Transcription of x1 x2 + 3x3 = 9 x1 7x2x3 = 2 x 5x = 15 Solution: 4 3 9 x 1 ...
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x1 x2 + 3x3 = 9 x1 7x2x3 = 2 x 5x = 15 Solution: 4 3 9 x 1 ...
Write a vector equation that is equivalent to the given system ofequations:4x1+x2+ 3x3= 9x1 7x2 2x3= 28x1+ 6x2 5x3= 15 Solution: x1 418 +x2 1 76 +x3 3 2 5 = 9215 ) Determine ifbis a linear combination ofa1,a2, 1 22 a2= 055 a3= 208 b= 511 7 Solution: To see ifbis a linear combination ofa1,a2, anda3we must find a solutionto the equationb=x1a1+x2a2+x3a3 Rewrite as a augmented matrix 1 0 2 5 2 5 0 112 5 8 7 Now we will row reduce:R2: 2R1+R2 1 0 2 20 5 4 12 5 8 7 R3:R3 2R1 1 0 5 20 5 4 10 5 4 11 1R3: R2+R3 1 0 5 20 5 4 10 0 0 12 Since 06= 12 we have thatbis not a linear combination ofa1,a2, ) LetA= 2 0 6 1 8 51 2 1 , letb= 1033 , and letWbe the setof all linear combinations of the columns IsbinWb. Show that the third column ofAis inWSolution: a. First leta1= 2 11 a2= 08 2 a3= 651 We want to findx1,x2, andx3such thatb=x1a1+x2a2+x3a3As in we will consider the Augmented matrixB= [a1,a2,a3,b]and row reduce. NowB= 2 0 6 10 1 8 5 31 2 1 3 R1: (1/2)R1 1 0 3 5 1 8 5 31 2 1 3 R2:R1+R2 1 0 3 50 8 8 81 2 1 3 2R2: (1/8)R2R3:R3 R1 1 0 3 50 1 1 10 2 2 2 R3:R3+ 2R2 1 0 3 50 1 1 10 0 0 0 From this we can see that there are an infinite number of solutions .
R 3: −R 2 +R 3 1 0 5 2 0 5 4 1 0 0 0 −12 Since 0 6= −12 we have that b is not a linear combination of a 1, a 2, and a 3. 1.3.26) Let A = 2 0 6 −1 8 5
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