a) The vector (-3,1,0,0,0)T is a basis vector for the null space of B. b) The ve
ID: 2980587 • Letter: A
Question
a) The vector (-3,1,0,0,0)T is a basis vector for the null space of B.
b) The vector (4,0,k,-1,0)Tis a basis vector for the null space of B, k being any real scalar.
c) The vector (-2,0,0,0,1)T is a basis vector for the null space of B.
d) A linear combination of (4,0,0,-1,0)T and (-2,0,0,0,1)T can form a basis vector for the null space of B.
e) The Nullity of B is 3.
f) None of the above.
PS: the answer is not the combination of a,d or just f
26-5-24 1, and ref(B) 0 0 0 1-1 005010 0 0 0 0 0 62 0010 0100 3000 1000 24 10 4 0 0 8 -2-550 25 3606 1202Explanation / Answer
E nullity of baisi is 3
basis for the nullspace is a linear combination of the special solutions. The special solutions are the columns which do not have pivots in them. So if we row reduce the matrix, we get
And if we put it in Reduced row echelon form it becomes....
So the special solutions ......................
And these three vectors form a basis for the nullspace.
This answer also reveals that the dimension of the nullspace is 3.
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