(1 point) In this problem, answer \"True\" or \"False\" for each question Note:
ID: 3116448 • Letter: #
Question
(1 point) In this problem, answer "True" or "False" for each question Note: there is no partial credit for this problem. You must answer all parts correctly to receive credit. You will not be shown the correct answers for individual parts. 1.If u, u2, us are linearly dependent vectors in R', then (u u, u ) can be extended to a basis (ul , u, u, v) of R True False 2, For each n there is only 1 magic square of dimension n × n. True False 3. If A is a square matrix then its column space and row space have the same basis. True False 4. If A is a matrix of size 5 × 4 and has a null space of dimension 1, then the matrix has rank 3. True FalseExplanation / Answer
1.The dimension of R4 is 4, hence any basis for R4 will contain 4 vectors. If the set {u1,u2,u3} is linearly dependent, then adding of any vector u4 to it will make the set {u1,u2,u3,u4} also linearly dependent. Hence this set cannot be a basis for R4. The statement is False.
2. As may be observed from the following counter example, the statement is False. A =
8
1
6
3
5
7
4
9
2
is a 3x3 magic square. Also, B =
2
7
6
9
5
1
4
3
8
is also a 3x3 magic square.
3. The statement is False. For a square matrix ( or for any other matrix also), the dimensions of its row space and column space are same, but the basis need not be same. For example, if A =
1
2
3
4
4
3
2
1
2
3
4
5
5
4
3
2
Then the RREF of A is
1
0
-1
-2
0
1
2
3
0
0
0
0
0
0
0
0
Here a basis for Row(A) is {(1,0,-1,-2),(0,1,2,3)} while a basis for Col(A) is {(1,0,0,0)T,(0,1,0,0)T}.
4. The statement is True. As per the rank-nullity theorem, the nullity of amatrix +its rank = number of its colunmns. Here, nullity ( i.e. dimension of its null-space)of a 5x4 matrix is 1. Then its rank = 4-1 = 3.
8
1
6
3
5
7
4
9
2
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