{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch.1 Solutions Pg36

# Ch.1 Solutions Pg36 - Chapter 1 ISM Linear Algebra 1 a[15 4...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 1 ISM: Linear Algebra 1 a [15 4 The reduced row echelon form of this matrix is I] 1 [1512 . Thus we deduce that a a 15 2c the ﬁrst merchant has 4 coins, the second has 12, and the third is the richest, with 21} coins. 63. For each of the three statements, we set up an equation of the form [initial amount of grass} + [grass growth) = {grass consumed by cows}, or [#fo'idd-SF + (#affiefds}l#0fdays}y = (#arwwsneardaysiz. For the ﬁrst statement, this produces the equation a: + 21; = 62, or m + 2y — 62 = 1]. Similarly, we obtain the equations 43+ 161: — 28.2 = [1 and 23 + 1111: — 15.2 = 1] for the 12—350 other two statements. Flnm this information, we write the matrix 4 16 _23 [1 , 2111—1550 1 a —5 E a ,, 5, whichreducesto ﬂ 1 _% I] .Thusoursolutionsareoftheform[ :|= |:%t:|, _ z t 1] I] D :l] where t is an arl:Iitra.1.";I,r positive real number. 1.3 1. a. No solution, since the last row indicates I] = 1. b. The unique solution is a: = 5, y = 6. c. Inﬁnitely many solutions; the ﬁrst variable can be chosen freely. 2. The rank is 3 since each row contains a. leading one. DEF 1 3. This matrix has rank 1 since its rref is |:|] I] DEF |——l 3B ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online