1. Consider the matrix 1 4569 3 -21 -1 1 0-1 2-1 2 35 78 (a) Find a basis B for
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Question
1. Consider the matrix 1 4569 3 -21 -1 1 0-1 2-1 2 35 78 (a) Find a basis B for the null space of A. Hint: you need to verify that the vectors you propose actually form a basis for the null space. (Recall (1) the null space of A consists of all x e R with Ax = 0, and (2) the matrix equation Ax = 0 is equivalent to a certain system of linear equations.) (b) Does a linear combination of the vectors in B belong to the null space of A? - Explain! (c) From among the standard basis vectors e,..., e, for R5 select suitable vectors to supplement B from part (a) to a basis of RExplanation / Answer
a). The null space of A is the set of solutions to the equation AX = 0. To solve this equation, we will reduce A to its RREF as under:
Add -3 times the 1st row to the 2nd row
Add 1 times the 1st row to the 3rd row
Add -2 times the 1st row to the 4th row
Multiply the 2nd row by -1/14
Add -4 times the 2nd row to the 3rd row
Add 5 times the 2nd row to the 4th row
Add -4 times the 2nd row to the 1st row
Then the RREF of A is
1
0
1
2
1
0
1
1
1
2
0
0
0
0
0
0
0
0
0
0
Now, if X = (x,y,z,w,u)T, then the equation AX = 0 is equivalent to x+z+2w +u = 0 or, x = - z-2w -u and y+z+w+2u = 0 or, y = - z-w-2u . Then X = (- z-2w -u, - z-w-2u,z,w,u)T = z(-1,-1,1,0,0)T+w(-2,-1,0,1,0)T+u(-1,-2,0,0,1)T. Thus, B = {(-1,-1,1,0,0)T,(-2,-1,0,1,0)T,(-1,-2,0,0,1)T} is a basis for Null(A).
(b). Yes, any linear combination of the vectors in B belongs to Null(A). If, the vectors in B are denoted by v1,v2,v3, and if v = av1+bv2+cv3 be a linear combination of v1,v2,v3, where a,b, c are arbitrary real numbers, then Av = A(av1+bv2+cv3) = aAv1+bAv2+cAv3 = a.0+b.0+vc.0 = 0. Hence v = av1+bv2+cv3 Null(A).
(c ). Let M = [v1,v2,v3] =
-1
-2
-1
-1
-1
-2
1
0
0
0
1
0
0
0
1
Then the RREF of M is
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
Thus, if e4 and e5 are added to B, we get a basis for R5.
1
0
1
2
1
0
1
1
1
2
0
0
0
0
0
0
0
0
0
0
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