → 0 0 , x2 y2 → x2 − x1 y2 − y1 , x3 y3 → x3 − x1 y3 − y1 Now we need to compute the area of a triangle which has one vertex at the origin. From the answer for exercise 29 we know that the area of R is 1 det 2 x2 − x1 x3 − x1 y2 − y1 y3 − y1 . On the other hand x1 y1 1 x2 y2 1 x3 y3 1 N R2 = R2 = R1 NR3 =R3 −R1 = x1 y1 1 x2 − x1 y2 − y1 0 x3 − x1 y3 − y1 0 = det x2 − x1 y2 − y1 x3 − x1 y3 − y1 = det x2 − x1 x3 − x1 y2 − y1 y3 − y1 [The right most equality follows from det AT = det A.] The result now follows by taking one half the absolute value of everything in the line above. ...
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