Thank you Page I of 6 MAT 343 ExAM 2. FALL 2017 Calculators are allowed for this
ID: 3185880 • Letter: T
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Page I of 6 MAT 343 ExAM 2. FALL 2017 Calculators are allowed for this test. No notes or books are allowed. Please write your solutions in the spaces provided. Show all work to receive full credit. Provide full and clear justifications when requested-if it does ot say, "explain" or ""justify. ·then none is needed. The exam is out of 100 points 1. Consider the following matrix: 1 2 34 OIO 12 2 231 rnd: | oO6I 3 -1 2 0 4 3-1 I a. (6 points) Describe the null space of matrix A and state a basis for Nul A. Clearly indicate what calculations were done using technology. to Bea (Nul A) (2 points) The vectors in NulA are in RP where p (2 points) Collectively the set of vectors in NulA is a (circle one): point, line, plane, 3-space, 4-space (2 points) State the dimension of the null space. i.e.. dim(Nula)Explanation / Answer
1.The null space of A is the set of solutions to the equation AX = 0. In view of the RREF of A, if X= (x,y,z,w)T then this equation is equivalent to x+z = 0 or, x = -z , y+z = 0 or, y = -z, and w= 0 so that X = (-z,-z,z,0)T = z(-1,-1,1,0)T. Hence, {(-1,-1,1,0)T} is a basis for the null space of A.
The vectors in Nul A are in Rp, where p = 4 ( as NulA has 4-vectors).
Nul A, being thev span of one vector, is a line.
The dimension of Nul A ie. dim(NulA) is 1.
b. It is apparent from the RREF of A that its 1st, 2nd and 4th columns are linearly independent and the 3rd column is a linear combination of the first 2 columns. Hence, {(1,2,-1,3,-4)T,((2,0,2,-1,3)T,(4,-1,2,0,1)T} is a basis for col(A).
The vectors in col(A) are in Rq, where q = 4 ( as col(A) has 5-vectors).
Collectively, the set of vectors in col(A), being the span of three 5-vectors is a 5-space.
dim(col(A)) = 3.
2. a. T: Rm ?Rn defined by T(X) = AX , where m = 4 and n = 5 ( as the row vectors of A are 4 vectors and as A has 5 columns).
b. SAince the columns of A are not linearly independent, hence T is not one-to-one.
c. Since the columns of A do not span R5, hence T is not onto.
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