Unformatted text preview: 24. 25. 2B. 21'. 28. 29. ISM: Linear Algebra Section 1.2 Property a holds by Step 2 of the Gauss—Jordan algorithm [page 1?].
Property b holds by Step 3 of the Gauss—Jordan algorithm.
Property c holds by Steps 1 and 4 of the algorithm. Yes; each elementary row operation is reversible, that is, it can be “rmdone.” For example,
the operation of row swapping can be undone by swapping the same rows again. The operation of dividing a row by a scalar can be reversed by multiplying the same row by
the same scalar. Yes; if A is transformed into B by a sequence of elementary row operations, then we can
recover A from B by applying the inverse operations in the reversed order [compare with Exercise 24). Yes, by Exercise 25, since rref[A] is obtained from A by a sequenoe of elementary row
operations. 1 2 3
No; whatever elementary row operations you apply to 4 5 B , you cannot make the
’F S 9 last column equal to zero. 611171+ﬂ121'2 +  * + ElmTn = bl
Suppose (cl, £32., . . . ,cn] is a solution of the system {121171 + 62232 +  .. + £12,113“, = I}; . To keep the notation simple, suppose we add .1: times the ﬁrst equation to the second; then the seoond equation of the new system will be [£121 +ka11]1:1 + n+{o2n+ka1njarn =
£32 + lat1. We have to verify that [c1,c2,.. ., on} is a solution of this new equation. Indeed, {r121 +
Jamaal + + (n.2,. + halal2,. = «21:21 + + 0211311 + Maura + + amen] = b; + as. We have shown that any solution of the “old” system is also a solution of the “new.” To see that, anversely, any solution of the new system is also a solution of the old system,
note that elementary row operations are reversible [oompare with Exercise 24]; we can
obtain the old system by subtracting in: times the ﬁrst equation from the second equation of the new system. Since the number of oxygen atoms remains constant, we must have 2n. + tr = 2r: + 3d. 2n. + tr = 2: +
Considering hydrogen and nitrogen as well, we obtain the system 21'? = c +
a. = c + Dr 19 ...
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 Spring '09
 Derksen
 Algebra

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