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5 Chapter 1 ISM: Linear Algebra {1 51 {1.1 35. 37. ﬁ:}=—5+13r—Im'+3:3 Figure 111]: for Problem 1.233. 1: —3r + t
These vectors are of the form 1; = 1" , where r and t are arbitrary real
2 t numbers. II+I2+$3+34 2']
We need to solve the system $1 + 352 + 333 + 4134 = D ,
2:1 —— 9:32 + 933 + 73.1 = 1‘1 + 0.253.; = I]
which reduces to 2:2 — 1.53,; = I] .
$3 + 2.2534 2 I]
$1 41.25:
The solutions are of the form :2 = 122;: , where t is an arbitraryr real number.
3 — .
£4 E . Writing the equation ii: Ilﬁl —— 3:217; + 33173 in terms of its components, we obtain the system
$1 + 2172 + 4273 = —8
411+ 51': + 553 = —l.
17131 +8172 + 933 = Q
5171+ 332 + 273 = 1.5 The system has the unique solution 1:1 2 3, $2 = 3, and $3 = —4.
Compare with the solution of Exercise 1.1.21. 31 = [1.232 + [1.333 + 320 $1 — {1.31:2 — [1.333 = 33‘]
The diagram tells us that $2 = H.111 + [1.433 + 9'] or —ll.11.‘1 + $2 — [1.45:3 as .
$3 = {1.2271 + 0.532 + 150 —ﬂ.21‘1 — [1.532 + $3 151] 22 ...
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 Spring '09
 Derksen
 Algebra

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