In this problem, answer \"True\" or \"False\" for each question. Note: there is
ID: 3116445 • Letter: I
Question
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, u are linearly dependent vectors in R4, then (u, u2, u,) can be extended to a basis (u,u, u, v) of R4 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 statement is False. If the set {u1,u2,u3} is linearly dependent, then the extended set {u1,u2,u3,v} is also linearly dependent and hence it cannot form a basis for R4.
2. The statement is False. A counter example will prove this. A =
8
1
6
3
5
7
4
9
2
and AT =
8
3
4
1
5
9
6
7
2
are both 3x3 magic squares and A AT.
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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