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Unformatted text preview: Math 330 Chapter 3 Quizzes Section 3.1  Take home assigned February 18, 2009 Suppose that A and the product AB are as given below. Determine B A = bracketleftBigg 1 1 3 bracketrightBigg AB = bracketleftBigg 2 2 0 0 6 0 1 bracketrightBigg SOLUTION: Since A is 2 × 2 and AB is 2 × 4 it must mean that B is 2 × 4. The 4 columns of AB come from come from linear combinations of the columns of A . Each column of B gives weights to produce the resulting column of AB . Let vectora 1 and vectora 2 denote the first and second columns respectively of A . Then we note the following about the columns vector c 1 ,vector c 2 ,vector c 3 , and vector c 4 of AB : • By inspection vector c 1 = 2 vectora 1 + 0 vectora 2 . • By inspection vector c 2 = 0 vectora 1 + 2 vectora 2 • By inspection vector 0 = vector c 3 = 0 vectora 1 + 0 vectora 2 • We can find the last column vectorx of B by solving the system Avectorx = vector c 4 : bracketleftBigg 1 1 3 bracketrightBiggbracketleftBigg b 14 b 24 bracketrightBigg = bracketleftBigg 1 bracketrightBigg = ⇒ bracketleftBigg 1 1 0 3 1 bracketrightBigg R 2 ← R 2 3 R 1 = ⇒ bracketleftBigg 1 1 0 3 1 bracketrightBigg R 2 ← R 2 3 = ⇒ bracketleftBigg 1 1 0 1 1 3 bracketrightBigg R 1 ← R 1+ R 2 = ⇒ bracketleftBigg 1 0 1 3 0 1 1 3 bracketrightBigg So we have that vector c 4 = 1 3 vectora 1 + 1 3 vectora 2 . Therefore, B = bracketleftBigg 2 0 0 1 3 0 2 0 1 3 bracketrightBigg Section 3.2 February 20, 2009 Without the use of technology, show all the steps needed to calculate A 1 : A = 1 1 1 1 1 2 1 0 1 SOLUTION: We triply augment A using I 3 and we reduce A to reduced Echelon form which (in this case) will be I 3 : 1 1 1 1 1 0 0 1 1 2 0 1 0 1 0 1 0 0 1 R 2 ← R 2+ R 1 & R 3 ← R 3+ R 1 = ⇒ 1 1 1 1 0 0 0 2 1 1 1 0 0 1 1 0 1 R 2 ↔ R 3 = ⇒ 1 1 1 1 0 0 0 1 1 0 1 0 2 1 1 1 0 R 3 ← R 3 2 R 2 = ⇒ 1 1 1 1 0 1 1 1 0 0 1 1 1 2 R 3 ← R 3 = ⇒ 1 1 1 1 0 1 0 1 1 0 0 1 1 1 2 This completes the reduction to Echelon form. We then continue to reduced Echelon form R 1 ← R 1 R 2 = ⇒ 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 2 R 1 ← R 1 R 3 = ⇒ 1 0 0 1 1 3 0 1 0 1 1 0 0 1 1 1 2 Therefore, A 1 =  1 1 3 1 1 1 1 2 Section 2.3 Takehome assigned February 23, 2009 Suppose that A is a 4 × 4 matrix with the following properties: • A is invertible. • A 1 2 3 4 = vectorv 1 = 1 1 1 1 • A 1 2 3 4 = vectorv 2 = 1 1 1 • A 4 3 2 1 = vectorv 3 = 1 1 1 1. 2 points Find a linear combination of vectorv 1 ,vectorv 2 ,vectorv 3 which produces...
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra

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