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'iddSF + (#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 ...
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 Spring '09
 Derksen
 Algebra

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