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**Unformatted text preview: **W2; Section 8.2 2- IAI = (-1)(2)(4) - (-1)(0)(1) - (3)(2)(4) + (3)(0)(1) + (1)(2)(1) - (1)(2)(1) = ‘32 4- W = (-4)(1)(0) - (-4)(1)(0) - (0)(0)(0) + (0)(1)(0) + (1)(0)(0) - (1)(1)(0) = 0 Section 8.5_ 2. Use row operations to reduce column one, then expand by the 15‘ column to get. 1 16 1 1 16 '2 —2 1 = 0 —4 ~11 ‘ :1 :1: 2“: :1; :12
[3 —1 4 0 —4 —14 63,775 Section 8.6 Mal-.8. 2. A‘1 does not exist since the reduced form of the matrix is AR 2 ( 1 1/4 > aé lg. 0 0
1 0 3
8. A—1 does not exist since the reduced form of the matrix is AR : 0 1 1 yé [3,
0 0 0
1 0 28/27
10. A—1 does not exist since the reduced form of the matrix is A3 = 0 1 14/9 aé I3.
0 0 0 iection 8.7 2.A-1=i 4 0
12 —1 3 1~ —10 —10 0
6. .4-12— —11 —95 36
120 3 15 12 SW14; 2. IAI = —3 51$ 0 so Cramer’s rule applies.
1’ 3 4 " _3 1 “>3 01 3 3 4. [AI = 108 aé 0 so Cramer’s rule applies.
63 7 165 55 243 x1=~~—~——— 108‘ 12’32:_i@:‘36’$3:‘ﬁ1:_ ' phlto Despite the numbers, the following are problems 11 and 12. 8. When we have ul, u2, u3, to determine linearly independence we can say:
clul + c2u2 + c3113 =0 multiplying both side by 111 112 and 113 cl <:ul,ul>' + c2 <u1,u2> + c3<ul,u3> = 0 cl <u2,ul> + c2 <u2,u2> + c3<u2,u3> = 0 cl <u3,ul> + c2 <u3,u2> + c3<u3,u3> = 0 then we will have: < 111,111 > < ul,u2 > < 241,143 > cl 0
< u2,ul > < 112,112 > < 112,213 > c2 = 0
< 113,111 > < 113,212 > < 113,113 > c3 0 Or AC=0 , now for ﬁnding C we can say
If , C can have a non-zero value so 111 ,u2,u3 are dependent. If .Al >0 or {AFC then inverse of A casts and C have to be zero which means ul,u2,u3
are independent . For problem number 8 : 2 l l
A: l 2 I therefore |A|=2.l: :l—lli :4-1: fl=4 soul,u2,u3 are independent
1 l 2
hereandrank is3 andul,u2,u3 are ourbasis
3 2 4 2 2 2 2 2 2 I
9.21: i i : therefore |A|=3_2 6—244 6+4.4 2=0 soul,u2,u3are dependent here andrank r’j Q , ﬂwmsim 1;,2. ...

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