Let and b=1-11, A= 036 4 2 -3 write a, ag, and ay for the columns of A, and let
ID: 3169176 • Letter: L
Question
Let and b=1-11, A= 036 4 2 -3 write a, ag, and ay for the columns of A, and let W - span(a, aj,as). Ax = b (o) Wirite the augmented matrix for the linear systen that corresponds to the matris equation (b) Solve the matrix equation from (a) and write the solution as a vector (c) ls b in {al,a2: as)? (d) How many vectors are in {al,a2, as)? (e) Is b in W? (f) How many vectors are in W? (g) Show that a is in W. (h) Show that a,- as is in W of A form a linearly independent set? Why or why not? (0) Do the columns 0) Let h be a real number and let u -1. For which value(s) h is u in span(a,, a2]? in 0). For what valuses) of h is (a, b, uj linearly dependent?Explanation / Answer
(a). The augmented matrix is B = [A|b] =
1
0
-2
4
0
3
6
-1
4
2
-3
4
(b). To solve the equation AX = b, we will reduce B to its RREF as under:
Add -4 times the 1st row to the 3rd row
Multiply the 2nd row by 1/3
Add -2 times the 2nd row to the 3rd row
Add -2 times the 3rd row to the 2nd row
Add 2 times the 3rd row to the 1st row
Then the RREF of B is
1
0
0
-56/3
0
1
0
67/3
0
0
1
-34/3
Thus, the solution is X = (-56/3,67/3,-34/3)T.
(c ). Since a1,a2,a3 and b are distinct vectors, b is not in {a1,a2,a3}.
(d). There are 3 vectors in {a1,a2,a3}.
(e ).It is apparent from the RREF of B that b = (-56/3)a+(67/3)a+(-34/3)a so that b, being a linear combination of a1,a2,a3, is in W = span{ a1,a2,a3}.
(f). W contains all linear combinations of a1,a2,a3. Hence,there are an infinite number of vectors in W.
(g).a2= 0a1+1a2+0a3. Hence a2 is in W.
(h). a1-a3 = 1a1+0a2-1a3. Hence a1-a3 is in W.
(i). It is apparent from the RREF of B that the columns of A form a linearly independent set.
(j). Let (1,-1,h)T= xa1+ya2 = x(1,0,4)T+y(0,3,2)T. Then x=1, 3y = -1 or, y = -1/3 and 4x+2y =h or, h = 4-2/3 =10/3. Thus, if h = 10/3 , then u is in span{a1,a2}.
(k). Let xa2+yb+zu = 0. The x(0,3,2)T+y(4,-1,4)T +z(1,-1,h)T= (0,0,0)T. Then 4y+z = 0 or, z = -4y, 3x-y-z = 0 or, 3x+3y = 0 or, x+y = 0 or, x = -y and 2x+4y +hz = 0 or, -2y+4y -4hy = 0 or, -2y-4hy = 0 or, y+2hy = 0 so that, (if y 0), then h = -1/2. However, y cannot be 0 as a2 and u are linearly independent, regardless of the value of h. Thus, the vectors a2,b and u are linearly dependent if h = -1/2.
1
0
-2
4
0
3
6
-1
4
2
-3
4
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